NCERT Solutions · Class 12 Chemistry Solutions

Dissolving a gas in a liquid is an exothermic process, so raising the temperature pushes the equilibrium back toward the free gas.When a gas dissolves in a liquid, its molecules go from a highly disordered, high-energy gaseous state into a more ordered state trapped among solvent molecules. This process releases heat, so it is written as\[\text{Gas} + \text{Solvent} \rightleftharpoons \text{Solution} + \text{Heat}, \qquad \Delta H_{\text{dissolution}} < 0 \]Here \(\displaystyle \Delta H_{\text{dissolution}} \) is the enthalpy change when the gas dissolves — negative because dissolution is exothermic.Le Chatelier's principle says that when a system at equilibrium is heated, the equilibrium shifts in the direction that absorbs the added heat — that is, in the endothermic direction. Since dissolution is exothermic, the reverse process (the gas coming back out of solution) is endothermic. So supplying heat by raising the temperature drives the equilibrium to the left, back toward undissolved gas, and less of it stays dissolved. This is exactly why the solubility of a gas falls as \(\displaystyle T \) rises — it is the same reasoning that makes solubility decrease when you increase the partial pressure of the gas above the liquid go down, only here the "stress" being relieved is heat instead of pressure.There is also a molecular-kinetic-energy way to see the same result. Gas molecules that are dissolved are held in the liquid only loosely, by weak intermolecular attractions with the solvent. Raising the temperature increases the average kinetic energy of these dissolved gas molecules (a rise in \(\displaystyle T \) means a rise in average molecular kinetic energy, by the kinetic theory of matter). With more kinetic energy, a larger fraction of the dissolved molecules have enough energy to overcome the attractive forces holding them in the liquid and escape back into the gas phase. The tendency of the dissolved gas to escape (its effective escaping tendency, or vapour pressure above the solution) increases with temperature, so the amount that stays in solution at equilibrium goes down.This is why, for example, a bottle of soda water goes flat much faster when warm than when cold, and why fish need more dissolved oxygen-rich (cooler) water — warm water simply cannot hold as much dissolved \(\displaystyle \mathrm{O_2} \) as cold water.Answer: Because dissolving a gas in a liquid is exothermic, heating the system (by Le Chatelier's principle) shifts the equilibrium in the endothermic direction — back toward the gas escaping the liquid — so gas solubility in liquids always decreases as temperature is raised.

A gas is only as soluble as the pressure pushing it in — more pressure above the liquid forces more gas into it.Henry's law: at a fixed temperature, the amount of a gas that dissolves in a liquid is directly proportional to the partial pressure of that gas above the liquid surface. In the most common statement, the partial pressure of the gas in the vapour phase, \(\displaystyle p \), is proportional to the mole fraction of the gas dissolved in the solution, \(\displaystyle x \):\[p = K_H \, x \]Here \(\displaystyle p \) is the partial pressure of the gas over the solution, \(\displaystyle x \) is the mole fraction of the dissolved gas in the solution, and \(\displaystyle K_H \) is the Henry's law constant — a number that depends on the identity of the gas and the solvent, and on temperature (it is not a universal constant like \(\displaystyle R \)).A step people get wrong: \(\displaystyle K_H \) is not "solubility" itself — it is inversely related to it. A gas with a high \(\displaystyle K_H \) in a given solvent is less soluble, because for the same mole fraction \(\displaystyle x \) you'd need a much bigger \(\displaystyle p \) to hold it there. Also, \(\displaystyle K_H \) increases with temperature, which is the direct statement of the everyday fact that gases become less soluble in liquids as the liquid warms up — dissolved gas is driven off when a glass of cold water is left to stand and warm to room temperature.Some important applications of Henry's law:Answer: Henry's law states that at constant temperature, \(\displaystyle p = K_H x \) — the partial pressure of a gas above a liquid is directly proportional to its mole fraction dissolved in the liquid, with \(\displaystyle K_H \) depending on the gas, the solvent, and temperature. Its key applications: raising \(\displaystyle \text{CO}_2 \) solubility to carbonate soft drinks, diluting nitrogen with helium in scuba tanks to prevent decompression sickness ("the bends") on ascent, and explaining why low atmospheric oxygen pressure at high altitude causes anoxia in climbers.

Henry's law says the partial pressure of a gas above a dilute solution is directly proportional to its mole fraction dissolved — and when the solvent amount is unchanged, that mole fraction is directly proportional to the mass of gas dissolved.Henry's law: \(\displaystyle p = K_H\,x \), where \(\displaystyle p\) is the partial pressure of the gas, \(\displaystyle x\) is its mole fraction in solution, and \(\displaystyle K_H\) is the Henry's law constant (fixed for a given gas–solvent pair at a given temperature).The mole fraction of dissolved ethane is\[x = \frac{n_{\text{ethane}}}{n_{\text{ethane}} + n_{\text{solvent}}} \approx \frac{n_{\text{ethane}}}{n_{\text{solvent}}} \]because the solution is dilute (\(\displaystyle n_{\text{ethane}} \ll n_{\text{solvent}}\)). The amount of solvent is the same in both cases described in the problem, so \(\displaystyle x\) is simply proportional to \(\displaystyle n_{\text{ethane}}\), the number of moles of ethane dissolved. And since \(\displaystyle n_{\text{ethane}} = \dfrac{w}{M}\) (mass divided by the molar mass \(\displaystyle M\) of ethane, which is the same constant in both cases), \(\displaystyle x\) is in turn directly proportional to the mass \(\displaystyle w\) of ethane dissolved — the molar mass cancels out, so you never actually need its value here.This is the step people skip: because \(\displaystyle M\) and \(\displaystyle n_{\text{solvent}}\) are identical in both scenarios, they cancel in the ratio, and the whole problem reduces to a simple ratio of masses.Combining \(\displaystyle p = K_H x\) with \(\displaystyle x \propto w\) gives, for the two states of the same solution,\[\frac{p_1}{p_2} = \frac{w_1}{w_2} \quad\Longrightarrow\quad p_2 = p_1 \times \frac{w_2}{w_1} \]Substituting the given values — \(\displaystyle p_1 = 1\ \text{bar}\) at \(\displaystyle w_1 = 6.56 \times 10^{-3}\ \text{g}\), and asking for \(\displaystyle p_2\) at \(\displaystyle w_2 = 5.00 \times 10^{-2}\ \text{g}\):\[p_2 = 1\ \text{bar} \times \frac{5.00 \times 10^{-2}\ \text{g}}{6.56 \times 10^{-3}\ \text{g}} \]\[p_2 = 1\ \text{bar} \times 7.6219\ldots = 7.6219\ldots\ \text{bar} \]Rounding to three significant figures, matching the precision of the given data:Answer: \(\displaystyle p_2 \approx 7.62\ \text{bar}\)

Deviation from Raoult's law is a mismatch between the A–B attraction in the mixture and the A–A, B–B attractions in the two pure liquids — and that same mismatch fixes the sign of \(\displaystyle Δ_{mix}H\).For an ideal solution, Raoult's law gives the partial vapour pressures \[p_A = p_A^{\circ}x_A, \qquad p_B = p_B^{\circ}x_B \] where \(\displaystyle p_A^{\circ}, p_B^{\circ}\) are the vapour pressures of the pure liquids and \(\displaystyle x_A, x_B\) are their mole fractions in solution. This holds only when A–A, B–B and A–B interactions are all essentially equal in strength — nothing changes on mixing, so \(\displaystyle Δ_{mix}H = 0\) and \(\displaystyle Δ_{mix}V = 0\).Positive deviation. Negative deviation. The point people get wrong is treating "deviation" as only about vapour pressure — the sign of \(\displaystyle Δ_{mix}H\) is not a separate fact to memorise, it is the same A–B-vs-(A–A, B–B) comparison read off in energy terms instead of escaping-tendency terms.Answer: A solution shows a positive deviation from Raoult's law when A–B forces are weaker than A–A and B–B forces (vapour pressure higher than the Raoult's-law line, \(\displaystyle Δ_{mix}H>0\), mixing endothermic — e.g. ethanol–acetone); it shows a negative deviation when A–B forces are stronger than A–A and B–B forces (vapour pressure lower than the Raoult's-law line, \(\displaystyle Δ_{mix}H<0\), mixing exothermic — e.g. chloroform–acetone).

Question (Q1.15, NCERT Class $\displaystyle 12$ Chemistry, Ch.1 Solutions): An aqueous solution of $\displaystyle 2$% non-volatile solute exerts a pressure of $\displaystyle 1.004$ bar at the normal boiling point of the solvent. What is the molar mass of the solute?Setup: $\displaystyle 2$% solution by mass = $\displaystyle 2$ g solute (w2) in $\displaystyle 100$ g solution, so solvent water w1 = $\displaystyle 98$ g, M1(water) = $\displaystyle 18$ g/mol.Normal boiling point of solvent means pure solvent's vapor pressure p1° = $\displaystyle 1$ atm = $\displaystyle 1.013$ bar. Solution's vapor pressure at that temperature is p1 = $\displaystyle 1.004$ bar.Raoult's Law (relative lowering of vapor pressure, dilute-solution approximation): (p1° − p1)/p1° = n2/n1 = (w2/M2)/(w1/M1) = w2·M1/(M2·w1)Substitute: ($\displaystyle 1.013$−$\displaystyle 1.004$)/$\displaystyle 1.013$ = ($\displaystyle 2$×$\displaystyle 18$)/(M2×$\displaystyle 98$) $\displaystyle 0.009$/$\displaystyle 1.013$ = $\displaystyle 36$/($\displaystyle 98$ M2) M2 = ($\displaystyle 2$×$\displaystyle 18$×$\displaystyle 1.013$)/($\displaystyle 98$×$\displaystyle 0.009$) = $\displaystyle 36.468$/$\displaystyle 0.882$ = $\displaystyle 41.35$ g/mol (rounded to $\displaystyle 4$ sig figs)Full worked solution written per the formatting spec (bold lead idea, named formulas/symbols, shown substitution, units carried, aside on the two common traps — mixing up $\displaystyle 2$% of solution vs. of solvent, and confusing the pure-solvent pressure $\displaystyle 1.013$ bar with the given solution pressure $\displaystyle 1.004$ bar — single rounding at the end, ending in "Answer: ...") was delivered as the final assistant message in the conversation.

In an ideal solution, each component's partial vapour pressure follows Raoult's law: \(\displaystyle p_i = x_i \, p_i^{\circ} \), where \(\displaystyle x_i\) is the MOLE fraction of that component in the liquid — not its mass fraction. The total vapour pressure is just the sum of the two partial pressures, \(\displaystyle p_{\text{total}} = p_1 + p_2\). So the real work here is converting grams to moles, then moles to mole fractions.Step $\displaystyle 1$: Molar massesHeptane is \(\displaystyle \text{C}_7\text{H}_{16} \): \[M_{\text{heptane}} = 7(12) + 16(1) = 100 \ \text{g mol}^{-1} \]Octane is \(\displaystyle \text{C}_8\text{H}_{18} \): \[M_{\text{octane}} = 8(12) + 18(1) = 114 \ \text{g mol}^{-1} \]Step $\displaystyle 2$: Moles of each component\[n_{\text{heptane}} = \frac{26.0 \ \text{g}}{100 \ \text{g mol}^{-1}} = 0.260 \ \text{mol} \]\[n_{\text{octane}} = \frac{35.0 \ \text{g}}{114 \ \text{g mol}^{-1}} = 0.30702 \ \text{mol} \]\[n_{\text{total}} = 0.260 + 0.30702 = 0.56702 \ \text{mol} \]Step $\displaystyle 3$: Mole fractionsThis is the step people skip — you need mole fraction, not the ratio of masses, because vapour pressure depends on the number of molecules escaping the surface, and moles count molecules while grams don't.\[x_{\text{heptane}} = \frac{0.260}{0.56702} = 0.4586 \]\[x_{\text{octane}} = \frac{0.30702}{0.56702} = 1 - 0.4586 = 0.5414 \]Step $\displaystyle 4$: Partial pressures, using Raoult's law\[p_{\text{heptane}} = x_{\text{heptane}} \, p^{\circ}_{\text{heptane}} = 0.4586 \times 105.2 \ \text{kPa} = 48.24 \ \text{kPa} \]\[p_{\text{octane}} = x_{\text{octane}} \, p^{\circ}_{\text{octane}} = 0.5414 \times 46.8 \ \text{kPa} = 25.34 \ \text{kPa} \]Step $\displaystyle 5$: Total vapour pressure of the mixture\[p_{\text{total}} = p_{\text{heptane}} + p_{\text{octane}} = 48.24 + 25.34 = 73.58 \ \text{kPa} \]The data (masses to three sig figs, pressures to three or four sig figs) supports rounding to three significant figures.Answer: \(\displaystyle p_{\text{total}} \approx 73.6 \ \text{kPa} \) (with heptane contributing about $\displaystyle 48.2$ kPa and octane about $\displaystyle 25.3$ kPa of that total).

Raoult's law for a solution of a non-volatile solute says the relative lowering of vapour pressure equals the mole fraction of the solute — not of the solvent.For a dilute solution, \[\frac{p^{0} - p_{s}}{p^{0}} = x_{2} \] where \(\displaystyle p^{0} \) is the vapour pressure of the pure solvent, \(\displaystyle p_{s} \) is the vapour pressure of the solution, and \(\displaystyle x_{2} \) is the mole fraction of the solute.Step $\displaystyle 1$ — moles of solvent. "$\displaystyle 1$ molal" means $\displaystyle 1$ mole of solute is dissolved in exactly $\displaystyle 1000$ g of solvent (water) — molality is defined per kilogram of solvent, not per kilogram of solution.\[n_{1} = \frac{1000\ \text{g}}{18\ \text{g mol}^{-1}} = 55.56\ \text{mol} \]Step $\displaystyle 2$ — mole fraction of the solute. With \(\displaystyle n_{2} = 1 \) mol of solute, \[x_{2} = \frac{n_{2}}{n_{1}+n_{2}} = \frac{1}{55.56+1} = \frac{1}{56.56} = 0.01768 \](The dilute-solution shortcut \(\displaystyle x_2 \approx n_2/n_1 \), which drops the solute from the denominator, gives the same value here to three significant figures — but the total-moles form above is the one that's always correct.)Step $\displaystyle 3$ — apply Raoult's law. \[p^{0} - p_{s} = p^{0}x_{2} = 12.3\ \text{kPa} \times 0.01768 = 0.2175\ \text{kPa} \] \[p_{s} = 12.3\ \text{kPa} - 0.2175\ \text{kPa} = 12.08\ \text{kPa} \]The pressure barely moves because the solution is so dilute — only one solute particle among roughly $\displaystyle 56$ particles total.Answer: The vapour pressure of the solution is ≈ $\displaystyle 12.08$ kPa (a drop of about $\displaystyle 0.22$ kPa from the pure solvent's $\displaystyle 12.3$ kPa).

A vapour-pressure drop is read through Raoult's law: the fractional lowering of vapour pressure equals the mole fraction of the solute, not the mass fraction or the percentage of solute added.Raoult's law for a dilute solution of a non-volatile solute states\[\frac{p_1^{\circ}-p_1}{p_1^{\circ}} = x_2 \]where \(\displaystyle p_1^{\circ}\) is the vapour pressure of the pure solvent (octane), \(\displaystyle p_1\) is the vapour pressure of the solution, and \(\displaystyle x_2\) is the mole fraction of the solute.Step $\displaystyle 1$: Turn "reduced to $\displaystyle 80$%" into a mole fraction.If the vapour pressure falls to $\displaystyle 80$% of its original value, \(\displaystyle p_1 = 0.80\,p_1^{\circ}\). So the fractional lowering is\[\frac{p_1^{\circ}-p_1}{p_1^{\circ}} = \frac{p_1^{\circ}-0.80\,p_1^{\circ}}{p_1^{\circ}} = 0.20 \]This $\displaystyle 0.20$ is directly the mole fraction of solute: \(\displaystyle x_2 = 0.20\).Step $\displaystyle 2$: Find the moles of octane (the solvent).Octane is \(\displaystyle \text{C}_8\text{H}_{18}\), so its molar mass is\[M_1 = 8(12) + 18(1) = 96 + 18 = 114\ \text{g mol}^{-1} \]It is not a coincidence that the given mass of octane, $\displaystyle 114$ g, equals its molar mass — that is exactly $\displaystyle 1$ mole:\[n_1 = \frac{w_1}{M_1} = \frac{114\ \text{g}}{114\ \text{g mol}^{-1}} = 1\ \text{mol} \]Step $\displaystyle 3$: Write the mole fraction in terms of moles and solve for the solute's moles.\[x_2 = \frac{n_2}{n_1+n_2} \]Substituting \(\displaystyle x_2 = 0.20\) and \(\displaystyle n_1 = 1\ \text{mol}\):\[0.20 = \frac{n_2}{1+n_2} \]\[0.20(1+n_2) = n_2 \]\[0.20 = n_2 - 0.20\,n_2 = 0.80\,n_2 \]\[n_2 = \frac{0.20}{0.80} = 0.25\ \text{mol} \]Step $\displaystyle 4$: Convert moles of solute to mass using its molar mass.The solute's molar mass is given as \(\displaystyle M_2 = 40\ \text{g mol}^{-1}\), so\[w_2 = n_2 \times M_2 = 0.25\ \text{mol} \times 40\ \text{g mol}^{-1} = 10\ \text{g} \]The step people skip is treating "$\displaystyle 80$%" as if it were the mole fraction itself — it is the remaining fraction of vapour pressure, so the mole fraction actually used in Raoult's law is the lost fraction, \(\displaystyle 1 - 0.80 = 0.20\).Answer: $\displaystyle 10$ g of the solute must be dissolved in $\displaystyle 114$ g of octane.

Question Q1.19 (Ch.1 Solutions): $\displaystyle 30$ g non-volatile solute in $\displaystyle 90$ g water gives vapour pressure $\displaystyle 2.8$ kPa at $\displaystyle 298$ K; adding $\displaystyle 18$ g more water (total $\displaystyle 108$ g water) gives $\displaystyle 2.9$ kPa. Find molar mass of solute and vapour pressure of pure water.Method: Raoult's Law p_solution = p1°·n1/(n1+n2) applied twice (n1=$\displaystyle 5.00$ mol water first case, n1'=$\displaystyle 6.00$ mol second case; n2 = $\displaystyle 30$/M2 mol solute, same in both cases). Solving the two simultaneous equations for the two unknowns (p1° and n2):$\displaystyle 2.8$ = p1°·$\displaystyle 5$/($\displaystyle 5$+n2) → p1° = $\displaystyle 2.8$ + $\displaystyle 0.560$ n2 $\displaystyle 2.9$ = p1°·$\displaystyle 6$/($\displaystyle 6$+n2) → p1° = $\displaystyle 2.9$ + $\displaystyle 0.4833$ n2Equate: $\displaystyle 0.0767$ n2 = $\displaystyle 0.1$ → n2 = $\displaystyle 1.304$ mol M2 = $\displaystyle 30$/$\displaystyle 1.304$ = $\displaystyle 23.0$ g/mol p1° = $\displaystyle 2.8$ + $\displaystyle 0.560$×$\displaystyle 1.304$ = $\displaystyle 3.53$ kPa (checked consistent via second equation)Full formatted worked solution (with bold lede, named formulas, substitutions, unit tracking, and asides on common mistakes — using proper Mathpix's soname of solvent mass total vs added mass, and exact vs dilute-approximation mole fraction) was written and delivered in the final response above, following SolveIt's NCERT Solutions house style (KaTeX \(\) / \[\] math, no markdown headings, ends with "Answer: ...").

Solved Q1.20 (NCERT Class $\displaystyle 12$ Chemistry, Ch.1 Solutions) — the sucrose/glucose freezing-point depression problem. Verified all arithmetic numerically with a Python check (mol masses: sucrose $\displaystyle 342$ g/mol, glucose $\displaystyle 180$ g/mol; $\displaystyle 5$% by mass means $\displaystyle 5$ g solute per $\displaystyle 100$ g solution, i.e. $\displaystyle 95$ g solvent). Key results: ΔTf(sucrose) = $\displaystyle 2.15$ K, molality = $\displaystyle 0.15389$ mol/kg, implied Kf(water) = $\displaystyle 13.97$ K·kg/mol from the given data; molality of $\displaystyle 5$% glucose = $\displaystyle 0.29240$ mol/kg; ΔTf(glucose) = $\displaystyle 4.085$ K (confirmed independently via the M1/M2 ratio shortcut: $\displaystyle 2.15$ × $\displaystyle 342$/$\displaystyle 180$ = $\displaystyle 4.085$ exactly); final freezing point = $\displaystyle 273.15$ − $\displaystyle 4.085$ = $\displaystyle 269.065$ K ≈ $\displaystyle 269.07$ K.Wrote the full worked solution in the required format (bold opening insight, named formulas, substitution shown, units carried, aside on the "$\displaystyle 5$% by mass = per $\displaystyle 100$ g solution not solvent" trap, KaTeX \(\) / \[\] delimiters, no markdown headings, ending in "Answer: ...").

Question: NCERT Class XII Chemistry, Chapter $\displaystyle 1$ (Solutions), Q1.21 — freezing-point depression problem to find atomic masses of A and B from molar masses of AB2 and AB4.Method: 1. ΔTf = Kf·m, with m = (w2/M2)/W1(kg). Rearranged: M2 = Kf·w2/(ΔTf·W1). 2. For AB2: w2=1g, ΔTf=2.3K, \(\displaystyle \mathrm{W_{1}}\)=0.020kg → M(AB2) = $\displaystyle 5.1$×$\displaystyle 1$/($\displaystyle 2.3$×$\displaystyle 0.020$) = $\displaystyle 5.1$/$\displaystyle 0.046$ = $\displaystyle 110.87$ g/mol ≈ $\displaystyle 110.9$ g/mol. 3. For AB4: w2=1.0g, ΔTf=1.3K, \(\displaystyle \mathrm{W_{1}}\)=0.020kg → M(AB4) = $\displaystyle 5.1$×$\displaystyle 1$/($\displaystyle 1.3$×$\displaystyle 0.020$) = $\displaystyle 5.1$/$\displaystyle 0.026$ = $\displaystyle 196.15$ g/mol ≈ $\displaystyle 196.2$ g/mol. 4. Set a = atomic mass A, b = atomic mass B: a+2b = $\displaystyle 110.9$ (i); a+4b = $\displaystyle 196.2$ (ii). 5. (ii)-(i): 2b = $\displaystyle 85.3$ → b = $\displaystyle 42.6$ u (more precisely $\displaystyle 42.64$ u using unrounded molar masses). 6. Substitute into (i): a = $\displaystyle 110.9$ - $\displaystyle 85.3$ = $\displaystyle 25.6$ u (more precisely $\displaystyle 25.59$ u).Full worked solution (formatted for KaTeX, with bold lead idea, named formulas, unit tracking, common-mistake asides, ending in Answer: ...) was written out in the response body.

Osmotic pressure obeys \(\displaystyle \pi = CRT \), the same form as the ideal-gas law — so at a fixed temperature, \(\displaystyle \pi \) is directly proportional to the molar concentration \(\displaystyle C\). That proportionality lets you find the new concentration without even needing a precise value of \(\displaystyle R\).Step $\displaystyle 1$: Get the concentration of the first solution from its mass.Glucose is \(\displaystyle \text{C}_6\text{H}_{12}\text{O}_6\), molar mass \[M = 6(12) + 12(1) + 6(16) = 180\ \text{g mol}^{-1} \]Moles of glucose in the first solution: \[n_1 = \frac{36\ \text{g}}{180\ \text{g mol}^{-1}} = 0.2\ \text{mol} \]This is dissolved to make $\displaystyle 1$ L of solution (not $\displaystyle 1$ L of water added — the volume given is already the solution volume, so no correction is needed here): \[C_1 = \frac{n_1}{V} = \frac{0.2\ \text{mol}}{1\ \text{L}} = 0.2\ \text{mol L}^{-1} \]Step $\displaystyle 2$: Write the van't Hoff equation for both solutions.\[\pi_1 = C_1RT, \qquad \pi_2 = C_2RT \]Both solutions are at the same temperature \(\displaystyle T = 300\ \text{K}\), and \(\displaystyle R\) is the same constant in both equations, so dividing one by the other cancels them: \[\frac{\pi_1}{\pi_2} = \frac{C_1}{C_2} \quad\Rightarrow\quad C_2 = C_1 \times \frac{\pi_2}{\pi_1} \]This is the step people skip — they try to solve each equation separately for \(\displaystyle C\) using \(\displaystyle R\), and then the answer depends on which value of \(\displaystyle R\) ($\displaystyle 0.0821$ L atm/mol K vs. $\displaystyle 0.0831$ L bar/mol K) they plug in. Taking the ratio makes \(\displaystyle R\) and \(\displaystyle T\) drop out entirely, so the answer doesn't depend on that choice at all.Step $\displaystyle 3$: Substitute the numbers.\[C_2 = 0.2\ \text{mol L}^{-1} \times \frac{1.52\ \text{bar}}{4.98\ \text{bar}} \]\[C_2 = 0.2\ \text{mol L}^{-1} \times 0.30522 = 0.061044\ \text{mol L}^{-1} \]Rounding to three significant figures (matching the precision of the given pressures): \[C_2 \approx 0.0610\ \text{mol L}^{-1} \]Answer: The concentration is approximately \(\displaystyle 0.0610\ \text{mol L}^{-1}\) (i.e. \(\displaystyle 6.10 \times 10^{-2}\ \text{M}\)).

The type of intermolecular force between two substances is decided by what each one IS — non-polar, polar, or ionic — not by memorising a list of names.Three categories cover every pair below: (i) n-hexane \(\displaystyle \mathrm{(C_{6}H_{14})}\) and n-octane \(\displaystyle \mathrm{(C_{8}H_{18})}\) Both are saturated hydrocarbons — only C–C and C–H bonds, both of very low polarity, arranged with no net molecular dipole. Neither molecule carries a permanent dipole, so the attraction between them can only be the fleeting, induced kind. Most important interaction: van der Waals forces of the London dispersion type.(ii) \(\displaystyle \mathrm{I_{2}}\) and \(\displaystyle \mathrm{CCl_{4}}\) \(\displaystyle \mathrm{I_{2}}\) is a homonuclear diatomic molecule — no dipole possible. \(\displaystyle \mathrm{CCl_{4}}\) has four polar C–Cl bonds, but they point symmetrically to the corners of a tetrahedron, so their dipole moments cancel and the molecule as a whole is non-polar. This is the same situation as (i): two species with zero permanent dipole, attracting only through instantaneous induced dipoles — which is exactly why iodine dissolves readily in \(\displaystyle CCl_{4}\). Most important interaction: van der Waals forces of the London dispersion type (induced dipole–induced dipole).(iii) \(\displaystyle \mathrm{NaClO_{4}}\) and water \(\displaystyle \mathrm{NaClO_{4}}\) is an ionic compound: it exists as separate \(\displaystyle \mathrm{Na^{+}}\) and \(\displaystyle \mathrm{ClO_{4}^{-}}\) ions, not as neutral molecules. Water is a bent, polar molecule with a permanent dipole (\(\displaystyle \delta-\) on O, \(\displaystyle \delta+\) on the H atoms). The positive \(\displaystyle \mathrm{Na^{+}}\) ion is pulled toward the \(\displaystyle \delta-\) oxygen end of water molecules, and the negative \(\displaystyle \mathrm{ClO_{4}^{-}}\) ion is pulled toward the \(\displaystyle \delta+\) hydrogen ends. This ion-to-permanent-dipole attraction is what hydrates the ions and pulls the solid into solution.This is the step people get wrong: because water is "polar," it is tempting to call this dipole–dipole — but one of the two partners here is a full ion, not a neutral polar molecule, so it must be named ion–dipole, not dipole–dipole.Most important interaction: ion–dipole interaction.(iv) methanol \(\displaystyle \mathrm{(CH_{3}OH)}\) and acetone \(\displaystyle \mathrm{(CH_{3}COCH_{3})}\) Methanol has a permanent dipole from its O–H bond; acetone has a permanent dipole from its C=O bond. Neither species is an ion, so ion–dipole is ruled out, and neither is non-polar, so dispersion is not the dominant term — each molecule's permanent, fixed dipole aligns with the other's. Most important interaction: dipole–dipole interaction.(v) acetonitrile \(\displaystyle \mathrm{(CH_{3}CN)}\) and acetone \(\displaystyle \mathrm{(C_{3}H_{6}O)}\) Acetonitrile's \(\displaystyle C\equiv N\) bond is strongly polarised (N is far more electronegative than C), giving the molecule a large permanent dipole; acetone again has a permanent dipole from its C=O bond. As in (iv), both partners are neutral polar molecules with fixed dipoles, not ions and not non-polar. Most important interaction: dipole–dipole interaction.Answer: (i) n-hexane–n-octane: London dispersion forces (van der Waals). (ii) \(\displaystyle \mathrm{I_{2}^{-}CCl_{4}}\): London dispersion forces (van der Waals). (iii) \(\displaystyle NaClO_{4}\)–water: ion–dipole interaction. (iv) methanol–acetone: dipole–dipole interaction. (v) acetonitrile–acetone: dipole–dipole interaction.

Solubility is a contest between the interactions you must break and the interactions you get to form — "like dissolves like" really means "match the polarity." n-Octane, \(\displaystyle CH_3(CH_2)_6CH_3\), is a long nonpolar hydrocarbon chain. It has no permanent dipole and no O–H or N–H bond, so the only force it can offer a solute is weak London dispersion. A solute dissolves well in it only if the solute's own interactions are also just dispersion-type — anything the solute holds together with dipole forces, hydrogen bonds, or ionic bonds has to be broken for nothing better in return, which costs energy the mixing can't pay back.Go through each solute and ask what force holds it together in the pure state:Cyclohexane, \(\displaystyle C_6H_{12}\): a nonpolar, saturated ring — molecules are held to each other purely by dispersion forces, exactly like n-octane. Mixing breaks weak octane–octane and cyclohexane–cyclohexane contacts and replaces them with equally weak octane–cyclohexane contacts. Nothing is lost in the swap, so cyclohexane is freely (fully) miscible with n-octane.\(\displaystyle CH_3CN\) (acetonitrile): the \(\displaystyle C{\equiv}N\) group carries a large dipole, so liquid acetonitrile is held together by dipole–dipole attractions — stronger than dispersion, but not directional/networked the way hydrogen bonds are. Dissolving it in octane means giving up those dipole–dipole contacts for weak dispersion contacts — unfavorable, but only moderately so, so \(\displaystyle CH_3CN\) is somewhat soluble.\(\displaystyle CH_3OH\) (methanol): polar and a hydrogen-bond donor and acceptor. Methanol molecules are locked into an extensive, directional hydrogen-bonded network with each other — a much stronger self-attraction than acetonitrile's simple dipole–dipole pull. n-Octane has no O or N atom to hydrogen-bond back with, so almost the entire network must be broken with no compensation. This is the step people get wrong: methanol "looks less polar" than an ionic salt, so it's tempting to rank it near cyclohexane — but it is the hydrogen bonding, not just the presence of a dipole, that makes it harder to dissolve in a nonpolar solvent than \(\displaystyle CH_3CN\).KCl: an ionic lattice held by full-charge electrostatic (ion–ion) attraction — the strongest interaction of the four by a wide margin. n-Octane, having no dipole, cannot solvate \(\displaystyle K^+\) or \(\displaystyle Cl^-\) at all (no ion–dipole stabilization). None of the lattice energy is recovered, so KCl is essentially insoluble in n-octane.Ranking by how badly the solute's own interactions mismatch n-octane's (dispersion-only) character — worst mismatch first — gives the increasing polarity order:\[\text{cyclohexane} < CH_3CN < CH_3OH < KCl \]Solubility in the nonpolar solvent runs the opposite way, from least to most soluble:Answer: increasing solubility in n-octane: \(\displaystyle KCl < CH_3OH < CH_3CN < \) cyclohexane — ranked by how closely each solute's own bonding (ionic → hydrogen-bonded → dipole–dipole → dispersion-only) matches n-octane's purely dispersion-force character.

Molarity is moles of solute per litre of SOLUTION — so the density is there to turn a MASS of lake water into a VOLUME of lake water, and nothing else.The trap in this question is the phrase "per kg of water". The water here is the lake water — the same lake water whose density \(\displaystyle 1.25\ \mathrm{g\ mL^{-1}}\) you were just handed. So \(\displaystyle 1\ \mathrm{kg}\) of lake water is \(\displaystyle 1\ \mathrm{kg}\) of solution, containing \(\displaystyle 92\ \mathrm{g}\) of \(\displaystyle \mathrm{Na^{+}}\) dissolved in it.Step $\displaystyle 1$ — moles of \(\displaystyle \mathrm{Na^{+}}\).The defining relation for amount of substance is \[n=\frac{m}{M} \] where \(\displaystyle n\) is the amount in mol, \(\displaystyle m\) the mass in g, and \(\displaystyle M\) the molar mass in \(\displaystyle \mathrm{g\ mol^{-1}}\). For the sodium ion, \(\displaystyle M(\mathrm{Na^{+}})=23\ \mathrm{g\ mol^{-1}}\) (an electron weighs about \(\displaystyle 1/42000\) of a sodium atom, so losing one changes the molar mass in the fifth decimal place — ignore it).\[n(\mathrm{Na^{+}})=\frac{92\ \mathrm{g}}{23\ \mathrm{g\ mol^{-1}}}=4.00\ \mathrm{mol} \]Aside — per mole, not per gram. Molarity counts particles, not grams. Every gram figure in this question has to pass through \(\displaystyle M\) before it can enter the molarity expression.Step $\displaystyle 2$ — volume of that $\displaystyle 1$ kg of lake water.Density is mass per unit volume, \[\rho=\frac{m}{V}\qquad\Longrightarrow\qquad V=\frac{m}{\rho} \] with \(\displaystyle \rho\) in \(\displaystyle \mathrm{g\ mL^{-1}}\), \(\displaystyle m\) in g, \(\displaystyle V\) in mL. Substituting the \(\displaystyle 1\ \mathrm{kg}=1000\ \mathrm{g}\) of lake water:\[V=\frac{1000\ \mathrm{g}}{1.25\ \mathrm{g\ mL^{-1}}}=800\ \mathrm{mL}=0.800\ \mathrm{L} \]Aside — the density belongs to the solution. Students often divide by \(\displaystyle 1.00\ \mathrm{g\ mL^{-1}}\) here, reasoning "it's water". It is not pure water; it is a salty solution that is $\displaystyle 25$% denser, and \(\displaystyle 1\ \mathrm{kg}\) of it occupies only \(\displaystyle 800\ \mathrm{mL}\), not \(\displaystyle 1000\ \mathrm{mL}\). Using the wrong density would hand you \(\displaystyle 4\ \mathrm{M}\).Step $\displaystyle 3$ — molarity.\[M=\frac{n(\text{solute})}{V(\text{solution in L})} \] where \(\displaystyle n\) is in mol and \(\displaystyle V\) is the volume of the whole solution, in litres.\[M(\mathrm{Na^{+}})=\frac{4.00\ \mathrm{mol}}{0.800\ \mathrm{L}}=5.00\ \mathrm{mol\ L^{-1}} \]Rounded once, at the end, to three significant figures (the data — \(\displaystyle 92\ \mathrm{g}\), \(\displaystyle 1.25\ \mathrm{g\ mL^{-1}}\), \(\displaystyle 1\ \mathrm{kg}\) — supports three): \(\displaystyle 5.00\ \mathrm{M}\).Two things worth knowing about this question.First, "m" versus "M". The printed answer is often typeset as "$\displaystyle 5$ m", which reads as $\displaystyle 5$ molal. It is not: molality is \(\displaystyle \mathrm{mol\ kg^{-1}}\) of solvent and never uses a density, whereas this calculation used the density to get a volume. The quantity asked for and computed is molarity, \(\displaystyle 5\ \mathrm{mol\ L^{-1}}\), symbol M. Lower-case m is an OCR/typography slip, not a different result.Second, the reading that gives \(\displaystyle 4.58\ \mathrm{M}\). If you insist that "kg of water" means \(\displaystyle 1\ \mathrm{kg}\) of pure solvent, then the solution mass becomes \(\displaystyle 1000+92=1092\ \mathrm{g}\), its volume \(\displaystyle 1092/1.25=873.6\ \mathrm{mL}\), and the molarity \(\displaystyle 4.00/0.8736=4.58\ \mathrm{M}\). That reading collapses on its own terms: a solution cannot hold \(\displaystyle \mathrm{Na^{+}}\) without counter-anions, whose mass is not given, so the solution mass would be unknowable and the problem unsolvable. The intended — and the only workable — reading is that the \(\displaystyle 1\ \mathrm{kg}\) is \(\displaystyle 1\ \mathrm{kg}\) of lake water. The clean \(\displaystyle 92/23=4\) exactly, and \(\displaystyle 1000/1.25=800\) exactly, confirm it.Answer: \(\displaystyle 5.00\ \mathrm{M}\) — the molarity of \(\displaystyle \mathrm{Na^{+}}\) in the lake water is \(\displaystyle 5.00\ \mathrm{mol\ L^{-1}}\).

Question: Q1.27, NCERT Class XII Chemistry, Chapter $\displaystyle 1$ (Solutions) — "If the solubility product of CuS is $\displaystyle 6$ × $\displaystyle 10$^-$\displaystyle 16$, calculate the maximum molarity of CuS in aqueous solution."Solved via: CuS(s) ⇌ Cu2+(aq) + S2-(aq), so [Cu2+]=[S2-]=s and Ksp = s^$\displaystyle 2$ ($\displaystyle 1$:$\displaystyle 1$ stoichiometry). s = sqrt($\displaystyle 6$×$\displaystyle 10$^-$\displaystyle 16$) = sqrt($\displaystyle 6$)×$\displaystyle 10$^-$\displaystyle 8$ = $\displaystyle 2.449$×$\displaystyle 10$^-$\displaystyle 8$, rounded to $\displaystyle 2.45$×$\displaystyle 10$^-$\displaystyle 8$ mol/L. This s equals the molar solubility/maximum molarity of CuS directly.Full worked solution written in the required KaTeX-compatible format (inline \(\) and display \[\] math, bold lead line naming the idea, named formula with symbols defined, substitution shown, unit carried through, an aside on why Ksp=s^$\displaystyle 2$ here ($\displaystyle 1$:$\displaystyle 1$ ion stoichiometry — flagged as the step people rush), single rounding at the end to $\displaystyle 3$ sig figs, ending with "Answer: ...") is provided in my response above.

Question (NCERT Class $\displaystyle 12$ Chemistry, Ch.1 Solutions, Q1.28): Calculate the mass percentage of aspirin \(\displaystyle \mathrm{(C_{9}H_{8}O_{4})}\) in acetonitrile \(\displaystyle \mathrm{(CH_{3}CN)}\) when $\displaystyle 6.5$ g of \(\displaystyle \mathrm{C_{9}H_{8}O_{4}}\) is dissolved in $\displaystyle 450$ g of CH3CN.Mass % = (mass of solute)/(mass of solution) × $\displaystyle 100$, where mass of solution = mass of solute + mass of solvent = $\displaystyle 6.5$ g + $\displaystyle 450$ g = $\displaystyle 456.5$ g (the molecular formula of aspirin is irrelevant here since this is a mass-to-mass ratio, not a mole calculation).Mass % = ($\displaystyle 6.5$/$\displaystyle 456.5$) × $\displaystyle 100$ = 1.4240...% ≈ $\displaystyle 1.424$% (≈$\displaystyle 1.42$% to $\displaystyle 3$ sig figs).Full worked solution (KaTeX-formatted, per house style) has been written above with a bold lead line on the solute-vs-solution-mass pitfall, formula/symbol definitions, explicit substitution, and a closing Answer: line.

Molality is moles of solute per kilogram of solvent — never per kilogram or litre of solution. So the given \(\displaystyle 1.5\times10^{-3}\,m\) tells you directly how much water carries a fixed number of moles of nalorphene; you find the moles needed from the dose, use molality to get the mass of water that must carry them, and only then add the solute's own (tiny) mass back on to get the mass of solution.Step $\displaystyle 1$ — molar mass of nalorphene, \(\displaystyle C_{19}H_{21}NO_3\).Using \(\displaystyle C=12,\ H=1,\ N=14,\ O=16\) g mol\(\displaystyle ^{-1}\):\[M = 19(12) + 21(1) + 14 + 3(16) = 228 + 21 + 14 + 48 = 311\ \text{g mol}^{-1} \]Step $\displaystyle 2$ — moles of nalorphene in one dose.The dose is \(\displaystyle 1.5\) mg \(\displaystyle = 1.5\times10^{-3}\) g. Moles are mass divided by molar mass:\[n_{\text{solute}} = \frac{1.5\times10^{-3}\ \text{g}}{311\ \text{g mol}^{-1}} = 4.823\times10^{-6}\ \text{mol} \]Step $\displaystyle 3$ — mass of water (solvent) that gives this molality.Molality is defined as\[m = \frac{n_{\text{solute}}}{w_{\text{solvent}}\,(\text{kg})} \]where \(\displaystyle n_{\text{solute}}\) is moles of solute and \(\displaystyle w_{\text{solvent}}\) is the mass of solvent in kilograms — not the mass of solution. Rearranging for the solvent mass:\[w_{\text{solvent}} = \frac{n_{\text{solute}}}{m} = \frac{4.823\times10^{-6}\ \text{mol}}{1.5\times10^{-3}\ \text{mol kg}^{-1}} = 3.215\times10^{-3}\ \text{kg} = 3.215\ \text{g of water} \]This is the point where it's easy to stop one step too early: \(\displaystyle 3.215\) g is the mass of water needed to carry the dose at this molality, not yet the mass of the solution the patient is given.Step $\displaystyle 4$ — mass of solution.Mass of solution is mass of solvent plus mass of solute:\[w_{\text{solution}} = w_{\text{solvent}} + w_{\text{solute}} = 3.215\ \text{g} + 1.5\times10^{-3}\ \text{g} = 3.217\ \text{g} \]The solute's own mass barely moves the total here — \(\displaystyle 1.5\) mg is negligible next to \(\displaystyle 3.215\) g of water — but it is the correct thing to add, and rounding to three significant figures (matching the data) gives:Answer: ≈ $\displaystyle 3.22$ g of the \(\displaystyle 1.5\times10^{-3}\,m\) aqueous nalorphene solution must be administered to deliver the $\displaystyle 1.5$ mg dose.

Question: Q1.30, NCERT Class $\displaystyle 12$ Chemistry, Ch.1 Solutions — "Calculate the amount of benzoic acid \(\displaystyle \mathrm{(C_{6}H_{5}COOH)}\) required for preparing $\displaystyle 250$ mL of $\displaystyle 0.15$ M solution in methanol."Worked solution (as written for the student-facing page):Molarity tells you moles of solute per litre of solution, not moles per litre of solvent — methanol here is just the solvent, and the $\displaystyle 250$ mL given is the total solution volume.Step $\displaystyle 1$ — find the moles of benzoic acid needed.Molarity, \(\displaystyle M = \dfrac{n}{V} \), where \(\displaystyle n\) is moles of solute and \(\displaystyle V\) is the volume of solution in litres.Rearranging for \(\displaystyle n\): \[n = M \times V \]Substitute \(\displaystyle M = 0.15\ \text{mol L}^{-1}\) and \(\displaystyle V = 250\ \text{mL} = 0.250\ \text{L}\): \[n = 0.15\ \text{mol L}^{-1} \times 0.250\ \text{L} = 0.0375\ \text{mol} \]Step $\displaystyle 2$ — convert moles to mass using the molar mass.Benzoic acid, \(\displaystyle C_6H_5COOH\), has molecular formula \(\displaystyle C_7H_6O_2\). Its molar mass is the sum of atomic masses: \[M_{\text{molar}} = 7(12.01) + 6(1.008) + 2(16.00)\ \text{g mol}^{-1} \] \[M_{\text{molar}} = 84.07 + 6.05 + 32.00 = 122.12\ \text{g mol}^{-1} \]Mass, \(\displaystyle m\), comes from \[m = n \times M_{\text{molar}} \]Substituting: \[m = 0.0375\ \text{mol} \times 122.12\ \text{g mol}^{-1} = 4.5795\ \text{g} \]The data ($\displaystyle 0.15$ M, $\displaystyle 250$ mL) supports three significant figures, so round only at this last step.Answer: $\displaystyle 4.58$ g of benzoic acid

Freezing-point depression measures the number of particles in solution, and here that number changes because the three acids dissociate to different extents.The depression in freezing point of any solute follows \[\Delta T_f = i K_f m \] where \(\displaystyle \mathrm{i\) is the van't Hoff factor (the actual number of particles produced per formula unit dissolved), \(\displaystyle K_f\) is the molal depression constant of water (fixed — same solvent throughout), and \(\displaystyle m\) is the molality. "Same amount" of each acid means the same moles dissolved in the same mass of water, so the nominal \(\displaystyle m^{-}}\) and \(\displaystyle K_f\) — are identical for all three solutions.Since \(\displaystyle K_f\) and the nominal \(\displaystyle m\) don't change across the three cases, any difference in \(\displaystyle \Delta T_f\) has to come entirely from \(\displaystyle i\).Each of these is a weak monoprotic acid that only partly ionizes, \[\mathrm{HA} \rightleftharpoons \mathrm{H^+} + \mathrm{A^-} \] If \(\displaystyle \alpha\) is the degree of dissociation, then out of $\displaystyle 1$ mole of acid taken, \(\displaystyle (1-\alpha)\) mole stays as \(\displaystyle \mathrm{HA}\) and \(\displaystyle \alpha\) mole each of \(\displaystyle \mathrm{H^+}\) and \(\displaystyle \mathrm{A^-}\) appear, giving \(\displaystyle (1+\alpha)\) moles of particles in total: \[i = 1 + \alpha \] The point people slip on here: \(\displaystyle i\) is not automatically $\displaystyle 2$ just because the molecule can split into two ions — it depends on how much of it actually splits, which is set by the acid's strength (\(\displaystyle K_a\)), not by the formula alone.Now compare the three acids. Replacing the three H atoms of the \(\displaystyle \mathrm{CH_3}-\) group with atoms that pull electron density away (the \(\displaystyle -I\), or electron-withdrawing inductive, effect) makes it easier for the molecule to lose \(\displaystyle \mathrm{H^+}\): once the proton leaves, that same electron-withdrawal spreads out (stabilizes) the negative charge left on the carboxylate ion. A more stable conjugate base means a larger \(\displaystyle K_a\) — the acid ionizes more readily, so \(\displaystyle \alpha\), and hence \(\displaystyle i\), is larger. Because \(\displaystyle i\) rises steadily through this series while \(\displaystyle K_f\) and the nominal \(\displaystyle m\) stay fixed, \(\displaystyle \Delta T_f = iK_fm\) rises in step with it — smallest for acetic acid, largest for trifluoroacetic acid.Answer: \(\displaystyle \Delta T_f\) increases from acetic acid to trichloroacetic acid to trifluoroacetic acid because the electron-withdrawing (\(\displaystyle -I\)) effect of the halogen substituents strengthens in that order (Cl weaker than F), increasing each acid's degree of dissociation \(\displaystyle \alpha\) and therefore its van't Hoff factor \(\displaystyle i = 1+\alpha\); with \(\displaystyle K_f\) and the nominal molality unchanged, a larger \(\displaystyle i\) directly gives a larger \(\displaystyle \Delta T_f = iK_fm\).

This is a WEAK ACID, so it doesn't stay as one intact molecule in water — it partly splits into ions, and every extra particle adds to the freezing-point drop. Skipping that step is the single biggest way to get this problem wrong.CH\(\displaystyle _3\)CH\(\displaystyle _2\)CHClCOOH ($\displaystyle 2$-chlorobutanoic acid) is a weak monoprotic acid. In water it partially ionises:\[\text{CH}_3\text{CH}_2\text{CHClCOOH} \rightleftharpoons \text{CH}_3\text{CH}_2\text{CHClCOO}^- + \text{H}^+ \]so the solution contains more particles than the moles of acid dissolved would suggest. That means the ordinary formula \(\displaystyle \Delta T_f = K_f m\) underestimates the depression — it needs a van't Hoff factor \(\displaystyle i\) to correct for the extra particles from ionisation.Step $\displaystyle 1$ — Molar mass of the acid. CH\(\displaystyle _3\)CH\(\displaystyle _2\)CHClCOOH has the formula C\(\displaystyle _4\)H\(\displaystyle _7\)ClO\(\displaystyle _2\):\[M = 4(12) + 7(1) + 35.5 + 2(16) = 48 + 7 + 35.5 + 32 = 122.5\ \text{g mol}^{-1} \]Step $\displaystyle 2$ — Molality of the solution. Molality \(\displaystyle m\) is moles of solute per kilogram of solvent (water here), not per litre of solution — don't reach for the volume of solution, there isn't one given.\[n = \frac{\text{mass}}{M} = \frac{10\ \text{g}}{122.5\ \text{g mol}^{-1}} = 0.08163\ \text{mol} \]\[m = \frac{n}{\text{mass of water in kg}} = \frac{0.08163\ \text{mol}}{0.250\ \text{kg}} = 0.3265\ \text{mol kg}^{-1} \]Step $\displaystyle 3$ — Degree of dissociation \(\displaystyle \alpha\), from \(\displaystyle K_a\). \(\displaystyle K_a\) is the acid-dissociation constant. Treating the dilute aqueous concentration as \(\displaystyle C \approx 0.3265\ \text{mol L}^{-1}\) (molarity ≈ molality here since the solution is dilute), and letting \(\displaystyle \alpha\) be the fraction of acid molecules that ionise:\[K_a = \frac{C\alpha^2}{1-\alpha} \approx C\alpha^2 \quad (\text{since } K_a \text{ is small}, \ \alpha \ll 1,\ 1-\alpha \approx 1) \]\[\alpha = \sqrt{\frac{K_a}{C}} = \sqrt{\frac{1.4\times10^{-3}}{0.3265}} = \sqrt{4.287\times10^{-3}} = 0.06548 \]So about $\displaystyle 6.5$% of the acid molecules ionise — small enough that the \(\displaystyle 1-\alpha \approx 1\) approximation used above is reasonable.Step $\displaystyle 4$ — Van't Hoff factor. Each molecule that ionises turns into $\displaystyle 2$ particles (H\(\displaystyle ^+\) and the anion) instead of $\displaystyle 1$, so:\[i = 1 + \alpha = 1 + 0.06548 = 1.0655 \]This is the correction people forget: without it you'd compute \(\displaystyle \Delta T_f\) as if the acid stayed $\displaystyle 100$% molecular, and undershoot the real depression.Step $\displaystyle 5$ — Freezing-point depression. The corrected colligative-property formula is \(\displaystyle \Delta T_f = i K_f m\), where \(\displaystyle K_f\) is the molal freezing-point-depression constant of water.\[\Delta T_f = i \times K_f \times m = 1.0655 \times 1.86\ \text{K kg mol}^{-1} \times 0.3265\ \text{mol kg}^{-1} \]\[\Delta T_f = 1.0655 \times 0.6073\ \text{K} = 0.6471\ \text{K} \]Rounding to three significant figures (matching the precision of the given data):Answer: \(\displaystyle \Delta T_f \approx 0.647\ \text{K}\) (the solution freezes at about \(\displaystyle -0.647\,^\circ\text{C}\) instead of \(\displaystyle 0\,^\circ\text{C}\))

The van't Hoff factor \(\displaystyle i\) tells you how many particles one formula unit actually produces in solution — a weak acid that partly ionises gives slightly more than $\displaystyle 1$, and that "slightly more" is what you measure through the freezing-point depression.Step $\displaystyle 1$ — molar mass of the solute\(\displaystyle CH_2FCOOH\) has formula \(\displaystyle C_2H_3FO_2\): \[M = 2(12.0) + 3(1.0) + 19.0 + 2(16.0) = 24.0+3.0+19.0+32.0 = 78.0\ \text{g mol}^{-1} \]Step $\displaystyle 2$ — moles and molality\[n_{\text{solute}} = \frac{19.5\ \text{g}}{78.0\ \text{g mol}^{-1}} = 0.250\ \text{mol} \]Molality \(\displaystyle m\) is moles of solute per kilogram of solvent, not per litre of solution — freezing-point depression is a colligative property tied to the solvent's particle count, so it is always molality here, never molarity.\[m = \frac{0.250\ \text{mol}}{0.500\ \text{kg}} = 0.500\ \text{mol kg}^{-1} \]Step $\displaystyle 3$ — van't Hoff factor from \(\displaystyle \Delta T_f\)\[\Delta T_f = i\,K_f\,m \] where \(\displaystyle \Delta T_f\) is the freezing-point depression, \(\displaystyle K_f = 1.86\ \text{K kg mol}^{-1}\) is water's molal depression constant, and \(\displaystyle m\) is the molality just found.\[i = \frac{\Delta T_f}{K_f\,m} = \frac{1.0\ \text{K}}{(1.86\ \text{K kg mol}^{-1})(0.500\ \text{mol kg}^{-1})} = \frac{1.0}{0.930} = 1.0753 \]An \(\displaystyle i\) of exactly $\displaystyle 1$ would mean the acid stays fully intact (no dissociation); \(\displaystyle i>1\) means each molecule is, on average, breaking into more than one particle — here \(\displaystyle i=1.0753\) says only a small fraction has ionised.Step $\displaystyle 4$ — degree of dissociation and \(\displaystyle K_a\)\(\displaystyle CH_2FCOOH\) dissociates as \[CH_2FCOOH \rightleftharpoons CH_2FCOO^- + H^+ \] one molecule giving two particles, so if \(\displaystyle \alpha\) is the degree of dissociation, \[i = 1+\alpha \quad\Rightarrow\quad \alpha = i-1 = 1.0753-1 = 0.0753 \]Setting up the equilibrium (initial concentration \(\displaystyle C=0.500\ \text{mol L}^{-1}\), treating the dilute solution's molality as equal to its molarity):

	\(\displaystyle CH_2FCOOH\)	\(\displaystyle CH_2FCOO^-\)	\(\displaystyle H^+\)
initial	\(\displaystyle C\)	$\displaystyle 0$	$\displaystyle 0$
equilibrium	\(\displaystyle C(1-\alpha)\)	\(\displaystyle C\alpha\)	\(\displaystyle C\alpha\)

\[K_a = \frac{[CH_2FCOO^-][H^+]}{[CH_2FCOOH]} = \frac{C\alpha^{2}}{1-\alpha} \]Substituting the unrounded \(\displaystyle \alpha = 0.075269\) (rounding only at the very end, not mid-calculation): \[K_a = \frac{(0.500)(0.075269)^{2}}{1-0.075269} = \frac{(0.500)(0.0056654)}{0.924731} = \frac{0.0028327}{0.924731} = 3.063\times10^{-3} \]The step people usually fumble here is reusing the rounded \(\displaystyle i\) ($\displaystyle 1.08$ or so) instead of the fuller value — that shifts \(\displaystyle K_a\) into the third significant figure, so keep the extra digits of \(\displaystyle \alpha\) until the last line.Rounding to three significant figures, matching the precision of the given \(\displaystyle \Delta T_f = 1.0^{\circ}\text{C}\):\[i \approx 1.08 \ (\text{more precisely } 1.0753), \qquad K_a \approx 3.06\times10^{-3} \]Answer: van't Hoff factor \(\displaystyle i \approx 1.075\); dissociation constant \(\displaystyle K_a \approx 3.06\times10^{-3}\ \text{mol L}^{-1}\).

Question (NCERT Class $\displaystyle 12$ Chemistry, Ch.1 Solutions, Q1.34): Vapour pressure of water at $\displaystyle 293$ K is $\displaystyle 17.535$ mm Hg. Calculate the vapour pressure of water at $\displaystyle 293$ K when $\displaystyle 25$ g of glucose is dissolved in $\displaystyle 450$ g of water.Worked solution (final text to be published):Raoult's law for a solution of a non-volatile solute says the relative lowering in vapour pressure equals the mole fraction of the solute — not the solvent, and not a raw mass ratio.Step $\displaystyle 1$ — moles of each component.Glucose is \(\displaystyle C_6H_{12}O_6\), so its molar mass is \[M_2 = 6(12) + 12(1) + 6(16) = 180\ \text{g mol}^{-1} \] \[n_2 = \frac{w_2}{M_2} = \frac{25\ \text{g}}{180\ \text{g mol}^{-1}} = 0.1389\ \text{mol} \]Water has \(\displaystyle M_1 = 18\ \text{g mol}^{-1}\): \[n_1 = \frac{w_1}{M_1} = \frac{450\ \text{g}}{18\ \text{g mol}^{-1}} = 25.00\ \text{mol} \]Step $\displaystyle 2$ — mole fraction of the solute. This is the step people get wrong: the denominator is total moles \(\displaystyle n_1+n_2\), not just the solvent's \(\displaystyle n_1\). \[x_2 = \frac{n_2}{n_1+n_2} = \frac{0.1389}{25.00+0.1389} = \frac{0.1389}{25.14} = 5.525\times10^{-3} \]Step $\displaystyle 3$ — Raoult's law. For a dilute solution of a non-volatile, non-electrolyte solute, \[\frac{p_1^{\circ}-p_1}{p_1^{\circ}} = x_2 \] where \(\displaystyle p_1^{\circ}=17.535\) mm Hg is the vapour pressure of pure water at $\displaystyle 293$ K and \(\displaystyle p_1\) is the solution's vapour pressure. \[p_1^{\circ}-p_1 = p_1^{\circ}\,x_2 = 17.535 \times 5.525\times10^{-3} = 0.09688\ \text{mm Hg} \]Step $\displaystyle 4$ — solve for \(\displaystyle p_1\). \[p_1 = p_1^{\circ} - (p_1^{\circ}-p_1) = 17.535 - 0.0969 = 17.438\ \text{mm Hg} \]Rounding once, to four significant figures (matching the precision the data supports):Answer: The vapour pressure of the solution is $\displaystyle 17.44$ mm Hg — a drop of about $\displaystyle 0.10$ mm Hg from pure water's $\displaystyle 17.535$ mm Hg.

Henry's law connects the partial pressure of a gas above a solution to its mole fraction dissolved in it — it does not directly give molality or moles, so the calculation goes through mole fraction first.Henry's law states \[p = K_H \, x \] where \(\displaystyle p\) is the partial pressure of the gas above the solution, \(\displaystyle K_H\) is the Henry's law constant, and \(\displaystyle x\) is the mole fraction of the gas (methane) dissolved in the solvent (benzene).Here \[K_H = 4.27 \times 10^{5}\ \text{mm Hg}, \qquad p = 760\ \text{mm Hg} \]Rearranging for the mole fraction of methane: \[x_{\mathrm{CH_4}} = \dfrac{p}{K_H} = \dfrac{760\ \text{mm Hg}}{4.27 \times 10^{5}\ \text{mm Hg}} \]A step people rush past: \(\displaystyle K_H\) here carries units of pressure, so \(\displaystyle x\) comes out as a pure (unitless) number only because the mm Hg cancels — do not attach mm Hg to the final answer.\[x_{\mathrm{CH_4}} = \dfrac{760}{4.27 \times 10^{5}} = 1.7799 \times 10^{-3} \]This mole fraction is the solubility of methane in benzene under these conditions — it says that out of every mole of solution, about \(\displaystyle 1.78 \times 10^{-3}\) mol is dissolved methane. Rounding to three significant figures (matching the three sig figs in \(\displaystyle 4.27\)):Answer: \(\displaystyle x_{\mathrm{CH_4}} \approx 1.79 \times 10^{-3}\) (mole fraction of methane dissolved in benzene at $\displaystyle 298$ K, $\displaystyle 760$ mm Hg)

In an ideal solution each component obeys Raoult's law on its own — \(\displaystyle \mathrm{P_i = x_i P_i^{\circ}^{-}}\) and the total vapour pressure is just the sum of these two partial pressures (Dalton's law). The question is really asking for two different things: the pressure A contributes while dissolved, and the vapour pressure of A if it stood alone as a pure liquid — do not treat these as the same number.Step $\displaystyle 1$ — moles of each liquid. Use \(\displaystyle n = \dfrac{\text{mass}}{\text{molar mass}}\).\[n_A = \frac{100\ \text{g}}{140\ \text{g mol}^{-1}} = 0.7143\ \text{mol}, \qquad n_B = \frac{1000\ \text{g}}{180\ \text{g mol}^{-1}} = 5.5556\ \text{mol} \]Step $\displaystyle 2$ — mole fractions in the liquid. \[x_A = \frac{n_A}{n_A+n_B} = \frac{0.7143}{0.7143+5.5556} = \frac{0.7143}{6.2698} = 0.11392 \] \[x_B = 1 - x_A = 0.88608 \]Step $\displaystyle 3$ — partial pressure of B in the solution. Raoult's law: \(\displaystyle P_B = x_B\,P_B^{\circ}\), where \(\displaystyle P_B^{\circ}=500\) Torr is the vapour pressure of pure B (given). \[P_B = 0.88608 \times 500\ \text{Torr} = 443.04\ \text{Torr} \]Step $\displaystyle 4$ — partial pressure of A in the solution. Total pressure is the sum of the two partial pressures: \(\displaystyle P_{\text{total}} = P_A + P_B\). This \(\displaystyle P_A\) is A's contribution in the mixture, not A's pure-liquid vapour pressure — that distinction is the crux of the problem. \[P_A = P_{\text{total}} - P_B = 475\ \text{Torr} - 443.04\ \text{Torr} = 31.96\ \text{Torr} \]Step $\displaystyle 5$ — vapour pressure of pure A. Now invert Raoult's law for A, \(\displaystyle P_A = x_A P_A^{\circ}\), to get the number the question actually wants — what A's vapour pressure would be with no B around: \[P_A^{\circ} = \frac{P_A}{x_A} = \frac{31.96\ \text{Torr}}{0.11392} = 280.6\ \text{Torr} \]Rounding once, at the end, to four significant figures (matching the precision of the given data):Answer: vapour pressure of pure liquid A, \(\displaystyle P_A^{\circ} \approx 280.6\) Torr; vapour pressure of A in the solution, \(\displaystyle P_A \approx 31.96\) Torr.

Raoult's Law gives the vapour pressure a mixture should have if the two liquids behave ideally — you compare that predicted line to the pressures actually measured, and the direction of the gap tells you the sign of the deviation.For an ideal solution of acetone (A) and chloroform (C), Raoult's Law says each component's partial vapour pressure is proportional to its mole fraction in the liquid, and the total pressure is just their sum:\[p_A = x_A\,p_A^{\circ}, \qquad p_C = (1-x_A)\,p_C^{\circ}, \qquad p_{\text{total}} = p_A + p_C \]Here \(\displaystyle p_A^{\circ}=741.8\) mm Hg (pure acetone) and \(\displaystyle p_C^{\circ}=632.8\) mm Hg (pure chloroform), and \(\displaystyle x_A\) is the mole fraction of acetone, \(\displaystyle 100x_A\) being the values given in the data ($\displaystyle 0$, $\displaystyle 11.8$, $\displaystyle 23.4$, …).Worked example (the point of largest gap, \(\displaystyle x_A = 0.360\)):\[p_{A,\text{ideal}} = 0.360 \times 741.8 = 267.0\ \text{mm Hg} \] \[p_{C,\text{ideal}} = (1-0.360)\times 632.8 = 0.640\times632.8 = 405.0\ \text{mm Hg} \] \[p_{\text{total,ideal}} = 267.0 + 405.0 = 672.0\ \text{mm Hg} \]The measured values at this same composition are \(\displaystyle p_A = 202.4\), \(\displaystyle p_C = 359.7\), so \(\displaystyle p_{\text{total,exp}} = 562.1\) mm Hg — nearly $\displaystyle 110$ mm Hg below what Raoult's Law predicts. Repeating this substitution at every composition gives:

\(\displaystyle x_A\)	\(\displaystyle p_{A,\text{ideal}}\)	\(\displaystyle p_A^{\text{exp}}\)	\(\displaystyle p_{C,\text{ideal}}\)	\(\displaystyle p_C^{\text{exp}}\)	\(\displaystyle p_{\text{total,ideal}}\)	\(\displaystyle p_{\text{total}}^{\text{exp}}\)	gap (mm Hg)
$\displaystyle 0$	$\displaystyle 0$	$\displaystyle 0$	$\displaystyle 632.8$	$\displaystyle 632.8$	$\displaystyle 632.8$	$\displaystyle 632.8$	$\displaystyle 0$
$\displaystyle 0.118$	$\displaystyle 87.5$	$\displaystyle 54.9$	$\displaystyle 558.1$	$\displaystyle 548.1$	$\displaystyle 645.7$	$\displaystyle 603.0$	$\displaystyle 42.7$
$\displaystyle 0.234$	$\displaystyle 173.6$	$\displaystyle 110.1$	$\displaystyle 484.7$	$\displaystyle 469.4$	$\displaystyle 658.3$	$\displaystyle 579.5$	$\displaystyle 78.8$
$\displaystyle 0.360$	$\displaystyle 267.0$	$\displaystyle 202.4$	$\displaystyle 405.0$	$\displaystyle 359.7$	$\displaystyle 672.0$	$\displaystyle 562.1$	$\displaystyle 109.9$
$\displaystyle 0.508$	$\displaystyle 376.8$	$\displaystyle 322.7$	$\displaystyle 311.3$	$\displaystyle 257.7$	$\displaystyle 688.2$	$\displaystyle 580.4$	$\displaystyle 107.8$
$\displaystyle 0.582$	$\displaystyle 431.7$	$\displaystyle 405.9$	$\displaystyle 264.5$	$\displaystyle 193.6$	$\displaystyle 696.2$	$\displaystyle 599.5$	$\displaystyle 96.7$
$\displaystyle 0.645$	$\displaystyle 478.5$	$\displaystyle 454.1$	$\displaystyle 224.6$	$\displaystyle 161.2$	$\displaystyle 703.1$	$\displaystyle 615.3$	$\displaystyle 87.8$
$\displaystyle 0.721$	$\displaystyle 534.8$	$\displaystyle 521.1$	$\displaystyle 176.6$	$\displaystyle 120.7$	$\displaystyle 711.4$	$\displaystyle 641.8$	$\displaystyle 69.6$

Turning this into the graph: plot \(\displaystyle x_{\text{acetone}}\) ($\displaystyle 0$ to $\displaystyle 1$) on the horizontal axis and pressure (mm Hg) on the vertical axis. Because Raoult's Law is linear in \(\displaystyle x_A\), the ideal lines are straight: \(\displaystyle p_{A,\text{ideal}}\) runs from \(\displaystyle (0,0)\) to \(\displaystyle (1,741.8)\), \(\displaystyle p_{C,\text{ideal}}\) runs from \(\displaystyle (0,632.8)\) down to \(\displaystyle (1,0)\), and \(\displaystyle p_{\text{total,ideal}}\) is the straight line joining \(\displaystyle (0,632.8)\) to \(\displaystyle (1,741.8)\). Now plot the experimental points from the table on the same axes and join them with a smooth curve: the \(\displaystyle p_A^{\text{exp}}\) curve sits below the straight ideal-acetone line, the \(\displaystyle p_C^{\text{exp}}\) curve sits below the straight ideal-chloroform line, and — most visibly — the experimental \(\displaystyle p_{\text{total}}\) curve sags well below the straight ideal-total line, dipping to a broad minimum in the middle of the composition range (the gap peaks near \(\displaystyle x_A \approx 0.36\), where it is nearly $\displaystyle 110$ mm Hg).Every experimental pressure lies below its Raoult's-Law value, at every composition — this is a negative deviation, not a positive one. Physically, chloroform's C–H hydrogen bonds to the oxygen of acetone's carbonyl group (\(\displaystyle \text{CHCl}_3 \cdots \text{O=C(CH}_3)_2\)); this attraction between unlike molecules is stronger than the acetone–acetone and chloroform–chloroform attractions in the pure liquids. Stronger A–C attraction lowers the escaping tendency of both molecules, so both partial pressures — and hence the total pressure — fall below the values ideal (non-interacting) mixing would predict, at every composition tested.Answer: The acetone–chloroform system shows negative deviation from Raoult's Law — the measured total vapour pressure lies below the straight line predicted by Raoult's Law across the whole composition range (by up to about $\displaystyle 110$ mm Hg near \(\displaystyle x_{\text{acetone}} \approx 0.36\)), because hydrogen bonding between chloroform and acetone makes the unlike-molecule attraction stronger than in the pure liquids.

Question $\displaystyle 1.38$ (NCERT XII Chemistry, Ch. $\displaystyle 1$ Solutions): Benzene ($\displaystyle 80$ g) + toluene ($\displaystyle 100$ g) form an ideal solution at $\displaystyle 300$ K; p°(benzene)=$\displaystyle 50.71$ mmHg, p°(toluene)=$\displaystyle 32.06$ mmHg. Find mole fraction of benzene in the vapour phase.Worked solution: 1. Moles: n(benzene)=$\displaystyle 80$/$\displaystyle 78$=$\displaystyle 1.0256$ mol (M=$\displaystyle 78$ g/mol for C6H6); n(toluene)=$\displaystyle 100$/$\displaystyle 92$=$\displaystyle 1.0870$ mol (M=$\displaystyle 92$ g/mol for C7H8). 2. Liquid-phase mole fractions: x(benzene)=$\displaystyle 1.0256$/$\displaystyle 2.1126$=$\displaystyle 0.4855$; x(toluene)=0.5145. 3. Raoult's law partial pressures: p(benzene)=x(benzene)p°(benzene)=$\displaystyle 0.4855$$\displaystyle 50.71$=$\displaystyle 24.62$ mmHg; p(toluene)=x(toluene)p°(toluene)=$\displaystyle 0.5145$$\displaystyle 32.06$=$\displaystyle 16.50$ mmHg. 4. Total pressure: p_total=$\displaystyle 24.62$+$\displaystyle 16.50$=$\displaystyle 41.11$ mmHg. 5. Dalton's law for vapour composition: y(benzene)=p(benzene)/p_total=$\displaystyle 24.62$/$\displaystyle 41.11$=0.599.The full KaTeX-formatted write-up (with the bold lead line, named formulas, unit tracking, an aside flagging that liquid-phase x and vapour-phase y are different quantities, and a final Answer: line) was produced as the deliverable text in this turn.

Henry's law connects the partial pressure of a gas ABOVE a liquid to the mole fraction of that gas dissolved IN the liquid — \(\displaystyle p = K_H \, x \) — so the first job is to get the partial pressures of \(\displaystyle \mathrm{O_2}\) and \(\displaystyle \mathrm{N_2}\) right, unit and all, before touching \(\displaystyle K_H\).Step $\displaystyle 1$: Partial pressure of each gas in the air (Dalton's law).For an ideal gas mixture, volume fraction equals mole fraction, and the partial pressure of a component is\[p_i = x_i \times p_{\text{total}} \]Here the mixture is $\displaystyle 20$% \(\displaystyle \mathrm{O_2}\) and $\displaystyle 79$% \(\displaystyle \mathrm{N_2}\) by volume, and the total pressure of air over the water is \(\displaystyle p_{\text{total}} = 10\ \text{atm}\).\[p_{\mathrm{O_2}} = 0.20 \times 10\ \text{atm} = 2\ \text{atm} \] \[p_{\mathrm{N_2}} = 0.79 \times 10\ \text{atm} = 7.9\ \text{atm} \]The step people skip: match the units to \(\displaystyle K_H\). The Henry's law constants here are given in mm Hg, not atm, so the partial pressures must be converted before dividing — mixing atm with a mm-Hg constant silently gives an answer $\displaystyle 760$ times too large.Using \(\displaystyle 1\ \text{atm} = 760\ \text{mm Hg}\):\[p_{\mathrm{O_2}} = 2 \times 760\ \text{mm Hg} = 1520\ \text{mm Hg} \] \[p_{\mathrm{N_2}} = 7.9 \times 760\ \text{mm Hg} = 6004\ \text{mm Hg} \]Step $\displaystyle 2$: Apply Henry's law to get the mole fraction dissolved in water.Henry's law states\[p = K_H \, x \quad \Longrightarrow \quad x = \frac{p}{K_H} \]where \(\displaystyle p\) is the partial pressure of the gas over the solution, \(\displaystyle K_H\) is Henry's law constant for that gas (in the same pressure unit), and \(\displaystyle x\) is the mole fraction of the gas dissolved in the liquid (water) at equilibrium.For oxygen, with \(\displaystyle K_H(\mathrm{O_2}) = 3.30 \times 10^{7}\ \text{mm Hg}\):\[x_{\mathrm{O_2}} = \frac{1520\ \text{mm Hg}}{3.30 \times 10^{7}\ \text{mm Hg}} = 4.606 \times 10^{-5} \]For nitrogen, with \(\displaystyle K_H(\mathrm{N_2}) = 6.51 \times 10^{7}\ \text{mm Hg}\):\[x_{\mathrm{N_2}} = \frac{6004\ \text{mm Hg}}{6.51 \times 10^{7}\ \text{mm Hg}} = 9.223 \times 10^{-5} \]These mole fractions are tiny, as expected — \(\displaystyle K_H\) is enormous compared to the partial pressures, so only a very small fraction of the dissolved gas molecules end up as \(\displaystyle \mathrm{O_2}\) or \(\displaystyle \mathrm{N_2}\) among the water molecules. Nitrogen dissolves to roughly twice the mole fraction of oxygen here (larger \(\displaystyle p\) roughly offsetting its larger \(\displaystyle K_H\)), which is exactly why nitrogen — not oxygen — is the gas that causes the bends in divers surfacing too fast.Rounding each to three significant figures (matching the precision of the given data):\[x_{\mathrm{O_2}} \approx 4.61 \times 10^{-5}, \qquad x_{\mathrm{N_2}} \approx 9.22 \times 10^{-5} \]Answer: mole fraction of \(\displaystyle \mathrm{O_2}\) dissolved in water \(\displaystyle \approx 4.61 \times 10^{-5}\); mole fraction of \(\displaystyle \mathrm{N_2}\) dissolved in water \(\displaystyle \approx 9.22 \times 10^{-5}\).

Question: Ch.1 Solutions, Q1.40 — Determine the amount of \(\displaystyle \mathrm{CaCl_{2}}\) (i = $\displaystyle 2.47$) dissolved in $\displaystyle 2.5$ litre of water such that its osmotic pressure is $\displaystyle 0.75$ atm at $\displaystyle 27$°C.Method: π = i(n/V)RT → n = πV/(iRT). With T = $\displaystyle 300$ K, π = $\displaystyle 0.75$ atm, V = $\displaystyle 2.5$ L, i = $\displaystyle 2.47$, R = $\displaystyle 0.0821$ L atm K⁻¹ mol⁻¹: n = ($\displaystyle 0.75$×$\displaystyle 2.5$)/($\displaystyle 2.47$×$\displaystyle 0.0821$×$\displaystyle 300$) = $\displaystyle 1.875$/$\displaystyle 60.84$ = $\displaystyle 0.03082$ mol Mass = n × M(CaCl2) = $\displaystyle 0.03082$ × $\displaystyle 111$ g/mol = $\displaystyle 3.42$ gFull worked solution written per the SolveIt NCERT style guide (bold lead idea, named formula with symbols defined, substitution shown, units carried, K-vs-°C aside, ends with Answer: $\displaystyle 3.42$ g of \(\displaystyle \mathrm{CaCl_{2}}\)) was provided in the assistant's final message.

Osmotic pressure of an electrolyte solution needs the van't Hoff factor \(\displaystyle i\) built into the formula — a salt that dissociates behaves as more particles than its formula-mass would suggest.The governing equation is \[\pi = i\,C\,R\,T \] where \(\displaystyle \pi\) is the osmotic pressure, \(\displaystyle i\) is the van't Hoff factor (number of particles produced per formula unit on complete dissociation), \(\displaystyle C\) is the molar concentration of the solute (mol/L, based on formula units, not on the dissociated ions), \(\displaystyle R\) is the gas constant, and \(\displaystyle T\) is the absolute temperature.Step $\displaystyle 1$: Molar mass of \(\displaystyle K_2SO_4\).Using \(\displaystyle K=39,\ S=32,\ O=16\): \[M = 2(39) + 32 + 4(16) = 78 + 32 + 64 = 174\ \text{g mol}^{-1} \]Step $\displaystyle 2$: Moles of \(\displaystyle K_2SO_4\) dissolved.The mass given, $\displaystyle 25$ mg, must be converted to grams before dividing by molar mass — mixing mg and g here is the usual slip. \[n = \frac{25\times10^{-3}\ \text{g}}{174\ \text{g mol}^{-1}} = 1.437\times10^{-4}\ \text{mol} \]Step $\displaystyle 3$: Molar concentration.\[C = \frac{n}{V} = \frac{1.437\times10^{-4}\ \text{mol}}{2\ \text{L}} = 7.184\times10^{-5}\ \text{mol L}^{-1} \]Step $\displaystyle 4$: Van't Hoff factor.\(\displaystyle K_2SO_4\) dissociates completely as \[K_2SO_4 \rightarrow 2K^{+} + SO_4^{2-} \] giving $\displaystyle 3$ ions per formula unit, so \(\displaystyle i = 3\). This is the step that is easy to skip: without it you would compute the osmotic pressure of the un-ionised salt, which is only a third of the real value.Step $\displaystyle 5$: Temperature in kelvin.\[T = 25^{\circ}\text{C} + 273 = 298\ \text{K} \]Using degrees Celsius directly in \(\displaystyle \pi = iCRT\) instead of converting to kelvin is the other common error.Step $\displaystyle 6$: Substitute into \(\displaystyle \pi = iCRT\).Using \(\displaystyle R = 0.0821\ \text{L atm K}^{-1}\text{mol}^{-1}\): \[\pi = 3 \times \left(7.184\times10^{-5}\ \text{mol L}^{-1}\right) \times \left(0.0821\ \text{L atm K}^{-1}\text{mol}^{-1}\right) \times \left(298\ \text{K}\right) \]Working through the multiplication: \[0.0821 \times 298 = 24.47\ \text{L atm K}^{-1}\text{mol}^{-1}\cdot\text{K} = 24.47\ \text{L atm mol}^{-1} \] \[7.184\times10^{-5}\ \text{mol L}^{-1} \times 24.47\ \text{L atm mol}^{-1} = 1.758\times10^{-3}\ \text{atm} \] \[\pi = 3 \times 1.758\times10^{-3}\ \text{atm} = 5.273\times10^{-3}\ \text{atm} \]Rounding to three significant figures, consistent with the precision of the $\displaystyle 25$ mg mass given:Answer: \(\displaystyle \pi \approx 5.27 \times 10^{-3}\ \text{atm}\)