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.

Raoult's Law says the vapour pressure of the solution equals the vapour pressure of the pure solvent times the solvent's mole fraction — and the relative drop in vapour pressure equals the solute's mole fraction. Two vapour-pressure readings on the same amount of solute (only water is added between them) give two equations for two unknowns: the molar mass of the solute and the vapour pressure of pure water.Setting up the relationFor a solution of a non-volatile solute in a volatile solvent, \[p_1 = p_1^{\circ}\,x_1 \qquad \Rightarrow \qquad \frac{p_1^{\circ}-p_1}{p_1^{\circ}} = x_2 = \frac{n_2}{n_1+n_2} \] where \(\displaystyle p_1\) is the vapour pressure of the solution, \(\displaystyle p_1^{\circ}\) is the vapour pressure of pure water (unknown), \(\displaystyle n_1\) is moles of water (solvent), and \(\displaystyle n_2\) is moles of solute. The solute is non-volatile, so \(\displaystyle n_2\) is the same number of moles in both readings — only \(\displaystyle n_1\) changes when water is added. That is the fact that lets two readings pin down two unknowns.Reading $\displaystyle 1$: $\displaystyle 90$ g water, \(\displaystyle p_1 = 2.8\) kPaMoles of water: \[n_1 = \frac{90\ \text{g}}{18\ \text{g mol}^{-1}} = 5\ \text{mol} \] Let \(\displaystyle n_2\) mol be the (unknown, fixed) amount of solute. Then \[\frac{p_1^{\circ}-2.8}{p_1^{\circ}} = \frac{n_2}{5+n_2} \] Rearranging (this is the step people skip, and it's the one that makes the algebra tractable): \[p_1^{\circ}\left(1-\frac{n_2}{5+n_2}\right)=2.8 \quad\Rightarrow\quad p_1^{\circ}\cdot\frac{5}{5+n_2}=2.8 \quad\Rightarrow\quad p_1^{\circ}=\frac{2.8(5+n_2)}{5} \qquad (1) \]Reading $\displaystyle 2$: $\displaystyle 108$ g water, \(\displaystyle p_1 = 2.9\) kPaAfter adding $\displaystyle 18$ g water, the water totals \(\displaystyle 90+18=108\) g — mass of solution, not just solvent, is a common slip here, but Raoult's Law only ever needs the solvent mass. Moles of water now: \[n_1' = \frac{108\ \text{g}}{18\ \text{g mol}^{-1}} = 6\ \text{mol} \] The same \(\displaystyle n_2\) mol of solute is still dissolved (nothing was removed), so \[\frac{p_1^{\circ}-2.9}{p_1^{\circ}} = \frac{n_2}{6+n_2} \quad\Rightarrow\quad p_1^{\circ}=\frac{2.9(6+n_2)}{6} \qquad (2) \]Solving the two equations for \(\displaystyle n_2\)Set ($\displaystyle 1$) equal to ($\displaystyle 2$): \[\frac{2.8(5+n_2)}{5} = \frac{2.9(6+n_2)}{6} \] Multiply both sides by $\displaystyle 30$ to clear denominators: \[6\times 2.8(5+n_2) = 5\times 2.9(6+n_2) \] \[16.8(5+n_2) = 14.5(6+n_2) \] \[84 + 16.8\,n_2 = 87 + 14.5\,n_2 \] \[2.3\,n_2 = 3 \quad\Rightarrow\quad n_2 = \frac{3}{2.3} = 1.304\ \text{mol} \]Part (i): molar mass of the soluteThe $\displaystyle 30$ g of solute corresponds to this \(\displaystyle n_2\): \[M_2 = \frac{w_2}{n_2} = \frac{30\ \text{g}}{1.304\ \text{mol}} = 23.0\ \text{g mol}^{-1} \]Part (ii): vapour pressure of pure waterSubstitute \(\displaystyle n_2 = 1.304\) mol back into equation ($\displaystyle 1$): \[p_1^{\circ} = \frac{2.8(5+1.304)}{5} = \frac{2.8\times 6.304}{5} = \frac{17.65}{5} = 3.530\ \text{kPa} \] Checking with equation ($\displaystyle 2$) as well: \[p_1^{\circ} = \frac{2.9(6+1.304)}{6} = \frac{2.9\times 7.304}{6} = \frac{21.18}{6} = 3.530\ \text{kPa} \] Both readings agree, confirming the value.Answer: molar mass of the solute \(\displaystyle M_2 \approx 23.0\ \text{g mol}^{-1}\); vapour pressure of pure water at $\displaystyle 298$ K, \(\displaystyle p_1^{\circ} \approx 3.53\ \text{kPa}\).

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: ...").