NCERT Solutions · Class 12 Chemistry Solutions

A solution is a homogeneous mixture of two or more chemically non-reacting substances, mixed in a single phase, whose relative composition can be varied within certain limits.Because it is homogeneous, a solution looks and behaves the same throughout — you cannot see the individual components, and a small sample taken from anywhere in it has the same composition as the whole. The component present in the larger proportion is called the solvent, and the component(s) present in smaller proportion is the solute.A short aside people get wrong: "solute" and "solvent" are defined by relative amount, not by physical state. A solid dissolved in a liquid is the usual picture, but a gas can be the solvent too (as in air), and even two liquids form a solute–solvent pair by whichever is present in excess.Classifying the types. A solution is classified according to the physical state of its solute and of its solvent. Each of the two components — solute and solvent — can independently be a solid, a liquid, or a gas, so the number of possible combinations is \[3 \times 3 = 9 \] This gives nine types of solutions: **Answer: A solution is a homogeneous mixture of two or more non-reacting substances with variable but limited composition. Since the solute and the solvent can each independently be a gas, a liquid, or a solid, there are \(\displaystyle 3 \times 3 = 9\) possible types of solutions — gas–gas, liquid–gas, solid–gas, gas–liquid, liquid–liquid, solid–liquid, gas–solid, liquid–solid, and solid–solid — each illustrated above with an example (air; moist air; naphthalene vapour in air; soda water; ethanol in water; salt in water; \(\displaystyle \mathrm{H_2} \) in Pd; mercury amalgam; and brass, respectively).

A solid solution is any solute dissolved in a solid solvent — the solute does not have to be a solid itself. A classic case with a gaseous solute is hydrogen gas absorbed into solid palladium.Solutions are classified by the physical states of the solute and the solvent, not just by whether the mixture "looks solid." Nine combinations are possible (gas-in-gas, gas-in-liquid, gas-in-solid, liquid-in-solid, solid-in-solid, and so on), and the state that decides the type of solution is always the solvent's state, since the solvent is the component present in larger amount that determines the phase of the final mixture.Here the solvent is solid palladium metal, and the solute is hydrogen gas, \(\displaystyle \text{H}_2 \). When palladium is exposed to hydrogen gas, the small \(\displaystyle \text{H}_2 \) molecules diffuse into the gaps between the metal atoms in the palladium lattice and get trapped there. The bulk material that results is still a solid — it retains the metallic lattice of palladium — so this is a solid solution (solid solvent) with a gas as the solute.A short aside on the trap here: it is tempting to think "solid solution" must mean solid dissolved in solid, like an alloy. That is only one member of the family — the defining feature is the solvent's state being solid, and the solute's state can be gas, liquid, or solid.Other accepted examples of the same gas-in-solid type include hydrogen gas dissolved in solid platinum, and traces of gases occluded in various metals during their formation.Answer: Hydrogen gas dissolved in solid palladium (H\(\displaystyle _2\) in Pd) — a solid solution with a gaseous solute.

Mole fraction, molality, molarity and mass percentage are the four ways to state "how much solute is in how much solution" — they differ only in what you put in the denominator: total moles, mass of solvent, volume of solution, or mass of solution.(i) Mole fractionMole fraction is the fraction of moles a component contributes to the total moles present — it never mixes in a mass or a volume at all. For a solution with components \(\displaystyle 1, 2, 3, \dots\), the mole fraction of component \(\displaystyle i\) is\[x_i = \frac{n_i}{n_1 + n_2 + n_3 + \cdots} \]where \(\displaystyle n_i\) is the number of moles of component \(\displaystyle i\), and the denominator is the total number of moles of every component in the solution (solute + solvent together).Because it is a ratio of moles to moles, mole fraction has no unit — it is a pure number between $\displaystyle 0$ and $\displaystyle 1$, and the mole fractions of all components of a solution always add up to $\displaystyle 1$: \[x_1 + x_2 + x_3 + \cdots = 1. \](ii) MolalityMolality is defined on the mass of the solvent, not the mass or volume of the solution — that is the detail people mix up with molarity. For a solute dissolved in a solvent of mass \(\displaystyle w_1\),\[\text{Molality } (m) = \frac{n_{\text{solute}}}{w_{\text{solvent}} \text{ (in kg)}} \]where \(\displaystyle n_{\text{solute}}\) is the number of moles of solute and \(\displaystyle w_{\text{solvent}}\) is the mass of the solvent expressed in kilograms (not grams — a common slip). Its unit is \(\displaystyle \text{mol kg}^{-1}\), often written "molal" (symbol \(\displaystyle m\)).Because molality is built from masses only, it does not change with temperature (mass doesn't expand or contract when heated) — unlike molarity, below.(iii) MolarityMolarity is defined on the volume of the whole solution after mixing, never on the volume of solvent used to prepare it. For a solute dissolved to make a solution of volume \(\displaystyle V\),\[\text{Molarity } (M) = \frac{n_{\text{solute}}}{V_{\text{solution}} \text{ (in L)}} \]where \(\displaystyle n_{\text{solute}}\) is the number of moles of solute and \(\displaystyle V_{\text{solution}}\) is the total volume of the solution in litres. Its unit is \(\displaystyle \text{mol L}^{-1}\), also written "molar" (symbol \(\displaystyle M\)).Since volume expands with temperature, molarity does change (slightly) with temperature, even though the amount of solute and solvent hasn't changed — this is the reason molality is preferred over molarity in precise physical-chemistry work.(iv) Mass percentageMass percentage puts the solute's mass over the mass of the whole solution — solute plus solvent together — not over the solvent's mass alone. If \(\displaystyle w_1\) is the mass of solute and \(\displaystyle w_2\) is the mass of solvent,\[\text{Mass \%} = \frac{w_1}{w_1 + w_2} \times 100 \]where \(\displaystyle w_1 + w_2\) is the total mass of the solution. Mass percentage is dimensionless (a pure number quoted as a percentage) — it carries no unit because it is a mass-to-mass ratio.Answer: Mole fraction \(\displaystyle x_i = \dfrac{n_i}{\sum n}\) (no unit); Molality \(\displaystyle m = \dfrac{n_{\text{solute}}}{w_{\text{solvent}}(\text{kg})}\) in \(\displaystyle \text{mol kg}^{-1}\); Molarity \(\displaystyle M = \dfrac{n_{\text{solute}}}{V_{\text{solution}}(\text{L})}\) in \(\displaystyle \text{mol L}^{-1}\); Mass percentage \(\displaystyle =\dfrac{w_{\text{solute}}}{w_{\text{solute}}+w_{\text{solvent}}}\times 100\) (no unit).

Molarity is moles of solute per litre of solution — so the way in is to pick a fixed volume of solution and find out how much acid and how much total mass sit inside it.Take exactly \(\displaystyle 1000\ \text{mL} = 1\ \text{L}\) of the acid solution as the sample to work with. Everything below is computed for that one litre.Step $\displaystyle 1$ — mass of the solution, from density. Density \(\displaystyle \rho\) is mass per unit volume: \(\displaystyle \rho = \dfrac{\text{mass of solution}}{\text{volume of solution}}\).\[\text{mass of solution} = \rho \times V = 1.504\ \text{g mL}^{-1} \times 1000\ \text{mL} = 1504\ \text{g} \]Step $\displaystyle 2$ — mass of \(\displaystyle HNO_3\) in that solution, from the mass percent. "$\displaystyle 68$% by mass" means $\displaystyle 68$ g of \(\displaystyle HNO_3\) sit in every $\displaystyle 100$ g of solution — not $\displaystyle 100$ g of water. This is the step people get wrong: the $\displaystyle 68$% is a fraction of the total solution mass, never of the solvent alone.\[\text{mass of } HNO_3 = \frac{68}{100} \times 1504\ \text{g} = 1022.72\ \text{g} \]Step $\displaystyle 3$ — convert that mass to moles. Molar mass of \(\displaystyle HNO_3\): \(\displaystyle 1(\text{H}) + 14(\text{N}) + 3\times16(\text{O}) = 63\ \text{g mol}^{-1}\).\[n(HNO_3) = \frac{\text{mass}}{\text{molar mass}} = \frac{1022.72\ \text{g}}{63\ \text{g mol}^{-1}} = 16.234\ \text{mol} \]Step $\displaystyle 4$ — molarity. Molarity \(\displaystyle M = \dfrac{n(\text{solute})}{V(\text{solution in L})}\), and the volume was chosen to be exactly $\displaystyle 1$ L, so the moles found above already are the molarity.\[M = \frac{16.234\ \text{mol}}{1\ \text{L}} = 16.234\ \text{mol L}^{-1} \]The data ($\displaystyle 68$%, $\displaystyle 1.504$ g mL⁻¹) carries four significant figures, so round there.Answer: The molarity of the concentrated nitric acid is \(\displaystyle 16.23\ \text{mol L}^{-1}\) ($\displaystyle 16.23$ M).

A "$\displaystyle 10$% w/w" solution is defined by mass, not volume — $\displaystyle 10$ g of glucose sits in every $\displaystyle 100$ g of solution (solute + solvent together), not per $\displaystyle 100$ g of water.So take $\displaystyle 100$ g of solution as the basis: Step $\displaystyle 1$: Convert masses to moles.\[n = \frac{\text{given mass}}{\text{molar mass}} \]Molar mass of glucose, \(\displaystyle \text{C}_6\text{H}_{12}\text{O}_6 = 6(12) + 12(1) + 6(16) = 180 \text{ g mol}^{-1} \).\[n_{\text{glucose}} = \frac{10 \text{ g}}{180 \text{ g mol}^{-1}} = 0.05556 \text{ mol} \]\[n_{\text{water}} = \frac{90 \text{ g}}{18 \text{ g mol}^{-1}} = 5.000 \text{ mol} \]Step $\displaystyle 2$: Molality — moles of solute per kilogram of solvent only.This is the step people get wrong: molality uses the mass of water (the solvent), never the mass of the whole solution.\[\text{molality} = \frac{n_{\text{glucose}}}{\text{mass of solvent in kg}} \]\[= \frac{0.05556 \text{ mol}}{0.090 \text{ kg}} = 0.6173 \text{ mol kg}^{-1} \]Step $\displaystyle 3$: Mole fractions — moles of each component over total moles.\[x_{\text{glucose}} = \frac{n_{\text{glucose}}}{n_{\text{glucose}} + n_{\text{water}}} = \frac{0.05556}{0.05556 + 5.000} = \frac{0.05556}{5.0556} = 0.01099 \]\[x_{\text{water}} = 1 - x_{\text{glucose}} = 1 - 0.01099 = 0.9890 \](Check: mole fractions of all components must add to $\displaystyle 1$ — they do.)Step $\displaystyle 4$: Molarity needs the volume of solution, which is where the density comes in.Density tells you how much space the $\displaystyle 100$ g of solution occupies:\[V = \frac{\text{mass of solution}}{\text{density}} = \frac{100 \text{ g}}{1.2 \text{ g mL}^{-1}} = 83.33 \text{ mL} = 0.08333 \text{ L} \]\[\text{molarity} = \frac{n_{\text{glucose}}}{V_{\text{solution in L}}} = \frac{0.05556 \text{ mol}}{0.08333 \text{ L}} = 0.6667 \text{ mol L}^{-1} \]Notice molarity ($\displaystyle 0.667$ mol L⁻¹) and molality ($\displaystyle 0.617$ mol kg⁻¹) come out close but not equal here — they are only the same when the solution is dilute enough that $\displaystyle 1$ L of solution has a mass of about $\displaystyle 1$ kg; the density of $\displaystyle 1.2$ g mL⁻¹ is what pulls them apart.Rounding each result to three significant figures (matching the $\displaystyle 10$%, $\displaystyle 1.2$ g mL⁻¹ data given):Answer: molality = $\displaystyle 0.617$ mol kg⁻¹; mole fraction of glucose = $\displaystyle 0.0110$, mole fraction of water = $\displaystyle 0.989$; molarity = $\displaystyle 0.667$ mol L⁻¹.

A mixture reacting with acid means you must add up the acid each component consumes separately — Na₂CO₃ needs two moles of HCl per mole, NaHCO₃ needs only one.Step $\displaystyle 1$: Find the moles of each carbonate from the total mass.Molar masses: \[M(Na_2CO_3) = 2(23) + 12 + 3(16) = 106\ \text{g mol}^{-1} \] \[M(NaHCO_3) = 23 + 1 + 12 + 3(16) = 84\ \text{g mol}^{-1} \]Let the equimolar amount of each be \(\displaystyle x \) mol. Since the mixture is equimolar, the total mass is\[106x + 84x = 1\ \text{g} \] \[190x = 1\ \text{g} \] \[x = \frac{1}{190}\ \text{mol} = 5.263 \times 10^{-3}\ \text{mol} \]So there is \(\displaystyle x \) mol of \(\displaystyle Na_2CO_3\) and \(\displaystyle x \) mol of \(\displaystyle NaHCO_3\) in the $\displaystyle 1$ g sample.Step $\displaystyle 2$: Write the balanced reactions with HCl and count moles of acid needed.Complete neutralisation (all the way to \(\displaystyle CO_2\) and water, which is what "react completely" means) proceeds as:\[Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2 \] \[NaHCO_3 + HCl \rightarrow NaCl + H_2O + CO_2 \]One mole of \(\displaystyle Na_2CO_3\) consumes $\displaystyle 2$ mol HCl, and one mole of \(\displaystyle NaHCO_3\) consumes $\displaystyle 1$ mol HCl. This factor of $\displaystyle 2$ is the step people skip — treating both salts as needing one mole of acid each undercounts the acid required for the carbonate.Total moles of HCl required: \[n(HCl) = 2x + x = 3x = 3 \times \frac{1}{190} = \frac{3}{190}\ \text{mol} = 1.579 \times 10^{-2}\ \text{mol} \]Step $\displaystyle 3$: Convert moles of HCl into a volume using the given molarity.Molarity is moles of solute per litre of solution: \[M = \frac{n}{V} \quad \Rightarrow \quad V = \frac{n}{M} \]Here \(\displaystyle n(HCl) = \dfrac{3}{190}\ \text{mol} \) and \(\displaystyle M = 0.1\ \text{mol L}^{-1} \):\[V = \frac{3/190\ \text{mol}}{0.1\ \text{mol L}^{-1}} = \frac{3}{19}\ \text{L} = 0.15789\ \text{L} \]Converting to millilitres ($\displaystyle 1$ L = $\displaystyle 1000$ mL):\[V = 0.15789\ \text{L} \times 1000\ \text{mL L}^{-1} = 157.9\ \text{mL} \]Answer: $\displaystyle 157.9$ mL of $\displaystyle 0.1$ M HCl are required.

Mass percentage compares the mass of solute to the mass of the whole solution — so before you can find the percentage of the mixture, you first need the actual mass of solute sitting inside each of the two solutions you're combining.A solution labelled "$\displaystyle 25$% by mass" means $\displaystyle 25$ g of solute is present in every $\displaystyle 100$ g of solution. So for each portion, find the solute mass: \[\text{mass of solute} = \text{mass percentage} \times \text{mass of solution} \]Solute from the first solution ($\displaystyle 300$ g, $\displaystyle 25$%): \[m_1 = \frac{25}{100} \times 300\ \text{g} = 75\ \text{g} \]Solute from the second solution ($\displaystyle 400$ g, $\displaystyle 40$%): \[m_2 = \frac{40}{100} \times 400\ \text{g} = 160\ \text{g} \]When the two solutions are mixed, both the solute and the solution mass simply add — nothing evaporates or reacts.Total mass of solute: \[m_{\text{solute}} = 75\ \text{g} + 160\ \text{g} = 235\ \text{g} \]Total mass of solution (this is the step people get wrong — you must add the full $\displaystyle 300$ g and $\displaystyle 400$ g, not just the solute masses, and you must not use only one of the original solvent masses): \[m_{\text{solution}} = 300\ \text{g} + 400\ \text{g} = 700\ \text{g} \]Now apply the mass percentage formula to the combined solution: \[\text{mass \%} = \frac{m_{\text{solute}}}{m_{\text{solution}}} \times 100 \]Substituting the totals: \[\text{mass \%} = \frac{235\ \text{g}}{700\ \text{g}} \times 100 = 33.5714...\% \]Rounding to four significant figures (matching the precision of the given data):Answer: $\displaystyle 33.57$% by mass

Molality is built on the mass of the solvent alone, while molarity is built on the volume of the whole solution — the two use different "how much liquid" numbers, and that's exactly where this problem catches people.Step $\displaystyle 1$ — moles of ethylene glycol.Molar mass, \(\displaystyle M \), of \(\displaystyle C_2H_6O_2\): add up $\displaystyle 2$ carbons, $\displaystyle 6$ hydrogens, $\displaystyle 2$ oxygens. \[M = 2(12) + 6(1) + 2(16) = 62\ \text{g mol}^{-1} \]Number of moles, \(\displaystyle n = \dfrac{\text{given mass}}{M} \): \[n = \frac{222.6\ \text{g}}{62\ \text{g mol}^{-1}} = 3.590\ \text{mol} \]Step $\displaystyle 2$ — molality.Molality, \(\displaystyle m = \dfrac{n_{\text{solute}}}{w_{\text{solvent}}\,(\text{kg})} \), where \(\displaystyle n_{\text{solute}} \) is moles of glycol and \(\displaystyle w_{\text{solvent}} \) is the mass of water only, in kilograms.The $\displaystyle 200$ g given is water, the solvent — not the mass of the finished solution. Converting to kg: \[w_{\text{solvent}} = 200\ \text{g} = 0.200\ \text{kg} \]Substituting: \[m = \frac{3.590\ \text{mol}}{0.200\ \text{kg}} = 17.95\ \text{mol kg}^{-1} \]Step $\displaystyle 3$ — volume of the solution, from its density.Molarity needs the volume of the whole solution, not the solvent volume — so first find the total solution mass, then use density to convert mass to volume.Total solution mass = mass of glycol + mass of water: \[m_{\text{soln}} = 222.6\ \text{g} + 200\ \text{g} = 422.6\ \text{g} \]Density, \(\displaystyle \rho = \dfrac{m_{\text{soln}}}{V} \), so \(\displaystyle V = \dfrac{m_{\text{soln}}}{\rho} \): \[V = \frac{422.6\ \text{g}}{1.072\ \text{g mL}^{-1}} = 394.2\ \text{mL} = 0.3942\ \text{L} \]Step $\displaystyle 4$ — molarity.Molarity, \(\displaystyle M_c = \dfrac{n_{\text{solute}}}{V_{\text{soln}}\,(\text{L})} \): \[M_c = \frac{3.590\ \text{mol}}{0.3942\ \text{L}} = 9.107\ \text{mol L}^{-1} \]Rounding to three significant figures (matching the precision of the density data): \[M_c \approx 9.11\ \text{mol L}^{-1} \]Answer: molality ≈ $\displaystyle 17.95$ mol kg⁻¹; molarity ≈ $\displaystyle 9.11$ mol L⁻¹

"Parts per million" is just a percentage with a bigger denominator — $\displaystyle 15$ ppm by mass means $\displaystyle 15$ g of \(\displaystyle CHCl_3\) sit inside every \(\displaystyle 10^6\) g of solution, not every $\displaystyle 100$ g.Part (i): percent by massPercent by mass is defined as \[\% \text{ by mass} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100 \] where the mass of solute is the \(\displaystyle CHCl_3\) and the mass of solution is the whole contaminated water sample.Since $\displaystyle 15$ ppm means $\displaystyle 15$ g of \(\displaystyle CHCl_3\) per \(\displaystyle 10^6\) g of solution, \[\% \text{ by mass} = \frac{15 \text{ g}}{10^{6}\text{ g}} \times 100 = 1.5 \times 10^{-3}\,\% \]Part (ii): molalityMolality needs two different numbers than the percentage did — moles of solute, and the mass of the solvent (water) alone, in kilograms, not the mass of the whole solution. \[m = \frac{n_{\text{solute}}}{w_{\text{solvent}}\,(\text{kg})} \]Moles of \(\displaystyle CHCl_3\). First get its molar mass: \[M(CHCl_3) = 12.0 + 1.0 + 3(35.5) = 119.5 \text{ g mol}^{-1} \] (carbon + hydrogen + three chlorines). With $\displaystyle 15$ g of \(\displaystyle CHCl_3\) in the sample, \[n(CHCl_3) = \frac{15 \text{ g}}{119.5 \text{ g mol}^{-1}} = 0.12552 \text{ mol} \]Mass of the solvent. Molality is per kilogram of solvent, and here that is easy to miss: the solvent is the water alone, which is the total solution mass minus the $\displaystyle 15$ g of solute, not the \(\displaystyle 10^6\) g of solution itself. \[w_{\text{water}} = 10^{6}\text{ g} - 15\text{ g} = 999\,985 \text{ g} = 999.985 \text{ kg} \] Because the contamination is only $\displaystyle 15$ g out of a million, this is $\displaystyle 0.0015$% away from \(\displaystyle 1.000\times10^3\) kg — small enough that using \(\displaystyle 1.000\times10^3\) kg does not change the final answer at this precision, so that is the value used below.Substitute. \[m = \frac{0.12552 \text{ mol}}{1.000\times10^{3}\text{ kg}} = 1.2552\times10^{-4} \text{ mol kg}^{-1} \]Rounding once, to three significant figures: \[m \approx 1.26\times10^{-4} \text{ mol kg}^{-1} \]Answer: (i) \(\displaystyle 1.5\times10^{-3}\,\%\) by mass; (ii) molality \(\displaystyle \approx 1.26\times10^{-4}\) mol kg⁻¹ (i.e., \(\displaystyle 1.26\times10^{-4}\,m\)).

Molecular interaction is the ONLY thing that decides whether a mixture behaves ideally — in alcohol + water the mixed (alcohol–water) attraction is weaker than the pure (water–water and alcohol–alcohol) attraction, and that mismatch is what drives every non-ideal property this system shows.Start from Raoult's law for a two-component liquid mixture of components A and B: \[p_A = x_A\,p_A^{\circ}, \qquad p_B = x_B\,p_B^{\circ} \] 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 law is only obeyed when the A–B interaction in the mixture is essentially the same strength as the A–A and B–B interactions in the separate pure liquids — swapping a neighbour from "same molecule" to "other molecule" then costs no extra energy, so each molecule escapes into the vapour exactly as easily as it did in its own pure liquid.Water is held together by an extensive three-dimensional hydrogen-bonded network — every water molecule can donate two H-bonds and accept two more. Pure ethanol is also hydrogen-bonded, but each molecule has only one \(\displaystyle -\text{OH}\) group, and the bulky \(\displaystyle -\text{C}_2\text{H}_5\) group gets in the way of close packing, so the ethanol–ethanol network is looser and weaker than water's.The step people skip: mixing them does not simply average these two networks — it breaks them. When an ethanol molecule wedges itself between water molecules, it interrupts water's hydrogen-bond network at that point. The new alcohol–water hydrogen bond that forms in its place is not able to replace the full connectivity that a water–water bond had, because ethanol can only donate/accept through one \(\displaystyle -\text{OH}\), not two. So: \[\text{(A–B interaction in solution)} \;<\; \text{(A–A and B–B interaction in the pure liquids)} \]Because each molecule is now held less tightly than it was in its own pure liquid, molecules of both water and alcohol escape into the vapour phase more easily than Raoult's law predicts. This shows up as three linked, observable effects — all consequences of the same weakened interaction, not three separate facts: This is exactly why an alcohol–water mixture shows a minimum-boiling azeotrope (around $\displaystyle 95$% ethanol by mass, boiling below either pure component): at that composition the weakened intermolecular attraction pushes the vapour pressure to a maximum, and a liquid with maximum vapour pressure must boil at a minimum temperature.Answer: In an alcohol–water solution, the alcohol–water intermolecular attractive forces are weaker than the water–water and alcohol–alcohol forces present in the pure liquids (because ethanol's bulky, singly-hydroxylated molecules disrupt water's hydrogen-bond network without fully replacing it). This weaker net interaction makes molecules escape into the vapour more easily than an ideal solution would allow, so the solution shows positive deviation from Raoult's law, along with \(\displaystyle \Delta_{\text{mix}}H > 0\) and \(\displaystyle \Delta_{\text{mix}}V > 0\), and a minimum-boiling azeotrope at about $\displaystyle 95$% ethanol.