Easy Practice — Set 1

Quick Recap — Concentration Terms

• Molarity $M=\dfrac{\text{mol solute}}{\text{L solution}}$ (temperature-dependent).
• Molality $m=\dfrac{\text{mol solute}}{\text{kg solvent}}$; mole fraction $x_A=\dfrac{n_A}{n_A+n_B}$ (both temperature-independent).
• Mass % $=\dfrac{\text{mass solute}}{\text{mass solution}}\times100$; ppm $=$ mg solute per kg solution.
• Normality $=$ molarity $\times$ n-factor. $M\times V(\text{L})=$ moles.

Beyond-NCERT JEE Formulae

Learn each relation together with when it applies. Unless stated, every colligative formula below assumes a dilute solution of a non-volatile solute.

1. Raoult's law and relative lowering

For a binary mixture of two volatile liquids A and B the total vapour pressure is $p=p_A^\circ x_A+p_B^\circ x_B$, where $x_A$ and $x_B$ are the mole fractions in the liquid. The vapour is always richer in the more volatile component, with $y_A=\dfrac{p_A}{p}$.

For a non-volatile solute only the solvent evaporates, so the relative lowering of vapour pressure equals the mole fraction of solute: $$\dfrac{p^\circ-p}{p^\circ}=x_{solute}$$ When to use: getting the molar mass of a non-volatile solute from vapour-pressure data. For dilute solutions $x_{solute}\approx\dfrac{n_2}{n_1}=\dfrac{w_2/M_2}{w_1/M_1}$.

[JEE Tip] Relative lowering needs no $K$ constant; it works with mole fraction directly. If the solute dissociates or associates, use $i\,n_2$ in place of $n_2$.

2. Colligative properties (with the van't Hoff factor)

Elevation of boiling point $\Delta T_b=iK_b m$, depression of freezing point $\Delta T_f=iK_f m$, and osmotic pressure $\pi=iCRT$ with C in mol/L. The molar mass of a non-electrolyte from freezing-point data is $$M_2=\dfrac{K_f\,w_2\,1000}{\Delta T_f\,w_1}$$ with the masses in grams. The constants depend only on the solvent: for water $K_b=0.52$ and $K_f=1.86$ K kg/mol.

[JEE Tip] Osmotic pressure is measurable at very low concentration, so $\pi=iCRT$ gives the most accurate molar masses of proteins and polymers. Two isotonic solutions (equal $\pi$) obey $i_1C_1=i_2C_2$.

3. van't Hoff factor, dissociation and association

The factor compares the real particle count with the formula count: $$i=\dfrac{\text{observed colligative effect}}{\text{value calculated for }i=1}=\dfrac{\text{normal molar mass}}{\text{observed molar mass}}$$ For a solute that gives $n$ particles: dissociation, for example AB to A+ and B- so that $n=2$, gives $\alpha=\dfrac{i-1}{n-1}$; association, such as the dimerisation 2A to A2 with $n=2$, gives $\alpha=\dfrac{1-i}{1-1/n}$.

[JEE Tip] Dissociation raises the particle count, so $i>1$ and the observed molar mass drops below normal; association lowers the count, so $i<1$ and the observed molar mass rises above normal. Benzoic acid in benzene is the classic dimer, with $i\approx0.5$ and an apparent molar mass near 244 g/mol.

4. Henry's law and real solutions

Gas solubility follows $p=K_H x$, where $x$ is the mole fraction of dissolved gas. A larger $K_H$ means a less soluble gas, and $K_H$ rises with temperature, so gases are less soluble when hot. For deviations from Raoult's law, a positive deviation (with $p$ above the Raoult value) comes from weaker solute-solvent forces, as in ethanol and acetone, and gives a minimum-boiling azeotrope; a negative deviation (with $p$ below the Raoult value) comes from stronger solute-solvent forces, as in acetone and chloroform, and gives a maximum-boiling azeotrope.

[JEE Tip] An ideal solution obeys Raoult's law at every composition with $\Delta_{mix}H=0$ and $\Delta_{mix}V=0$. Azeotropes boil at a fixed temperature and cannot be separated by fractional distillation, a favourite one-mark trap.

Solved Examples — Beyond-NCERT Formulae

Example 1 — Molar mass from freezing-point depression. A 1.00 g sample of a non-volatile, non-electrolyte solute in 50.0 g of benzene lowers the freezing point by 0.40 K, with $K_f=5.12$ K kg/mol for benzene. Using $M_2=\dfrac{K_f\,w_2\,1000}{\Delta T_f\,w_1}$, $$M_2=\dfrac{5.12\times1.00\times1000}{0.40\times50.0}=\dfrac{5120}{20}=256$$ so the molar mass is 256 g/mol.

Example 2 — Molar mass from boiling-point elevation. Dissolving 1.80 g of a non-volatile solute in 100 g of water raises the boiling point by 0.052 K, with $K_b=0.52$ K kg/mol for water. Then $$M_2=\dfrac{K_b\,w_2\,1000}{\Delta T_b\,w_1}=\dfrac{0.52\times1.80\times1000}{0.052\times100}=\dfrac{936}{5.2}=180$$ so the molar mass is 180 g/mol, matching glucose.

Example 3 — Osmotic pressure and molar mass. A 1.0 g sample of a polymer in 100 mL of aqueous solution has an osmotic pressure of $4.926\times10^{-3}$ atm at 300 K, with $R=0.0821$ L atm/mol/K. From $\pi=\dfrac{w}{M_2 V}RT$, rearrange to $M_2=\dfrac{wRT}{\pi V}$: $$M_2=\dfrac{1.0\times0.0821\times300}{4.926\times10^{-3}\times0.100}=\dfrac{24.63}{4.926\times10^{-4}}=5.0\times10^{4}$$ The molar mass is $5.0\times10^{4}$ g/mol; the tiny yet measurable osmotic pressure is exactly why osmometry suits macromolecules.

Example 4 — van't Hoff factor and percent dissociation. A 0.10 m aqueous solution of a weak binary electrolyte, which ionises as AB to A+ and B- so that $n=2$, shows an observed freezing-point depression of 0.279 K, with $K_f=1.86$ K kg/mol. First the van't Hoff factor, $i=\dfrac{\Delta T_f}{K_f\,m}=\dfrac{0.279}{1.86\times0.10}=1.5$. Then the degree of dissociation from $\alpha=\dfrac{i-1}{n-1}$, $$\alpha=\dfrac{1.5-1}{2-1}=0.5$$ so the electrolyte is 50 percent dissociated.

Example 5 — Association of a dimer (benzoic acid). 2.44 g of benzoic acid (normal molar mass 122 g/mol) in 50 g of benzene gives an observed freezing-point depression of 1.024 K, with $K_f=5.12$ K kg/mol; in benzene the acid dimerises, 2A to A2, so $n=2$. The moles of acid are $\dfrac{2.44}{122}=0.0200$ and the molality is $\dfrac{0.0200}{0.050}=0.400$ mol/kg, so the depression expected without association is $\Delta T_f(\text{calc})=K_f m=5.12\times0.400=2.048$ K. Hence $$i=\dfrac{\Delta T_f(\text{obs})}{\Delta T_f(\text{calc})}=\dfrac{1.024}{2.048}=0.50$$ The degree of association is $\alpha=\dfrac{1-i}{1-1/n}=\dfrac{1-0.50}{1-1/2}=1.0$, i.e. essentially complete dimerisation, and the apparent molar mass is $\dfrac{122}{0.50}=244$ g/mol, twice the monomer value.

Example 6 — Two-component Raoult's law. An ideal solution is made from 2 mol of liquid A (with $p_A^\circ=100$ torr) and 3 mol of liquid B (with $p_B^\circ=60$ torr) at 300 K. The liquid mole fractions are $x_A=\dfrac{2}{5}=0.40$ and $x_B=0.60$, so the total vapour pressure is $$p=p_A^\circ x_A+p_B^\circ x_B=100\times0.40+60\times0.60=40+36=76$$ that is 76 torr. The vapour mole fraction of A is $y_A=\dfrac{p_A}{p}=\dfrac{40}{76}=0.53$, confirming the vapour is richer than the liquid in the more volatile component A.

Example 7 — Relative lowering of vapour pressure. At a fixed temperature the vapour pressure of pure water is 25.0 torr and falls to 24.0 torr when 12.5 g of a non-volatile solute is dissolved in 90 g of water. The relative lowering equals the solute mole fraction, $x_2=\dfrac{p^\circ-p}{p^\circ}=\dfrac{25.0-24.0}{25.0}=0.040$. With water $n_1=\dfrac{90}{18}=5.00$ mol, and since $x_2=\dfrac{n_2}{n_1+n_2}$, rearranging gives $n_2=\dfrac{n_1 x_2}{1-x_2}=\dfrac{5.00\times0.040}{0.96}=0.2083$ mol. Hence $$M_2=\dfrac{w_2}{n_2}=\dfrac{12.5}{0.2083}=60$$ so the molar mass is 60 g/mol, matching urea.

Example 8 — Henry's law solubility. For carbon dioxide in water at 298 K the Henry constant is $K_H=2.0\times10^{3}$ bar. To find the dissolved amount in 1 kg of water when the partial pressure of carbon dioxide is 4.0 bar, use $p=K_H x$ to get $x=\dfrac{p}{K_H}=\dfrac{4.0}{2.0\times10^{3}}=2.0\times10^{-3}$. In 1 kg of water there are $n_1=\dfrac{1000}{18}=55.6$ mol, so the dissolved amount is $n\approx x\,n_1=2.0\times10^{-3}\times55.6=0.11$ mol, that is $$0.11\times44=4.9$$ grams of carbon dioxide per kilogram of water.