Q.Two elements A and B form compounds having formula $AB_2$ and $AB_4$. When dissolved in 20 g of benzene ($C_6H_6$), 1 g of $AB_2$ lowers the freezing point by 2.3 K whereas 1.0 g of $AB_4$ lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg mol$^{-1}$. Calculate atomic masses of A and B.

Concept and Intuition

For a non-electrolyte, $\Delta T_f = K_f \cdot m$, where $m$ is molality. Both solutions here are 5% by mass (5 g solute per 100 g solution, so 95 g water), but sucrose ($\mathrm{C_{12}H_{22}O_{11}}$, $M\approx342$) is much heavier per mole than glucose ($\mathrm{C_6H_{12}O_6}$, $M\approx180$), so the same mass of glucose gives more moles — and a bigger depression. Since $K_f$ is the same for both solutions (same solvent), we can find the glucose depression as a ratio of the sugar depression, without needing water's textbook $K_f$ value.

Step-by-step reasoning

1. Depression for the sugar solution.

2. Ratio of molalities. For the same mass of solute (5 g) in the same mass of water (95 g), molality is inversely proportional to molar mass: …

Concept and Intuition

This question checks your understanding of four distinct topics in physical chemistry:

• Hypotonic vs. hypertonic solutions (osmosis)
• Freezing point depression (vapour pressure of solid vs. liquid solvent)
• Van’t Hoff factor (effect of dissociation on colligative properties)
• Molal boiling point elevation constant (its relation to enthalpy of vaporisation)

Each statement must be evaluated carefully — common pitfalls arise from mixing up definitions or misremembering proportionalities.

Step-by-step reasoning

1. Statement (A):

   “A hypotonic solution is more concentrated with respect to the other solution separated from it by a semipermeable membrane.”

   • A hypotonic solution has a lower solute concentration than the solution on the other side of the membrane.
   • Water moves from the hypotonic (less concentrated) side to the hypertonic (more concentrated) side.
   • Therefore, a hypotonic solution is less concentrated, not more. ➜ False.

2. Statement (B):

   “At the freezing point of a solution containing a non-volatile solute, the vapour pressure of the liquid solvent becomes less than that of the solid solvent.”

   • At the freezing point of a pure solvent, the vapour pressures of solid and liquid are equal.
   • For a solution, the vapour pressure of the liquid solvent is lowered by the solute (Raoult’s law).
   • At the new freezing point, the vapour pressure of the liquid solvent equals that of the solid solvent (both are lower than the pure solvent’s freezing point).
   • So the vapour pressure of the liquid is not less than that of the solid — they are equal at equilibrium. ➜ False.

3. Statement (C):

   “In case of dissociation of solute particles in a solution, the van't Hoff factor is more than unity because the observed molar mass has a lesser value.”

   • Van’t Hoff factor $i = \frac{\text{actual number of particles in solution}}{\text{number of formula units dissolved}}$.
   • For dissociation, $i > 1$ (e.g., NaCl → 2 ions, so $i \approx 2$).
   • Colligative properties depend on particle count. Observed molar mass $M_{\text{obs}} = \frac{M_{\text{theoretical}}}{i}$. …

Concept & Intuition

Freezing-point depression is a colligative property: it depends on the number of solute particles in solution, not their identity. When acetic acid dimerises in benzene, two molecules join to form one dimer, reducing the total particle count. The observed depression is smaller than expected for monomeric acid. By comparing the experimental van’t Hoff factor $i$ (actual particles per formula unit) with the ideal value of 1, we can calculate the fraction of acid that has dimerised.

Step-by-step solution

1. Calculate the expected freezing-point depression if no dimerisation occurred

   Mass of acetic acid = $0.4\ \mathrm{g}$

   Molar mass of $\mathrm{CH_3COOH}$ = $60\ \mathrm{g/mol}$

   Moles of acid = $\frac{0.4}{60} = 0.006667\ \mathrm{mol}$

   Mass of benzene = $40\ \mathrm{g} = 0.040\ \mathrm{kg}$

   Molality (if monomeric) = $\frac{0.006667}{0.040} = 0.1667\ \mathrm{mol/kg}$

   Using $\Delta T_f = K_f \cdot m$:

   $\Delta T_f(\text{ideal}) = 5.12 \times 0.1667 = 0.8533\ ^\circ\mathrm{C}$

2. Find the van’t Hoff factor $i$ from the observed depression

   Observed $\Delta T_f = 0.45\ ^\circ\mathrm{C}$

   The van’t Hoff factor is defined as:

This $i < 1$ confirms association (dimerisation).

3. Relate $i$ to the degree of dimerisation Let $\alpha$ be the fraction of acetic acid molecules that dimerise. For every 2 monomers that react, we get 1 dimer. Start with 1 mole of monomer:
   • Moles that dimerise = $\alpha$ → these form $\alpha/2$ moles of dimer
   • Moles that remain monomer = $1 - \alpha$ …

Using $\Delta T_b = i\,K_b\,m$:

For association of $n$ monomers into one particle, $i = 1 - \alpha\left(1 - \tfrac{1}{n}\right)$. For a dimer $n=2$: …

Concept

As the solution is cooled, pure ice crystallises out. Ice keeps forming until the remaining liquid water is concentrated enough that its freezing point equals the bath temperature. So we find how much liquid water is needed to give a freezing-point depression of $9.3\,^\circ\text{C}$; the rest of the original water has frozen.

Step 1 — Molality required for $\Delta T_f = 9.3\ \text{K}$

Step 2 — Moles of ethylene glycol (non-volatile solute, does not freeze) …

1. Take a 100 g basis of the solution

Because the strengths are given as % (w/w):

• mass of AB $= 5.85$ g
• mass of $\mathrm{XY_2} = 9.50$ g
• mass of water (solvent) $= 100 - 5.85 - 9.50 = 84.65$ g $= 0.08465$ kg

2. Moles of each solute

3. van't Hoff factors (partial ionisation)

For a solute giving $n$ particles with degree of ionisation $\alpha$:

AB $\rightarrow \mathrm{A^+ + B^-}$, so $n = 2$, $\alpha = 0.80$:

$\mathrm{XY_2}$ $\rightarrow \mathrm{X^{2+} + 2Y^-}$ (Y is monovalent, so X must be divalent), $n = 3$, $\alpha = 0.60$:

4. Total effective (particle) moles

5. Effective molality

6. Depression of freezing point …

Colligative properties (relative lowering of vapour pressure, boiling-point elevation, freezing-point depression, osmotic pressure) depend only on the number of solute particles, not their nature.

When a solute dissociates (e.g. $\mathrm{NaCl \to Na^+ + Cl^-}$), the particle count increases, so the observed colligative property is larger than expected (van't Hoff factor $i>1$). …

At 373 K (100 °C) the vapour pressure of pure water $P^0 = 760$ mm.

Relative lowering of vapour pressure = mole fraction of solute: …

Step 1 — The concept: metal-deficiency defect. In an oxide written $M_{0.96}O$ the metal sites are deficient (a cation vacancy defect, as in FeO / wüstite). The crystal must still be electrically neutral, so some cations are oxidised from $M^{2+}$ to $M^{3+}$ to make up the missing positive charge.

Step 2 — Set up the balance. Take $1$ mol of the oxide: it contains $0.96$ mol of metal and $1$ mol of $O^{2-}$ (total negative charge $= 2$).

Let

• $a$ = mol of $M^{2+}$,
• $b$ = mol of $M^{3+}$.

Then

$2a + 3b = 2 \qquad \text{(charge neutrality: total } + \text{ charge} = 2)$ …

$X$ dimerises: $2X \rightleftharpoons X_2$, degree of association $\alpha = 0.80$. The van't Hoff factor is

$i = 1 - \dfrac{\alpha}{2} = 1 - \dfrac{0.80}{2} = 0.60.$

Elevation of boiling point: $\Delta T_b = i\,K_b\,m$, with molality $m = \dfrac{2.5/M}{0.100} = \dfrac{25}{M}\ \text{mol kg}^{-1}$. …

The key idea: when a solute associates (forms larger molecules), the number of particles in solution is less than the number of formula units dissolved. The freezing point depression depends on the actual number of particles, so we first find the apparent molar mass from the given $\Delta T_f$, then use the degree of association to find the true molar mass.

For tetramerisation, 4 molecules of A combine to form one tetramer. If the degree of association is $\alpha$, the van’t Hoff factor $i$ is given by:

where $n = 4$ here. With $\alpha = 0.80$ (80% association), we can compute $i$ directly.

1. Calculate the van’t Hoff factor $i$

   For tetramerisation: $n = 4$, $\alpha = 0.80$.

So only 40% of the original number of particles remain, on average.

2. Find the apparent molar mass from the observed depression

   The freezing point depression is:

where $m$ is the molality of the solution if no association occurred (i.e., using the formula mass of A). Let the true molar mass of A be $M$ g/mol.

Molality (if no association) = $\frac{2.5 / M}{0.100} = \frac{25}{M}$ mol/kg.

Plug into the equation:

3. Solve for $M$

First compute $0.40 \times 1.86 = 0.744$.

Then $0.744 \times 25 = 18.6$.

So: …