Solutions Set-7

Q1. A non-volatile, non-electrolyte solute is dissolved in benzene. The vapour pressure of pure benzene at 25°C is 150.0 torr and the vapour pressure of the solution is 120.0 torr. Calculate the mole fraction of solute in the solution. (Use Raoult's law $P_{\text{solution}}=x_{\text{benzene}}P^\circ_{\text{benzene}}$.)

Q2. 0.0100 mol of a weak acid HA is dissolved in 100.0 g of water. The observed freezing point depression is $\Delta T_f=0.212^\circ\text{C}$. (Take $K_f(\text{water})=1.86\ \text{K kg mol}^{-1}$.) Assuming HA dissociates as $\mathrm{HA \rightleftharpoons H^+ + A^-}$ and that the van 't Hoff factor is $i=1+\alpha$, find the degree of dissociation $\alpha$.

Q3. For an ideal binary solution of volatile liquids A and B, $P^\circ_A=400\ \text{mmHg}$ and $P^\circ_B=200\ \text{mmHg}$. If the mole fraction of A in the liquid is $x_A=0.40$, calculate the mole fraction of A in the vapour phase $y_A$ using $P_A=x_AP^\circ_A$, $P_B=x_BP^\circ_B$ and $y_A=\dfrac{P_A}{P_A+P_B}$.

Q4. At 298 K, a solution prepared by dissolving 1.00 g of an ionic compound MX$_2$ (molar mass $=111.0\ \text{g mol}^{-1}$) in 100.0 g of water has the same osmotic pressure as a solution containing 0.100 mol of a non-electrolyte in 1.00 L of water. Assume volumes are additive and ideal behaviour. If MX$_2$ dissociates as $\mathrm{MX_2 \rightleftharpoons M^{2+} + 2X^-}$ and the van 't Hoff factor is $i=1+2\alpha$, calculate the degree of dissociation $\alpha$.

Q5. 20.0 g of solute A (molar mass $=50.0\ \text{g mol}^{-1}$) is dissolved in 200.0 g of water, and separately 10.0 g of solute B (molar mass $=100.0\ \text{g mol}^{-1}$) is dissolved in 100.0 g of water. The two solutions are then mixed (assume additive masses, no association/dissociation). Calculate the freezing point depression $\Delta T_f$ of the final mixture. (Take $K_f(\text{water})=1.86\ \text{K kg mol}^{-1}$.)

Q6. An ideal binary liquid mixture is prepared by mixing $1.00\ \mathrm{mol}$ of liquid A (pure vapour pressure $P_A^0 = 300.0\ \mathrm{mmHg}$) and $3.00\ \mathrm{mol}$ of liquid B (pure vapour pressure $P_B^0 = 100.0\ \mathrm{mmHg}$) at the same temperature. Using Raoult’s law $P_{\text{total}}=x_A P_A^0 + x_B P_B^0$, the total vapour pressure of the mixture is:

Q7. At a given temperature pure benzene has vapour pressure $P^0 = 150\ \text{mmHg}$. A non‑volatile solute (molar mass $M = 60.0\ \text{g\ mol}^{-1}$) is dissolved in $100.0\ \text{g}$ of benzene ($M_{\text{benzene}} = 78.0\ \text{g\ mol}^{-1}$) and the equilibrium vapour pressure of the solution is $P = 120\ \text{mmHg}$. Assuming ideal behaviour and non‑volatility of the solute, the mass of solute dissolved is approximately:

Q8. At $60^\circ\mathrm{C}$ pure liquids A and B have vapour pressures $P_A^0 = 400\ \text{mmHg}$ and $P_B^0 = 200\ \text{mmHg}$. An ideal solution has liquid mole fraction $x_A = 0.40$ ($x_B=0.60$). Using $y_A=\dfrac{x_A P_A^0}{x_A P_A^0 + x_B P_B^0}$, the mole fraction of A in the vapour phase $y_A$ and the component expected to distil first are:

Q9. $1.00\ \mathrm{g}$ of an electrolyte AB (formula unit AB, molar mass $58.0\ \mathrm{g\ mol^{-1}}$) is dissolved in $100.0\ \mathrm{g}$ of water. The observed freezing point depression is $\Delta T_f = 0.372\ \mathrm{K}$. Given $K_f(\mathrm{H_2O})=1.86\ \mathrm{K\;kg\;mol}^{-1}$ and that AB dissociates as AB $\rightleftharpoons$ A$^+$ + B$^-$ with degree of dissociation $\alpha$, use $\Delta T_f = i\,K_f\,m$ and $i=1+\alpha$ to find $\alpha$:

Q10. At $298\ \mathrm{K}$ pure solvent A has vapour pressure $P_A^0 = 400\ \mathrm{mmHg}$. In a solution with liquid mole fraction $x_A = 0.60$ the measured partial vapour pressure of A is $P_A = 210\ \mathrm{mmHg}$. Using $a_A=\dfrac{P_A}{P_A^0}=\gamma_A x_A$, calculate the activity coefficient $\gamma_A$ and indicate whether the solution shows positive or negative deviation from Raoult’s law and what this implies about A–B interactions:

Q11. 5.40 g of an unknown non‑volatile, non‑electrolyte solute was dissolved in 100.0 g of water and the freezing point was lowered by 1.00°C. (Given $K_f(\text{water})=1.86\ \text{K kg mol}^{-1}$.) Using $\Delta T_f=K_f\cdot m$ and $m=\dfrac{n}{\text{kg solvent}}$, the molar mass of the solute is closest to:

Q12. At 25°C the vapour pressure of pure water is $0.316\ \text{atm}$. A non‑volatile, non‑electrolyte solute lowers the vapour pressure of water to $0.290\ \text{atm}$. Assuming an ideal solution and that $1.000\ \text{kg}$ of water is the solvent, calculate the molality $m$ of the solution. (Use $x_{\text{solvent}}=\dfrac{p}{p^*}$ and $n_{\text{solvent}}=\dfrac{1000}{18}\ \text{mol}$.)

Q13. A polymer sample of mass $0.500\ \text{g}$ is dissolved in $250.0\ \text{mL}$ of solution and the measured osmotic pressure at $27^\circ\text{C}$ (300 K) is $0.082\ \text{atm}$. Using $\pi=CRT$ and $R=0.0821\ \text{L atm K}^{-1}\text{mol}^{-1}$, the molar mass of the polymer is closest to:

Q14. A compound of molar mass $60.0\ \text{g mol}^{-1}$ is dissolved (3.00 g) in $100.0\ \text{g}$ of benzene. ($K_f(\text{benzene})=5.12\ \text{K kg mol}^{-1}$.) The observed freezing point depression is $1.92\ \text{K}$. Assuming the compound partially dimerises in benzene and that the van't Hoff factor for dimerisation is $i=1-\dfrac{\alpha}{2}$ (where $\alpha$ is the degree of dimerisation), calculate $\alpha$.

Q15. Consider two aqueous solutions at the same temperature (ideal behaviour except for dissociation effects):
Solution I: $0.10\ \text{M}$ CaCl$_2$ with degree of dissociation $\alpha_{\text{Ca}}=0.60$.
Solution II: $0.15\ \text{M}$ AlCl$_3$ with degree of dissociation $\alpha_{\text{Al}}=0.40$.
For an electrolyte that produces $\nu$ ions on complete dissociation, $i=1+\alpha(\nu-1)$ and osmotic pressure $\pi\propto iC$. Which solution has the higher osmotic pressure?