Question

Vessel-1 contains \(w_2\) g of a non-volatile solute \(X\) dissolved in \(w_1\) g of water. Vessel-2 contains \(w_2\) g of another non-volatile solute \(Y\) dissolved in \(w_1\) g of water. Both the vessels are at the same temperature and pressure. The molar mass of \(X\) is 80\% of that of \(Y\). The van’t Hoff factor for \(X\) is 1.2 times that of \(Y\) for their respective concentrations. The elevation of boiling point for solution in Vessel-1 is \_\_\_\_\_ \% of the solution in Vessel-2.

Hint:

For colligative properties, compare molar mass, van’t Hoff factors, and solute concentrations to calculate the relative effects.

Answer 1

1. Expression for elevation of boiling point in Vessel-1 (\((\Delta T_b)_I\)): \[ (\Delta T_b)_I = i_X \times \frac{w_2}{M_X} \times \frac{1000}{w_1} \times K_B. \] 2. Expression for elevation of boiling point in Vessel-2 (\((\Delta T_b)_{II}\)): \[ (\Delta T_b)_{II} = i_Y \times \frac{w_2}{M_Y} \times \frac{1000}{w_1} \times K_B. \] 3. Ratio of elevations: \[ \frac{(\Delta T_b)_I}{(\Delta T_b)_{II}} = \frac{i_X}{i_Y} \times \frac{M_Y}{M_X}. \] 4. Substituting \(M_X = 0.8 M_Y\) and \(i_X = 1.2 i_Y\): \[ \frac{(\Delta T_b)_I}{(\Delta T_b)_{II}} = \frac{1.2}{1} \times \frac{1}{0.8}. \] \[ \frac{(\Delta T_b)_I}{(\Delta T_b)_{II}} = 1.5. \] 5. Percentage increase: \[ (\Delta T_b)_I = 150\% \text{ of } (\Delta T_b)_{II}. \]