Question

A solution was prepared by dissolving 2g of a non-volatile non electrolyte solid B in 100g of solvent A. At 354K, the vapour pressure of the solution became exactly equal to the atmospheric pressure of 1.0 atm. If the molar mass of A is 80 $g \text{mol}^{-1}$, the molar mass (in $g \text{mol}^{-1}$) of solid B is ____.

[Given: Gas constant, $R = 8.3 \text{J K}^{-1} \text{mol}^{-1}$; Vapour pressure of A at 353K = 1.0 atm, Standard entropy of vapourisation of A at 353K = 85 $\text{J K}^{-1} \text{mol}^{-1}$,]

Answer 1

55.1

Answer 2

Here's how to solve this problem using the concept of boiling point elevation:

1. Determine the boiling point elevation ($\Delta T_b$) from the boiling points of the pure solvent and the solution: $\Delta T_b = T_{solution} - T_{solvent} = 354 \text{ K} - 353 \text{ K} = 1 \text{ K}$

2. Calculate the molar enthalpy of vaporization ($\Delta H_{vap}$) of the solvent using the given standard entropy of vaporization ($\Delta S_{vap}$) and the boiling point of the solvent ($T_b$): $\Delta H_{vap} = T_b \cdot \Delta S_{vap} = 353 \text{ K} \cdot 85 \text{ J K}^{-1} \text{mol}^{-1} = 30005 \text{ J mol}^{-1}$

3. Calculate the molal elevation constant ($K_b$) using the formula: $K_b = \frac{R \cdot T_b^2 \cdot M_A}{1000 \cdot \Delta H_{vap}}$ Where:

   • $R = 8.3 \text{ J K}^{-1} \text{mol}^{-1}$
   • $T_b = 353 \text{ K}$
   • $M_A = 80 \text{ g mol}^{-1}$
   • $\Delta H_{vap} = 30005 \text{ J mol}^{-1}$

   $K_b = \frac{8.3 \cdot (353)^2 \cdot 80}{1000 \cdot 30005} \approx 2.7545 \text{ K kg mol}^{-1}$

4. Use the boiling point elevation equation to find the molality ($m$) of the solution: $\Delta T_b = K_b \cdot m$ $1 \text{ K} = 2.7545 \text{ K kg mol}^{-1} \cdot m$ $m \approx 0.3630 \text{ mol kg}^{-1}$

5. Use the definition of molality to calculate the molar mass of the solute ($M_B$): $m = \frac{\text{moles of solute}}{\text{kg of solvent}} = \frac{w_B / M_B}{w_A}$ Where:

   • $w_B = 2 \text{ g}$
   • $w_A = 100 \text{ g} = 0.1 \text{ kg}$

   $0.3630 = \frac{2 / M_B}{0.1}$ $M_B = \frac{2}{0.3630 \cdot 0.1} \approx 55.09 \text{ g mol}^{-1}$

Therefore, the molar mass of solid B is approximately 55.1 $g \text{mol}^{-1}$.