Calculate the molar mass of non volatile solute when 5 g of it is dissolved in 50 g solvent, boils at $119…

Calculate the molar mass of non volatile solute when 5 g of it is dissolved in 50 g solvent, boils at $119.6^{\circ} \mathrm{C}$. $\left[\mathrm{K}_{\mathrm{b}}=3.2 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right.$, boiling point of pure solvent $\left.=118^{\circ} \mathrm{C}\right]$.
  1. $180 \mathrm{~g} \mathrm{~mol}^{-1}$
  2. $210 \mathrm{~g} \mathrm{~mol}^{-1}$
  3. $200 \mathrm{~g} \mathrm{~mol}^{-1}$
  4. $190 \mathrm{~g} \mathrm{~mol}^{-1}$

Solution

$\begin{aligned} & \Delta \mathrm{T}_{\mathrm{b}}=\mathrm{T}_{\mathrm{b}}-\mathrm{T}_{\mathrm{b}}^0=119.6-118=1.6^{\circ} \mathrm{C}=1.6 \mathrm{~K} \\ & \Delta \mathrm{~T}_{\mathrm{b}}=\frac{1000 \mathrm{~K}_{\mathrm{b}} \mathrm{W}_2}{\mathrm{M}_2 \mathrm{~W}_1} \\ & \mathrm{M}_2=\frac{1000 \mathrm{~K}_{\mathrm{b}} \mathrm{W}_2}{\Delta \mathrm{~T}_{\mathrm{b}} \mathrm{W}_1}=\frac{1000 \times 3.2 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}{ }^{-1} \times 5 \mathrm{~g}}{1.6 \mathrm{~K} \times 50 \mathrm{~g}} \\ & \quad=200 \mathrm{~g} \mathrm{~mol}^{-1}\end{aligned}$

Asked in: MHT CET 2024 (02 May Shift 2)

Practice more Solutions questions on Aicharya