A solution of non volatile solute has boiling point elevation 0.5 K . Calculate molality of solution…

A solution of non volatile solute has boiling point elevation 0.5 K . Calculate molality of solution $\left[\mathrm{K}_{\mathrm{b}}=2.40 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right]$.
  1. $0.12 \mathrm{~mol} \mathrm{~kg}^{-1}$
  2. $0.21 \mathrm{~mol} \mathrm{~kg}^{-1}$
  3. $0.16 \mathrm{~mol} \mathrm{~kg}^{-1}$
  4. $0.28 \mathrm{~mol} \mathrm{~kg}^{-1}$.

Solution

$\begin{aligned} & \Delta \mathrm{T}_{\mathrm{b}}=\mathrm{K}_{\mathrm{b}} \times \mathrm{m} \\ & 0.5 \mathrm{~K}=2.40 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \times \mathrm{m} \\ \therefore \quad & \mathrm{m}=\frac{0.5}{2.40}=0.21 \mathrm{~mol} \mathrm{~kg}^{-1}\end{aligned}$

Asked in: MHT CET 2024 (09 May Shift 2)

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