A solution of non volatile solute has boiling point elevation 1.75 K . Calculate molality of solution…

A solution of non volatile solute has boiling point elevation 1.75 K . Calculate molality of solution $\left[\mathrm{K}_{\mathrm{b}}=3.5 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right]$
  1. $0.77 \mathrm{~mol} \mathrm{~kg}^{-1}$
  2. $0.69 \mathrm{~mol} \mathrm{~kg}^{-1}$
  3. $0.50 \mathrm{~mol} \mathrm{~kg}^{-1}$
  4. $0.35 \mathrm{~mol} \mathrm{~kg}^{-1}$

Solution

$\begin{aligned} & \Delta \mathrm{T}_{\mathrm{b}}=\mathrm{K}_{\mathrm{b}} \mathrm{m} \\ & 1.75 \mathrm{~K}=3.5 \mathrm{Kkg} \mathrm{mol}^{-1} \times \mathrm{m} \\ & \mathrm{m}=\frac{1.75}{3.5}=0.50 \mathrm{~mol} \mathrm{~kg}^{-1}\end{aligned}$

Asked in: MHT CET 2024 (03 May Shift 1)

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