What is the molal elevation constant if one gram mole of a nonvolatile solute is dissolved in $1…

What is the molal elevation constant if one gram mole of a nonvolatile solute is dissolved in $1 \mathrm{~kg}$ of ethyl acetate? $\left(\Delta \mathrm{T}_{\mathrm{b}}=x \mathrm{~K}\right)$
  1. $\quad x \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$
  2. $\frac{x}{2} \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$
  3. $\quad 2 x \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$
  4. $3 x \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$

Solution

$\begin{aligned} & \mathrm{M}_2=\frac{1000 \mathrm{~K}_{\mathrm{b}} \mathrm{W}_2}{\Delta \mathrm{T}_{\mathrm{b}} \mathrm{W}_1} \\ \therefore \quad & \mathrm{K}_{\mathrm{b}}=\frac{\mathrm{M}_2 \Delta \mathrm{T}_{\mathrm{b}} \mathrm{W}_1}{1000 \mathrm{~W}_2} \\ & \mathrm{Now}, \mathrm{W}_2=\text { one gram mole }=\mathrm{M}_2 \mathrm{~g} \\ & \mathrm{~W}_1=1 \mathrm{~kg}=1000 \mathrm{~g} \\ \therefore \quad & \mathrm{K}_{\mathrm{b}}=\frac{\mathrm{M}_2 \times x \times 1000}{1000 \times \mathrm{M}_2}=x \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\end{aligned}$

Asked in: MHT CET 2023 (13 May Shift 1)

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