The molal elevation boiling point constant for water is $0,513^{\circ} \mathrm{C} \mathrm{Kg}…

The molal elevation boiling point constant for water is $0,513^{\circ} \mathrm{C} \mathrm{Kg} \mathrm{mol}^{-1}$. Calculate boiling point of solution if 0.1 mole of sugar is dissolved in 200 g water?
  1. $100.513^{\circ} \mathrm{C}$
  2. $100.256^{\circ} \mathrm{C}$
  3. $100.0513^{\circ} \mathrm{C}$
  4. $100.025^{\circ} \mathrm{C}$

Solution

$\begin{aligned} & \Delta \mathrm{T}_{\mathrm{b}}=\mathrm{K}_{\mathrm{b}} \mathrm{m} \\ & \Delta \mathrm{T}_{\mathrm{b}}=0.513{ }^{\circ} \mathrm{C} \mathrm{kg} \mathrm{mol}^{-1} \times \frac{0.1 \mathrm{~mol}}{200 \times 10^{-3} \mathrm{~kg}} \\ & \mathrm{~T}-\mathrm{T}_{\mathrm{b}}=0.2565^{\circ} \mathrm{C} \\ & \mathrm{T}=\mathrm{T}_{\mathrm{b}}+0.2565=100+0.2565=100.256^{\circ} \mathrm{C}\end{aligned}$

Asked in: MHT CET 2024 (03 May Shift 2)

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