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MHT CET · Chemistry · Solutions

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. A \(100.513^{\circ} \mathrm{C}\)
  2. B \(100.256^{\circ} \mathrm{C}\)
  3. C \(100.0513^{\circ} \mathrm{C}\)
  4. D \(100.025^{\circ} \mathrm{C}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(100.256^{\circ} \mathrm{C}\)

Step-by-step Solution

Detailed explanation

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