The 0.1 molal aqueous solution of glucose boils at \(100.16^\circ \mathrm{C}\). The boiling point of 0.5…

The 0.1 molal aqueous solution of glucose boils at \(100.16^\circ \mathrm{C}\). The boiling point of 0.5 molal aqueous solution of sucrose will be
  1. $100.32^{\circ} \mathrm{C}$
  2. $100.80^{\circ} \mathrm{C}$
  3. $100.16^{\circ} \mathrm{C}$
  4. $100.62^{\circ} \mathrm{C}$

Solution

$\begin{aligned} & \Delta \mathrm{T}_{\mathrm{b}}=\mathrm{K}_{\mathrm{b}} \cdot \mathrm{m} \\ & \mathrm{T}_{\mathrm{b}}-\mathrm{T}_{\mathrm{b}}^{\circ}=\mathrm{K}_{\mathrm{b}} \cdot \mathrm{m} \\ & \mathrm{T}_{\mathrm{b}}-100^{\circ} \mathrm{C}=\mathrm{K}_{\mathrm{b}} \times 0.5 \\ & 100.16-100^{\circ} \mathrm{C}=\mathrm{K}_{\mathrm{b}} \times 0.1 \\ & \left.\mathrm{~K}_{\mathrm{b}}=\frac{0.16}{0.1}=1.6 \text { (From equation } 2\right) \\ & \mathrm{T}_{\mathrm{b}}-100^{\circ} \mathrm{C}=1.6 \times 0.5 \\ & \mathrm{~T}_{\mathrm{b}}=100+0.80=100.80^{\circ} \mathrm{C}\end{aligned}$

Asked in: MHT CET 2021 (21 Sep Shift 2)

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