The mole boiling point constant for water is ${0.513}^{\circ} Ckg {mol}^{- 1} . When 0.1 mole$ of sugar is dissolved in $200 ml$ f water, the solution boils under a pressure of one atmosphere at

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${101.025}^{\circ} C$

${100.256}^{\circ} C$

${100.0513}^{\circ} C$

${100.513}^{\circ} C$

answer is C.

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Detailed Solution

$\begin{array}{l} {ΔT}_{b} = K_{b} \times m = 0 .513 \times (\frac{0 .1}{200} \times 1000) \\ T_{b} - T_{b}^{o} = 0 {.2565}^{\circ} C, T_{b} = 100 {.256}^{\circ} C \end{array}$

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