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The molal elevation constant of water is 0.51. The boiling point of 0.1 molal aqueous glucose solution is nearly

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${101.0}^{\circ} C$

${100.05}^{\circ} C$

${100.1}^{\circ} C$

${100.2}^{\circ} C$

answer is B.

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Detailed Solution

$Δ T_{f} = K_{b} \times m$

$Δ T_{f} = 0.51 \times 0.1 = 0.051$

$Δ T_{f} =$ BP of solution – BP of solvent

0.051 = BP of solution – 100

BP of solution = 100+ 0.051 = 100.051

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