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Q.

Freezing point of sucrose solution is -0.186⁰ C. Boiling point of the same solution will be (K_b and K_f of water are 0.52 K kg mol^-1 and 1.86 K kg mol^-1)

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a

373.52 K

b

373.104 K

c

374.04 K

d

373.052 K

answer is B.

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Detailed Solution

$\frac{∆ T_{f}}{∆ T_{b}} = \frac{K_{f}}{K_{b}}; \frac{0.186}{∆ T_{b}} = \frac{1.86}{0.52}; ∆ T_{b} = 0.052 K; T_{b} = 373.052 K;$

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Freezing point of sucrose solution is -0.1860 C. Boiling point of the same solution will be (Kb and Kf of water are 0.52 K kg mol-1 and 1.86 K kg mol-1)