SolutionHard
Question
72 g of glucose is dissolved in 1 kg of water in a sauce pan. At what temperature will the solution boil at 1.013 bar pressure? The value of Kb for water is 0.52 K-kg mol−1.
Options
A.337.208 K
B.373.208°C
C.373.208 K
D.375.08 K
Solution
$\Delta T_{b} = K_{b}.m = 0.52 \times \frac{72}{180} = 0.208\text{ K}$
∴ B.P. of solution = 373.208 K
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