SolutionHard
Question
The molal boiling point elevation constant of water is 0.513°C kg mol−1. When 0.1 mole of sugar is dissolved in 200 g of water, the solution boils under a pressure of 1 atm at
Options
A.100.513°C
B.102.565°C
C.100.256°C
D.101.025°C
Solution
$\Delta T_{b} = K_{b}.m = 0.513 \times \frac{0.1}{200/1000} = 0.2565^{o}C$
∴ B.P. of solution = 100.2565°C
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