SolutionHard
Question
At 100°C, the vapour pressure of a solution of 4.0 g of solute in 100 g of water is 750 mm. The boiling point of the solution is (kb of water = 0.52 K-kg/mol)
Options
A.100°C
B.100.04°C
C.100.4°C
D.104.0°C
Solution
If molality of solution is ‘m’, then
$\frac{P^{o} - P}{P} = \frac{n_{1}}{n_{2}} \Rightarrow \frac{760 - 750}{750} = \frac{m}{1000/18} \Rightarrow m = \frac{20}{27}$
Now, $\Delta T_{b} = K_{b}.m = 0.52 \times \frac{20}{27} = 0.358\text{ K}$
∴ B.P. of solution = 100.385° C
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