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Question

The vapour pressure of an aqueous solution is found to be 750 torr at a temperature T, and the same solution shows an elevation in boiling point equal to 1.04 K. If T is the boiling point of pure water, then the atmospheric pressure should be (Kb of water = 0.52 K-kg/mol)

Options

A.760 torr
B.777 torr
C.746 torr
D.750 torr

Solution

$\Delta T_{b} = K_{b}.m \Rightarrow 1.04 = 0.52 \times m \Rightarrow m = 2$

Now, $\frac{P^{o} - P}{P} = \frac{n_{1}}{n_{2}} \Rightarrow \frac{P^{o} - 750}{750} = \frac{2}{1000/18} \Rightarrow P^{o} = 777\text{ torr}$

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