At 100°C, the vapour pressure of a solution of 4.0 g of solute in 100 g of water is 750 mm. The boil

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)

Accepted Answer

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, $\Delta T_{b} = K_{b}.m = 0.52 	imes \frac{20}{27} = 0.358	ext{ K}$∴ B.P. of solution = 100.385° C

Question

Options

Solution

More Solution Questions