Question

111.1g of the non-volatile solute, urea needs to be dissolved in 100 g of water, in order to decrease the vapour pressure of water by 25%. The molarity of the solution will be:

• A. 18.52 M
• B. 18.92 M
• C. 19.52 M
• D. None of the above

Answer 1

Answer: (A)

18.52 M

The depression in the vapour pressure of water is equal to the mole fraction of the solution.

$\frac{P^0-P}{P^0}=\frac{n}{n+N}$

Here n is the number of moles of urea and N is the number of moles of water. The number of moles is the ratio of mass to molar mass. Let M g/mol be the molar mass of urea.

$0.25=\frac{\frac{111.1}{M}}{\frac{100}{18}+\frac{111.1}{M}}$

$0.25=\frac{\frac{111.1}{M}}{5.556+\frac{111.1}{M}}$

$4=\frac{5.556+\frac{111.1}{M}}{\frac{111.1}{M}}$

$5.556+\frac{111.1}{M}=\frac{444.4}{M}$

$5.556=\frac{444.4-111.1}{M}$

$5.556M=333.3$

$M=60g/mol$

Number of moles of urea $\frac{111.1g}{60g/mol}=1.8516$ moles.

The density of water is 1g/mL. Hence 100 g of water corresponds to 100 mL or 0.1 L.

Molarity of urea solution is $\frac{1.8516moles}{0.1L}=18.52M$

Hence, the correct option is $\text{A}$