Question

0.161 mol of non-volatile, non-electrolyte solute is
dissolved in 1000 g water at 298 K, the vapour
pressure of solution is (vapour pressure of pure water
at 298 K is 23.51 mm Hg)​

Answer 1

Given:

0.161 mol of non-volatile, non-electrolyte solute is

dissolved in 1000 g water at 298 K.

To find:

Vapour Pressure of solution.

Calculation:

Let vapour pressure of pure water be ${P_{A}}^{\degree}$ and that of solution be $P_{A}$.

$\sf{ \therefore \: P_{A} = {P_{A}}^{ \degree} \times (\chi_{solvent} )}$

$\sf{ = > \: P_{A} = 23.51 \times \bigg(\dfrac{ \frac{1000}{18} }{0.161 + \frac{1000}{18} } \bigg)}$

$\sf{ = > \: P_{A} = 23.51 \times \bigg(\dfrac{55.55}{0.161 + 55.55 } \bigg)}$

$\sf{ = > \: P_{A} = 23.51 \times \bigg(\dfrac{55.55}{ 55.71} \bigg)}$

$\sf{ = > \: P_{A} = 23.44 \: \: mm \: of \: Hg }$

So, final answer is:

$\boxed{ \large{ \blue{\rm{ \: P_{A} = 23.44 \: \: mm \: of \: Hg }}}}$