Question

A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate:

1. molar mass of the solute
2. vapour pressure of water at 298 K.

Answer 1

(i) Let, the molar mass of the solute be M g mol⁻¹.

Now, the no. of moles of solvent (water), n₁ = ${\displaystyle \frac{90 \text{g}}{18 {\text{g mol}}^{-1}}}$ = 5 mol

And, the no. of moles of solute, n₂ = ${\displaystyle \frac{30 \text{g}}{{\text{M mol}}^{-1}}=\frac{30}{\text{M}}\text{mol}}$

p₁ = 2.8 kPa

Applying the relation,

${\displaystyle \frac{{\text{p}}_1^0-{\text{p}}_1}{{\text{p}}_1^0}=\frac{{\text{n}}_2}{{\text{n}}_1+{\text{n}}_2}}$

${\displaystyle \Rightarrow \frac{{\text{p}}_1^0-2.8}{{\text{p}}_1^0}=\frac{\frac{30}{\text{M}}}{5+\frac{30}{\text{M}}}}$

${\displaystyle \Rightarrow 1-\frac{2.8}{{\text{p}}_1^0}=\frac{\frac{30}{\text{M}}}{\frac{5\text{M}+30}{\text{M}}}}$

${\displaystyle \Rightarrow 1-\frac{2.8}{{\text{p}}_1^0}=\frac{30}{5\text{M}+30}}$

${\displaystyle \Rightarrow \frac{2.8}{{\text{p}}_1^0}=1-\frac{30}{5\text{M}+30}}$

${\displaystyle \Rightarrow \frac{2.8}{{\text{p}}_1^0}=\frac{5\text{M}+30-30}{5\text{M}+30}}$

${\displaystyle \Rightarrow \frac{{\text{p}}_1^0}{2.8}=\frac{5\text{M}+30}{5\text{M}}}$ ......(i)

After the addition of 18 g of water

${\displaystyle {\text{n}}_1=\frac{90+18 \text{g}}{18}}$ = 6 mol

p₁ = 2.9 kPa

Again, applying the relation,

${\displaystyle \frac{{\text{p}}_1^0-{\text{p}}_1}{{\text{p}}_1^0}=\frac{{\text{n}}_2}{{\text{n}}_1+{\text{n}}_2}}$

${\displaystyle \Rightarrow \frac{{\text{p}}_1^0-2.9}{{\text{p}}_1^0}=\frac{\frac{30}{\text{M}}}{\frac{6+30}{\text{M}}}}$

${\displaystyle \Rightarrow 1-\frac{2.9}{{\text{p}}_1^0} =\frac{\frac{30}{\text{M}}}{\frac{6\text{M}+30}{\text{M}}}}$

${\displaystyle \Rightarrow 1-\frac{2.9}{{\text{p}}_1^0}=\frac{30}{6\text{M}+30}}$

${\displaystyle \Rightarrow \frac{2.9}{{\text{p}}_1^0}=1-\frac{30}{6\text{M}+30}}$

${\displaystyle \Rightarrow \frac{2.9}{{\text{p}}_1^0}=\frac{6\text{M}+30-30}{6\text{M}+30}}$

${\displaystyle \Rightarrow \frac{2.9}{{\text{p}}_1^0}=\frac{6\text{M}}{6\text{M}+30}}$           

${\displaystyle \Rightarrow \frac{{\text{p}}_1^0}{2.9}=\frac{6\text{M}+30}{6\text{M}}}$ .......(ii)

Dividing equation (i) by (ii), we have:                 

${\displaystyle \frac{2.9}{2.8}=\frac{\frac{5\text{M}+30}{5\text{M}}}{\frac{6\text{M}+30}{6\text{M}}}}$

${\displaystyle \Rightarrow \frac{2.9}{2.8}\times \frac{6\text{M}+30}{6}=\frac{5\text{M}+30}{5}}$

${\displaystyle \Rightarrow }$ 2.9 × 5 × (6m + 30) = 2.8 × 5 × (5M + 30)

${\displaystyle \Rightarrow }$ 87M + 435 = 84M + 504

${\displaystyle \Rightarrow }$ 3M = 69

${\displaystyle \Rightarrow }$ M = 23u

Therefore, the molar mass of the solute is 23 g mol⁻¹.

(ii) Putting the value of ‘M’ in equation (i), we have:

${\displaystyle \frac{{\text{p}}_1^0}{2.8}=\frac{5\times 23+30}{5\times 23}}$

${\displaystyle \Rightarrow \frac{{\text{p}}_1^0}{2.8} =\frac{145}{115}}$

${\displaystyle \Rightarrow {\text{P}}_1^0=3.53}$

Hence, the vapour pressure of water at 298 K is 3.53 kPa.