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.
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:

i. molar mass of the solute

ii. vapour pressure of water at 298 K.

Solution:

(i) Let, the molar mass of the solute be $\mathrm{M} \mathrm{g} \mathrm{mol}^{-1}$

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

And, the no. of moles of solute, $n_{2}=\frac{30 \mathrm{~g}}{\mathrm{M} \mathrm{mol}^{-1}}=\frac{30}{\mathrm{M}} \mathrm{mol}$

$p_{1}=2.8 \mathrm{kPa}$

Applying the relation:

$\frac{p_{1}^{0}-p_{1}}{p_{1}^{0}}=\frac{n_{2}}{n_{1}+n_{2}}$

$\Rightarrow \frac{p_{1}^{0}-2.8}{p_{1}^{0}}=\frac{\frac{30}{\mathrm{M}}}{5+\frac{30}{\mathrm{M}}}$

$\Rightarrow 1-\frac{2.8}{p_{1}^{0}}=\frac{\frac{30}{\mathrm{M}}}{\frac{5 \mathrm{M}+30}{\mathrm{M}}}$

$\Rightarrow 1-\frac{2.8}{p_{1}^{0}}=\frac{30}{5 \mathrm{M}+30}$

$\Rightarrow \frac{2.8}{p_{1}^{0}}=1-\frac{30}{5 \mathrm{M}+30}$

$\Rightarrow \frac{2.8}{p_{1}^{0}}=\frac{5 \mathrm{M}+30-30}{5 \mathrm{M}+30}$

$\Rightarrow \frac{2.8}{p_{1}^{0}}=\frac{5 \mathrm{M}}{5 \mathrm{M}+30}$

$\Rightarrow \frac{p_{1}^{0}}{2.8}=\frac{5 \mathrm{M}+30}{5 \mathrm{M}}$    …(i)

After the addition of 18 g of water:

$n_{1}=\frac{90+18 \mathrm{~g}}{18}=6 \mathrm{~mol}$

$p_{1}=2.9 \mathrm{kPa}$

Again, applying the relation:

$\frac{p_{1}^{0}-p_{1}}{p_{1}^{0}}=\frac{n_{2}}{n_{1}+n_{2}}$

$\Rightarrow \frac{p_{1}^{0}-2.9}{p_{1}^{0}}=\frac{\frac{30}{\mathrm{M}}}{6+\frac{30}{\mathrm{M}}}$

$\Rightarrow 1-\frac{2.9}{p_{1}^{0}}=\frac{\frac{30}{\mathrm{M}}}{\frac{6 \mathrm{M}+30}{\mathrm{M}}}$

$\Rightarrow 1-\frac{2.9}{p_{1}^{0}}=\frac{30}{6 \mathrm{M}+30}$

$\Rightarrow \frac{2.9}{p_{1}^{0}}=1-\frac{30}{6 \mathrm{M}+30}

$\Rightarrow \frac{2.9}{p_{1}^{0}}=\frac{6 \mathrm{M}+30-30}{6 \mathrm{M}+30}$

$\Rightarrow \frac{2.9}{p_{1}^{0}}=\frac{6 \mathrm{M}}{6 \mathrm{M}+30}$

$\Rightarrow \frac{p_{1}^{0}}{2.9}=\frac{6 \mathrm{M}+30}{6 \mathrm{M}}$   …(ii)

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

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

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

$\Rightarrow 2.9 \times 5 \times(6 \mathrm{M}+30)=2.8 \times 6 \times(5 \mathrm{M}+30)$

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

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

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

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

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

$\frac{p_{1}^{0}}{2.8}=\frac{5 \times 23+30}{5 \times 23}$

$\Rightarrow \frac{p_{1}^{0}}{2.8}=\frac{145}{115}$

$\Rightarrow p_{1}^{0}=3.53$

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

 

 

 

 

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