Question

The vapour pressure of pure water at $25^\circ$ is 23.76 torr. The vapour pressure of a solution containing 5.40g of a non-volatile substance in 90.0 g water is 23.32 torr. The molecular weight of the solute is:
A.97.24 g/mol
B.24.29 g/mol
C.50.99 g/mol
D.57.24 g/mol

Answer 1

Hint:Vapour pressure, which is sometimes also called equilibrium vapour pressure is defined as the pressure exerted by a vapour which is in thermodynamic equilibrium with its condensed phases at a given temperature in a closed system.
-Whenever a non-volatile solute is added to a solvent there is a decline in the vapour pressure of that solvent. And this decline in the vapour pressure of the solvent is equal to the mole fraction of the solute.

Complete step by step answer:
-Let us assume that the molecular weight of solute is X.
Now, since the given weight of solute is 5.40 g, the number of moles of the solute is \[ \dfrac{5.40}{X}\]
-Similarly, the molecular weight of water is 18 g and its given weight is 90 g. Therefore, the number of moles of water = $ \dfrac{90}{18}$ = 5 moles.
-As we know, the mole fraction of solute is $ \dfrac{n_1}{n_1 + n_2}$, where ${{{n}}_{{1}}}$ is the number of moles of solute and ${{{n}}_{{2}}}$ is the number of moles of water.
-Putting all the values in the above formula we get,
Mole fraction of solute = $ \dfrac{\left ( \dfrac{5.40}{X} \right )}{\left ( \dfrac{5.40}{X}+5\right )}$ = $ \dfrac{5.40}{5.40 + 5X}$
-Relative lowering of vapour pressure is = $ \dfrac{P^0 - P}{P^0}$, where ${{{P}}^{{0}}}$ is the initial vapour pressure and P is the vapour pressure of solvent after the addition of non-volatile solute.
Thus, the relative lowering of vapour pressure = $ \dfrac{23.76 - 23.32}{23.76} = 0.01852$.
-As we are already aware, the relative lowering of vapour pressure is equal to the mole fraction of the non-volatile solute, therefore,
\[{{0}}{{.01852 = }}\dfrac{5.40}{5.40 + 5X}\]
\[ \Rightarrow {{5}}{{.40 + 5X = 291}}{{.6}}\]
\[ \Rightarrow {{5X = 286}}{{.2}}\]
\[ \therefore {{X = 57}}{{.24\text{g/mol}}}\]
-Thus the molecular weight of the non-volatile solute is 57.24 g/mol.
Hence, option D is the correct answer.

Note: The addition of non-volatile solute results in the reduction of the vapour pressure of the solvent because some of the surface is now occupied by the solute particles and there is less room for solvent particles. Due to this fewer solvent molecules are being able to escape the condensed phase.