The mass of a non-volatile, non-electrolyte solute (molar mass = 50 g mol$$^{-1}$$) needed to be dissolved in 114 g octane to reduce its vapour pressure by 75%, is:

According to Raoult’s law for a solution in which the solute is non-volatile and non-electrolyte, the vapour pressure of the solution is given by

$$P_{\text{solution}} = P^{0}_{\text{solvent}}\;x_{\text{solvent}}$$

where $$P^{0}_{\text{solvent}}$$ is the vapour pressure of the pure solvent and $$x_{\text{solvent}}$$ is the mole fraction of the solvent in the solution.

The statement “vapour pressure is reduced by 75 %” means the pressure falls to only 25 % (one-quarter) of the original value. Mathematically,

$$P_{\text{solution}} = 0.25\,P^{0}_{\text{solvent}}$$

Substituting this in Raoult’s law gives

$$0.25\,P^{0}_{\text{solvent}} = P^{0}_{\text{solvent}}\;x_{\text{solvent}}$$

On cancelling $$P^{0}_{\text{solvent}}$$ from both sides we obtain

$$x_{\text{solvent}} = 0.25$$

Now, let

$$n_{\text{solvent}} = \text{moles of octane}$$

$$n_{\text{solute}} = \text{moles of the non-volatile solute}$$

The mole fraction of the solvent is

$$x_{\text{solvent}} = \frac{n_{\text{solvent}}}{n_{\text{solvent}} + n_{\text{solute}}}$$

Setting this equal to 0.25, we write

$$\frac{n_{\text{solvent}}}{n_{\text{solvent}} + n_{\text{solute}}} = 0.25$$

Cross-multiplying gives

$$n_{\text{solvent}} = 0.25\,(n_{\text{solvent}} + n_{\text{solute}})$$

Expanding the right side,

$$n_{\text{solvent}} = 0.25\,n_{\text{solvent}} + 0.25\,n_{\text{solute}}$$

Rearranging all terms of $$n_{\text{solvent}}$$ to the left,

$$n_{\text{solvent}} - 0.25\,n_{\text{solvent}} = 0.25\,n_{\text{solute}}$$

So we have

$$0.75\,n_{\text{solvent}} = 0.25\,n_{\text{solute}}$$

Dividing both sides by 0.25,

$$3\,n_{\text{solvent}} = n_{\text{solute}}$$

Hence,

$$n_{\text{solute}} = 3\,n_{\text{solvent}}$$

The mass of octane given is 114 g. Its molar mass is also 114 g mol−1 (since C8H18 has molar mass $$8\times12 + 18\times1 = 96 + 18 = 114$$). Therefore,

$$n_{\text{solvent}} = \frac{114\;\text{g}}{114\;\text{g mol}^{-1}} = 1\;\text{mol}$$

Substituting $$n_{\text{solvent}} = 1$$ in the earlier relation,

$$n_{\text{solute}} = 3 \times 1 = 3\;\text{mol}$$

The molar mass of the solute is given as 50 g mol−1. Therefore, the required mass of solute is

$$\text{mass}_{\text{solute}} = n_{\text{solute}} \times M_{\text{solute}} = 3\;\text{mol} \times 50\;\text{g mol}^{-1} = 150\;\text{g}$$

Hence, the correct answer is Option C.