The vapour pressure of a solution is $2985 Nm^{-2}$ when 5 g of non-electrolyte added into 100 g of water at particular temperature.

The vapour pressure of pure water at this temperature is $3000 Nm^{-2}$. The molecular mass of the solute is.

179

The correct answer is Option (1) → 179.

To find the molecular mass of the solute, we will use Raoult's Law, which relates the vapor pressure of a solution to the mole fraction of the solvent.

Raoult’s Law states:

\(P_{\text{solution}} = P_{\text{solvent}} \times X_{\text{solvent}}\)

Where:

\( P_{\text{solution}} \) = vapor pressure of the solution (2985 N/m²)

\( P_{\text{solvent}} \) = vapor pressure of pure solvent (water) (3000 N/m²)

\(\chi _{\text{solvent}} \) = mole fraction of the solvent (water)

The reduction in vapor pressure is due to the presence of the solute. According to Raoult’s Law:

\(\chi _{\text{solvent}} = \frac{P_{\text{solution}}}{P_{\text{solvent}}}\)

Substitute the given values:

\(\chi _{\text{solvent}} = \frac{2985}{3000} = 0.995\)

The mole fraction of the solvent is given by:

\(\chi _{\text{solvent}} = \frac{n_{\text{solvent}}}{n_{\text{solvent}} + n_{\text{solute}}}\)
Where:

\( n_{\text{solvent}} \) = moles of solvent (water)

\( n_{\text{solute}} \) = moles of solute (unknown)

First, calculate the moles of water:

\(n_{\text{solvent}} = \frac{\text{mass of water}}{\text{molar mass of water}} = \frac{100 \, \text{g}}{18 \, \text{g/mol}} = 5.56 \, \text{mol}\)

Now, rearrange the mole fraction equation to solve for \( n_{\text{solute}} \):

\(0.995 = \frac{5.56}{5.56 + n_{\text{solute}}}\)

\(5.56 + n_{\text{solute}} = \frac{5.56}{0.995} = 5.588\)

\(n_{\text{solute}} = 5.588 - 5.56 = 0.028 \, \text{mol}\)

The molecular mass (\(M_{\text{solute}}\)) is given by:

\(M_{\text{solute}} = \frac{\text{mass of solute}}{\text{moles of solute}}\)

Substitute the known values:

\(M_{\text{solute}} = \frac{5 \, \text{g}}{0.028 \, \text{mol}} \approx 179 \, \text{g/mol}\)

The molecular mass of the solute is 179 g/mol.