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.
Answer & explanation
Correct answer: option 1
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.