Question

10 g of a non-volatile solute is dissolved in 200 g of water to make a solution. It has a vapour pressure of 31.84 mm Hg at 308 K.

Calculate the molar mass of the solute.

[Given : Vapour pressure of pure water at 308 K = 32 mm Hg]

Hint:

Use the relative lowering of vapour pressure, (p0 - p)/p0 = w2 M1/(M2 w1), with M1 = 18 for water.

Answer 1

Step 1: Understanding the Concept:
For a non-volatile solute, Raoult's law gives the relative lowering of vapour pressure as the mole fraction of the solute.
\[ \frac{p^\circ - p}{p^\circ} = \frac{n_2}{n_1 + n_2} \]
Here 1 is water and 2 is the solute.

Step 2: Data:
\( p^\circ = 32 \) mm Hg, \( p = 31.84 \) mm Hg, \( w_2 = 10 \) g, \( w_1 = 200 \) g, \( M_1 = 18 \) g/mol.
Moles of water: \( n_1 = 200/18 = 11.11 \) mol.

Step 3: Relative lowering:
\[ \frac{p^\circ - p}{p^\circ} = \frac{32 - 31.84}{32} = \frac{0.16}{32} = 0.005 \]

Step 4: Find the molar mass:
The solution is dilute, so \( n_2 \ll n_1 \) and \( n_1 + n_2 \approx n_1 \).
\[ \frac{p^\circ - p}{p^\circ} = \frac{w_2 \times M_1}{M_2 \times w_1} \]
\[ 0.005 = \frac{10 \times 18}{M_2 \times 200} \]
\[ M_2 = \frac{10 \times 18}{0.005 \times 200} = \frac{180}{1} = 180\ \mathrm{g\,mol^{-1}} \]

Final Answer:
The molar mass of the solute is 180 g/mol.
\[ \boxed{M_2 = 180\ \mathrm{g\,mol^{-1}}} \]