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The vapour pressure of pure benzene is 0.850 bar. On dissolving 0.5 g of a non-volatile non-electrolyte in 39.0 g of benzene (molar mass 78 g/mol), the vapour pressure falls to 0.845 bar. The molar mass of the solute is:
A170 g/mol
B85 g/mol
C78 g/mol
D340 g/mol
Answer & Solution
Correct answer: A. 170 g/mol
1. Use $(p_1^0 - p_1)/p_1^0 = (w_2 M_1)/(M_2 w_1)$ for dilute solutions.
2. $(0.850 - 0.845)/0.850 = (0.5 \times 78)/(M_2 \times 39)$.
3. $0.00588 = 39 / (39 M_2) = 1 / M_2$.
4. $M_2 = 1 / 0.00588 = 170$ g/mol.
_Source: NCERT Class 12 Chemistry Ch 1 "Solutions", p.15_
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