CBSE Class 12 · 2003 · All India Compartment · SolutionsShort Answer

1.0 g of a non-volatile solute was dissolved in 100 g of acetone (molar mass = 58 g mol

^{- 1}

) at 298 K. The vapour pressure of the solution was found to be 192.5 mm Hg. Calculate the molar mass of the solute. (Vapour pressure of pure acetone at 298 K is 195 mm Hg)

Answer: M ≈ 45 g mol⁻¹

Solution

Relative lowering of vapour pressure:

\frac{p ^{\circ} - p}{p ^{\circ}} = \frac{195 - 192.5}{195} = 0.01282 = x_{2}

Moles of acetone:

n_{1} = \frac{100}{58} = 1.724 mol

For a dilute solution,

x_{2} \approx \frac{n _{2}}{n _{1}}

n_{2} = 0.01282 \times 1.724 = 0.0221 mol

Molar mass:

M = \frac{1.0}{0.0221}

M \approx 45 g mol^{- 1}

CBSE Class 12 Board · 2003 · All India Compartment · Subjective

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