1.4 g of acetone dissolved in 100 g of benzene gave a solution which freezes at 277.12 K. Pure benzene freezes at 278.4 K. 2.8 g of a solid (A) dissolved in 100 g of benzene gave a solution which froze at 277.76 K. Calculate the molar mass of (A)

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Solution

For acetone + Benzene mixture:
$Δ T = 278.40 - 277.12 = 1.28$
$Δ T = \frac{K_{f} \times 1000 \times w}{m \times W}$
$1.28 = \frac{1000 \times K_{f} \times 1.4}{100 \times 58}$ .....(i)
For solute (A) + Benzene mixture (let m be the molar mass of A)
$(278.40 - 277.76) = \frac{1000 \times K_{f} \times 2.8}{100 \times m}$
or $0.64 = \frac{1000 \times K_{f} \times 2.8}{100 \times m}$ .....(ii)
By eqs.(i) and (ii), $m = 232$
The molar mass of (A) is 232 g/mol.

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