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Q.

The boiling point of benzene is 3 53.23 K. When 1.80 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. (K_b for benzene is 2.53 K kg mol^-1)

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a

58 g mol-1

b

49 g mol-1

c

85 g mol-1

d

94 g mol-1

answer is A.

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Detailed Solution

$The elevation ({ΔT}_{b}) in boiling point$
$= 354 .11 K - 353 .23 K = 0 .88 K$
${ΔT}_{b} = \frac{K_{f} \times w_{2} \times 10^{3}}{M_{2} \times w_{1}} and M_{2} = \frac{K_{f} \times w_{2} \times 10^{3}}{{ΔT}_{b} \times w_{1}}$
$∴ M_{2} = \frac{2 .53 Kkg {mol}^{- 1} \times 1 .8 g \times 1000 g {kg}^{- 1}}{0 .88 K \times 90 g}$
$\approx 58 g {mol}^{- 1}$
Therefore, molar mass of the solute, M₂ = 58 g mol^-1

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