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The normal boiling point of pure ethyl acetate is 77.06 °C. A solution of 50.0 g of naphthalene $(C_{10} H_{8})$ dissolved in 150 g of ethyl acetate boils at 84.27 °C. The boiling point elevation constant of ethyl acetate is

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$2.77 Kkg {mol}^{- 1}$

$1.88 Kkg {mol}^{- 1}$

$2.12 Kkg {mol}^{- 1}$

$1.54 Kkg {mol}^{- 1}$

answer is A.

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Detailed Solution

Molality of solution is

$\begin{aligned} m = \frac{m_{2} / M_{2}}{m_{1}} = \frac{(50 g) / (128 g {mol}^{- 1})}{0.150 kg} = 2.604 mol {kg}^{- 1} \\ K_{b} = \frac{Δ T_{b}}{m} = \frac{(84.27 - 77.06) K}{2.604 mol {kg}^{- 1}} = 2.77 Kkg {mol}^{- 1} \end{aligned}$

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