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The vapour pressure of a pure liquid A is 10.0 Torr. at 27 °C. One gram of B is dissolved in 20 g of A, the vapour pressure is lowered to 9.0 Torr. If the molar mass of A is $200 g {mol}^{- 1}$ , the molar mass of B is

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$85 g {mol}^{- 1}$

$75 g {mol}^{- 1}$

$100 g {mol}^{- 1}$

$115 g {mol}^{- 1}$

answer is C.

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Detailed Solution

$x_{2} = - \frac{Δ p}{p_{1}^{*}} = \frac{1 Torr}{10 Torr} = \frac{1}{10}$ Now $x_{2} ≃ \frac{(m_{2} / M_{2})}{(m_{1} / M_{1})}$

Hence, $\frac{1}{10} = \frac{(1 g / M_{2})}{(20 g / 200 g {mol}^{- 1})}$ This gives $M_{2} = 100 g {mol}^{- 1}$

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