A solution of30 g of an unknown compound in 116 g of solvent (molar mass $= 58 g {mol}^{- 1})$ has vapour pressure of 0.475 atm. If the vapour pressure of pure solvent is 0.5 atm, the molar mass of unknown compound is

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$200 g {mol}^{- 1}$

$250 g {mol}^{- 1}$

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

$300 g {mol}^{- 1}$

answer is D.

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Detailed Solution

$M_{2} = \frac{m_{2}}{(m_{1} / M_{l}) (- Δ p_{1} / p_{l}^{*})} M_{2} = \frac{30 g}{(116 g / 58 g {mol}^{- 1}) (0.025 atm / 0.5 atm)} M_{2} = 300 g {mol}^{- 1}$

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