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(B) Let the total mass of the solution be $100 \ g$. Mass of liquid $A = 28 \ g$,Mass of water $= 72 \ g$. Moles of $A$ $(n_A)$ $= \frac{28}{140} = 0.

Vedclass · Accepted Answer

$360$

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(B) Let the total mass of the solution be $100 \ g$. Mass of liquid $A = 28 \ g$,Mass of water $= 72 \ g$. Moles of $A$ $(n_A)$ $= \frac{28}{140} = 0.2 \ mol$. Moles of water $(n_w)$ $= \frac{72}{18} = 4 \ mol$. Total moles $= 0.2 + 4 = 4.2 \ mol$. Mole fraction of water $(x_w)$ $= \frac{4}{4.2} = \frac{20}{21}$. Mole fraction of $A$ $(x_A)$ $= \frac{0.2}{4.2} = \frac{1}{21}$. According to Raoult's Law for a solution of two volatile liquids: $P_{total} = P_w^\circ x_w + P_A^\circ x_A$. $160 = 150 	imes (\frac{20}{21}) + P_A^\circ 	imes (\frac{1}{21})$. $160 = \frac{3000}{21} + \frac{P_A^\circ}{21}$. $3360 = 3000 + P_A^\circ$. $P_A^\circ = 360 \ mm$.

An aqueous solution containing $28\%$ by mass of a liquid $A$ (molar mass $= 140 \ g/mol$) has a vapour pressure of $160 \ mm$ at $37 \ ^\circ C$. Find the vapour pressure of the pure liquid $A$ approximately .......... $mm$. (The vapour pressure of pure water at $37 \ ^\circ C$ is $150 \ mm$)

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