# 0.5 g of nonvolatile solute is dissolved in 100 g of ethyl acetate in 20C.the vapour pressure of the solution and pure ethyl acetate are 72 and 72.8 torr rezpectively at 20C. Calculate the molecular weight of the solute

Please find below the solution to the query posted by you.

We are given,

Mass of non-volatile solute,

*m*

_{1}= 0.5 g

Mass of solvent (ethyl acetate, CH

_{3}COOCH

_{2}CH

_{3}),

*m*

_{2}= 100 g

Molar mass of solvent,

*M*

_{2}= 88 g

*T*= 20

^{o}C = 293 K

Vapour pressure of pure ethyl acetate,

*p*

_{2}

^{o}= 72.8 torr

Vapour pressure of solution,

*p*

_{s}= 72 torr

Mole fraction of the solute can be determined by using the formula for relative lowering of vapour pressure$\frac{{\mathrm{p}}_{2}^{\mathrm{o}}-{\mathrm{p}}_{\mathrm{s}}}{{\mathrm{p}}_{2}^{\mathrm{o}}}=\frac{{\mathrm{m}}_{1}\times {\mathrm{M}}_{2}}{{\mathrm{M}}_{1}\times {\mathrm{m}}_{2}}\phantom{\rule{0ex}{0ex}}\frac{72.8-72}{72.8}=\frac{0.5\times 88}{{\mathrm{M}}_{1}\times 100}\phantom{\rule{0ex}{0ex}}{\mathrm{M}}_{1}=\frac{0.5\times 88\times 72.8}{0.8\times 100}=40.04\mathrm{g}{\mathrm{mol}}^{-1}$.

Hence, the molar mass of the non-volatile solute is 40.04 g mol

^{−1}.

Hope this information will clear your doubt regarding relative lowering of vapour pressure.

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