The vapour pressure of acetone (CH_3COCH_3) a | vedclass

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(D) According to Raoult's law for non-volatile solutes,the relative lowering of vapour pressure is given by: $\frac{P^o - P_s}{P^o} = \frac{n_2}{n_1 +

Vedclass · Accepted Answer

$64.38$

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(D) According to Raoult's law for non-volatile solutes,the relative lowering of vapour pressure is given by: $\frac{P^o - P_s}{P^o} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1}$ (for dilute solutions).Here,$P^o = 185\, torr$,$P_s = 183\, torr$,$W_2 = 1.2\, g$,$W_1 = 100\, g$,and $M_1 (	ext{acetone}) = 58\, g\, mol^{-1}$.Substituting the values: $\frac{185 - 183}{185} = \frac{1.2 / M_2}{100 / 58}$.$\frac{2}{185} = \frac{1.2 	imes 58}{100 	imes M_2}$.$M_2 = \frac{1.2 	imes 58 	imes 185}{2 	imes 100} = \frac{12876}{200} = 64.38\, g\, mol^{-1}$.

The vapour pressure of acetone $(CH_3COCH_3)$ at $20\,^{\circ}C$ is $185\, torr$. When $1.2\, g$ of a non-volatile substance was dissolved in $100\, g$ of acetone at $20\,^{\circ}C$,its vapour pressure was $183\, torr$. The molar mass $(g\, mol^{-1})$ of the substance is

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