The vapour pressure of acetone at 20,^{circ}C | vedclass

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(D) The relative lowering of vapour pressure is given by the formula: $\frac{P^o - P_s}{P_s} = \frac{w_2 	imes M_1}{w_1 	imes M_2}$Here,$P^o = 185\,

Vedclass · Accepted Answer

$64$

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(D) The relative lowering of vapour pressure is given by the formula: $\frac{P^o - P_s}{P_s} = \frac{w_2 	imes M_1}{w_1 	imes M_2}$Here,$P^o = 185\,torr$ (vapour pressure of pure solvent),$P_s = 183\,torr$ (vapour pressure of solution),$w_1 = 100\,g$ (mass of solvent,acetone),$M_1 = 58\,g\,mol^{-1}$ (molar mass of acetone,$CH_3COCH_3$),$w_2 = 1.2\,g$ (mass of solute),$M_2 = ?$ (molar mass of solute).Substituting the values:$\frac{185 - 183}{183} = \frac{1.2 	imes 58}{100 	imes M_2}$$\frac{2}{183} = \frac{69.6}{100 	imes M_2}$$M_2 = \frac{69.6 	imes 183}{2 	imes 100} = \frac{12736.8}{200} = 63.684 \approx 64\,g\,mol^{-1}$.

The vapour pressure of acetone 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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