In 108 , g of water,18 , g of a non-volatile | vedclass

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(D) The relative lowering of vapour pressure is given by the formula: $\frac{p_0 - p_s}{p_0} = \chi_2 = \frac{n_2}{n_1 + n_2}$.Given: $p_0 = 760 \, mm

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$228$

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(D) The relative lowering of vapour pressure is given by the formula: $\frac{p_0 - p_s}{p_0} = \chi_2 = \frac{n_2}{n_1 + n_2}$.Given: $p_0 = 760 \, mm \, Hg$ (at $100^{\circ} C$),$p_s = 750 \, mm \, Hg$,$w_2 = 18 \, g$,$w_1 = 108 \, g$,$M_1 = 18 \, g \, mol^{-1}$.$n_1 = \frac{108}{18} = 6 \, mol$.$n_2 = \frac{18}{M_2}$.Using the formula $\frac{760 - 750}{760} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1}$ (since $n_2$ is small compared to $n_1$):$\frac{10}{760} = \frac{18/M_2}{6}$.$\frac{1}{76} = \frac{3}{M_2}$.$M_2 = 3 	imes 76 = 228 \, g \, mol^{-1}$.

In $108 \, g$ of water,$18 \, g$ of a non-volatile compound is dissolved. At $100^{\circ} C$,the vapour pressure of the solution is $750 \, mm \, Hg$. Assuming that the compound does not undergo association or dissociation,the molar mass of the compound in $g \, mol^{-1}$ is

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