W g of a non-volatile electrolyte solid solu | vedclass

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(A) Given:$P^0 = 640 \ mm \ Hg$, $P_s = 600 \ mm \ Hg$.Relative lowering of vapor pressure: $\frac{P^0 - P_s}{P^0} = i \cdot X_2$.$\frac{640 - 600}{64

Vedclass · Accepted Answer

$\frac{1.3}{8} 	imes \frac{W}{M}$

Vedclass · Answer

(A) Given:$P^0 = 640 \ mm \ Hg$, $P_s = 600 \ mm \ Hg$.Relative lowering of vapor pressure: $\frac{P^0 - P_s}{P^0} = i \cdot X_2$.$\frac{640 - 600}{640} = i \cdot X_2 \implies i \cdot X_2 = \frac{40}{640} = \frac{1}{16}$ ... $(i)$.Elevation in boiling point: $\Delta T_b = T_b - T_b^0 = 375 - 373 = 2 \ K$.Formula: $\Delta T_b = i \cdot K_b \cdot m$.$2 = i \cdot 0.52 \cdot \frac{W 	imes 1000}{M 	imes 100} \implies 2 = i \cdot 5.2 \cdot \frac{W}{M} \implies i = \frac{2 \cdot M}{5.2 \cdot W} = \frac{M}{2.6 \cdot W}$ ... $(ii)$.Substituting $(ii)$ in $(i)$:$\left(\frac{M}{2.6 \cdot W}ight) \cdot X_2 = \frac{1}{16}$.$X_2 = \frac{2.6 \cdot W}{16 \cdot M} = \frac{1.3}{8} \cdot \frac{W}{M}$.

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