Calculate the molar mass of a nonvolatile sol | vedclass

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Question

(C) The relative lowering in vapour pressure is given by Raoult's Law: $\frac{P^\circ - P_s}{P^\circ} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1}$

Vedclass · Accepted Answer

$72 	ext{ g mol}^{-1}$

Vedclass · Answer

(C) The relative lowering in vapour pressure is given by Raoult's Law: $\frac{P^\circ - P_s}{P^\circ} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1}$ for dilute solutions.Given: $\frac{P^\circ - P_s}{P^\circ} = 0.016$, mass of solute $w_2 = 6.4 	ext{ g}$, mass of solvent $w_1 = 100 	ext{ g}$, molar mass of water $M_1 = 18 	ext{ g mol}^{-1}$.Formula: $\frac{P^\circ - P_s}{P^\circ} = \frac{w_2 	imes M_1}{M_2 	imes w_1}$.Substitute the values: $0.016 = \frac{6.4 	imes 18}{M_2 	imes 100}$.$M_2 = \frac{6.4 	imes 18}{0.016 	imes 100} = \frac{115.2}{1.6} = 72 	ext{ g mol}^{-1}$.

Calculate the molar mass of a nonvolatile solute if $6.4 \text{ g}$ of it dissolved in $100 \text{ g}$ water produces a relative lowering in vapour pressure of $0.016$ at $300 \text{ K}$.

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