x g of a solute is dissolved in y g of two | vedclass

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(C) According to Raoult's law,the relative lowering of vapor pressure is given by $\frac{P^o - P}{P^o} = \chi_{solute} = \frac{n_{solute}}{n_{solute}

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$M_A = 2 \ M_B$

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(C) According to Raoult's law,the relative lowering of vapor pressure is given by $\frac{P^o - P}{P^o} = \chi_{solute} = \frac{n_{solute}}{n_{solute} + n_{solvent}}$.Since $n_{solute} \ll n_{solvent}$,the expression simplifies to $\frac{P^o - P}{P^o} \approx \frac{n_{solute}}{n_{solvent}} = \frac{x/M_{solute}}{y/M_{solvent}} = \frac{x \cdot M_{solvent}}{y \cdot M_{solute}}$.For liquid $A$: $(\frac{P^o - P}{P^o})_A = \frac{x \cdot M_A}{y \cdot M_{solute}}$.For liquid $B$: $(\frac{P^o - P}{P^o})_B = \frac{x \cdot M_B}{y \cdot M_{solute}}$.Given that $(\frac{P^o - P}{P^o})_A = 2 	imes (\frac{P^o - P}{P^o})_B$,we have $\frac{x \cdot M_A}{y \cdot M_{solute}} = 2 	imes \frac{x \cdot M_B}{y \cdot M_{solute}}$.Simplifying this,we get $M_A = 2 \ M_B$.

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