The mole fraction of a solute in a binary sol | vedclass

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(A) Given:Mole fraction of solute $(x_2)$ = $0.1$Mole fraction of solvent $(x_1)$ = $1 - 0.1 = 0.9$Density $(d)$ = $2.0 \ g \ cm^{-3}$Molarity $(M)$ =

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$9$

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(A) Given:Mole fraction of solute $(x_2)$ = $0.1$Mole fraction of solvent $(x_1)$ = $1 - 0.1 = 0.9$Density $(d)$ = $2.0 \ g \ cm^{-3}$Molarity $(M)$ = Molality $(m)$We know that:$m = \frac{x_2 	imes 1000}{x_1 	imes M_1}$$M = \frac{1000 	imes d 	imes x_2}{x_1 M_1 + x_2 M_2}$Since $M = m$:$\frac{x_2 	imes 1000}{x_1 	imes M_1} = \frac{1000 	imes d 	imes x_2}{x_1 M_1 + x_2 M_2}$$\frac{1}{x_1 M_1} = \frac{d}{x_1 M_1 + x_2 M_2}$$x_1 M_1 + x_2 M_2 = d 	imes x_1 M_1$$0.9 M_1 + 0.1 M_2 = 2.0 	imes 0.9 M_1$$0.9 M_1 + 0.1 M_2 = 1.8 M_1$$0.1 M_2 = 0.9 M_1$$\frac{M_2}{M_1} = \frac{0.9}{0.1} = 9$Therefore,the ratio of molecular weights of the solute and the solvent is $9$.

The mole fraction of a solute in a binary solution is $0.1$. At $298 \ K$,molarity of this solution is same as its molality. Density of this solution at $298 \ K$ is $2.0 \ g \ cm^{-3}$. The ratio of molecular weights of the solute and the solvent $(M_{solute} / M_{solvent})$ is:

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