The mole fraction of a solute in a solution i | vedclass

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(B) Let the molecular weight of the solute be $M_1$ and the solvent be $M_2$.Given mole fraction of solute $X_1 = 0.1$,so mole fraction of solvent $X_

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(B) Let the molecular weight of the solute be $M_1$ and the solvent be $M_2$.Given mole fraction of solute $X_1 = 0.1$,so mole fraction of solvent $X_2 = 0.9$.For a solution,the relation between molarity $(M)$ and molality $(m)$ is given by $m = \frac{1000M}{1000d - MM_1}$,where $d$ is the density in $g \ cm^{-3}$.Since $M = m$,we have $1 = \frac{1000}{1000d - MM_1}$.$1000d - MM_1 = 1000$ $\Rightarrow 1000(2) - MM_1 = 1000$ $\Rightarrow MM_1 = 1000$.Since $M = \frac{n_1}{V(L)}$,and $n_1 = \frac{mass}{M_1}$,we have $M = \frac{mass}{M_1 	imes V(L)}$.Given $d = \frac{mass}{V} = 2000 \ g \ L^{-1}$,so $mass = 2000 	imes V$.$M = \frac{2000 	imes V 	imes w_1}{M_1 	imes V} = \frac{2000 w_1}{M_1}$ where $w_1$ is the mass fraction of solute.Using $X_1 = \frac{n_1}{n_1 + n_2} = 0.1$,we get $n_2 = 9n_1$.$\frac{mass_2}{M_2} = 9 	imes \frac{mass_1}{M_1} \Rightarrow \frac{mass_2}{mass_1} = 9 	imes \frac{M_2}{M_1}$.Since $M = m$,the mass of solvent is $1000 \ g$.From $MM_1 = 1000$,we find $n_1 = 1 \ mol$.Thus,$n_2 = 9 \ mol$.Mass of solvent $= 9 	imes M_2 = 1000 \ g \Rightarrow M_2 = \frac{1000}{9}$.Mass of solute $= 1 	imes M_1 = 1000 - (2000 - 1000) = 1000 \ g$ is incorrect; using $X_1 = 0.1$,$\frac{M_1}{M_2} = \frac{X_1}{X_2} 	imes \frac{n_2}{n_1} = \frac{0.1}{0.9} 	imes 9 = 1$ is wrong.Correct approach: $\frac{M_1}{M_2} = \frac{X_1}{X_2} 	imes \frac{n_2}{n_1}$. With $M=m$,$d = 1 + \frac{M M_1}{1000} = 2$. $M M_1 = 1000$. $n_1 = M 	imes 1 = 1000/M_1$. $n_2 = (2000 - 1000)/M_2 = 1000/M_2$. $X_1 = \frac{1000/M_1}{1000/M_1 + 1000/M_2} = 0.1$ $\Rightarrow \frac{M_2}{M_1+M_2} = 0.1$ $\Rightarrow M_2 = 0.1M_1 + 0.1M_2$ $\Rightarrow 0.9M_2 = 0.1M_1$ $\Rightarrow \frac{M_1}{M_2} = 9$.

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

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