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Q.
If mole fraction of a solute in $1\, kg$ benzene is $0.2$ then molality of solute is
Solutions
Solution:
Let number of moles of solute in solution $=x$
Moles of benzene in solution $=\frac{1000\, g }{78\, g\, mol ^{-1}}$
$=12.82$ moles
Mole fraction of solute $=\frac{x}{x+12.82}$
$\Rightarrow 0.2=\frac{x}{x+12.82}$
On solving, $x=3.2$
$\therefore$ Molality $(m)=\frac{\text { Number of moles of solute }}{\text { Mass of solvent (in kg) }}=\frac{3.2}{1}=3.2$