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  4. An X molal solution of a compound in benzene has mole fraction of solute = 0.2. The value of X is

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Q. An X molal solution of a compound in benzene has mole fraction of solute = 0.2. The value of X is

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Solution:

Mole fraction of solute $X_{2}=0.2$ Therefore, mole fraction of solvent $X_{1}=0.8$
Or $\frac{n_{2}}{n_{1}+n_{2}}=0.2$ and $ \frac{n_{1}}{n_{1}+n_{2}}=0.8$
or $\frac{n_{2}}{n_{1}}=\frac{0.2}{0.8}=\frac{1}{4}$
Now, if $n_{1}$ (solvent moles) $=1000/78=12.8$ moles
$n_{2}=12.8/4=3.2$ moles. Therefore, 3.2 moles of the compound are present in one kg of solvent benzene and so molality $=3.2$