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  4. The vapour pressure of pure liquid solvent is 0.80 atm. When a non-volatile substance B is added to the solvent, its vapour pressure drops to 0.6 atm. What is the mole fraction of component B in this solution?

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Q. The vapour pressure of pure liquid solvent is $0.80\, atm$. When a non-volatile substance $B$ is added to the solvent, its vapour pressure drops to $0.6\, atm$. What is the mole fraction of component $B$ in this solution?

J & K CETJ & K CET 2010Solutions

Solution:

According to Raoult's law, Relative lowering of vapour pressure $\propto$ mole fraction of solute or
$\frac{p^{o}-p_{s}}{p^{0B}}=\chi_{B}$
or $\frac{0.80-0.60}{0.80}=\chi_{B}$
$\therefore \chi_{B}=\frac{1}{4}=0.25$