The vapour pressure of pure benzene at `88^(@)C` is `957 mm` and that of toluene at the same temperature is `379.5 mm`. The composition of benzene-toluene mixture boiling at `88^(@)C` will be

Options

A. `chi_(benz)=0.341`
B. `chi_("benzene")=0.786`
C. `chi_(benz)=0.214`
D. `chi_("benzene")=0.659`

Answer

4

Explanation

Normal boiling point of solution is that temperature at which the vapor pressure of the solution becomes equal to 1 atmospheric pressure `(i. e., 760 mmHg)`. According to Daitons' law
`P_("total")=P_("benzene")+P_("toluene")`
Applying Raoult's law, we have
`P_("total")=P_("benzene")^(0) chi_("benzene")+P_("toluene")^(0) chi_("toluene")`
`=P_("toluene")^(0)+(P_("benzene")^(0)-P_("toluene")^(0)) chi_("benzene")`
`760 mm=380 mmHg+(957-380) mmHg chi_(benz)`
`380 mmHg=577 mmHg chi_("benzene")`
`chi_("benzene")=0.659`