Question

The vapour pressure of pure benzene at $88C$ is $960$ mm and that of toluene at the same temperature is 380 mm of benzene. At what mole fraction of benzene. the mixture will boil at ${88}^{o}C$?

• A. 0.655
• B. 0.345
• C. 0.05
• D. 0.25

Answer 1

The correct option is A 0.655

Given,

Vapour presence of pure benzene

$P_b^o=960mm⟶\left(1\right)$

Vapour presence of pure toluene

$P_t^o=380mm⟶\left(2\right)$

Also, Temperature=${88}^{o}C$

It is given that this mixture boils at ${88}^{o}C$

$\therefore$ It means that at ${88}^{o}C$

Total pressure of the mixture will be equal to atmospheric pressure.

$\therefore$ Total pressure of mixture at ${88}^{o}C=760$ $mm$ $Hg$ or $1atm⟶\left(3\right)$

Now we know

According to Dalton's Law and Raolt's Law

$P_{Total}=p_1^o{x}_1+p_2^o{x}_2$

$p_1^o,p_2^o$ are pure vapour pressures

$x_1,x_2$ are mole fractions

From (1), (2), (3)

$760=p_b^o\left(x_b\right)+p_t^o\left(x_t\right)$

$760=960\left(x_b\right)+380(1-x_b)$ $[\because x_b+x_t=1]$

$\Rightarrow 760=960x_b+380-380x_b$

$\therefore x_b=0.655$

Mole fraction of Benzene=$0.655$