JEE Main 2021 — Solutions Question with Solution

Vapour pressure of pure benzene, $$p_A^o$$ = 70 Torr

Vapour pressure of pure methyl benzene, $$p_B^o$$ = 20 Torr

This mixture is equimolar, so number of moles of benzene,

n_A = number of moles of methyl benzene, n_B

Mole fraction of benzene in vapour phase, y_A = $$\frac{p_A}{p_T}$$ .... (i)

where p_A is pressure of benzene in mixture.

$$p_A=\underset{Molefraction}{p_A^o{\chi }_A}=p_A^o\times \frac{n_A}{n_A+n_B}=p_A^o\times \frac{\frac{n}{A}}{2\frac{n}{A}}=\frac{p_A^o}{2}$$

p_T is total pressure,

$$p_T=p_A^o{\chi }_A+p_B^o{\chi }_B=p_A+p_B$$ (pressure of methyl benzene)

$$p_T=\frac{p_A^o}{2}+\frac{p_B^o}{2}=\frac{p_A^o+p_B^o}{2}$$

Putting in above equation (i),

$$y_A=\frac{\frac{70}{2}}{\frac{(70+20)}{2}}=\frac{70}{90}$$

y_A = 0.78 = 78 $$\times$$ 10^$$-$$2