QJEEM21Chac00: JEE Main 2021 Chemistry - Class 12 JEEM21Chac00 Solution: At 20 C, the vapour pressure of benzene is 70 torr and that of…

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 = ${{{p_A}} \over {{p_T}}}$ .... (i)

where p_A is pressure of benzene in mixture.

${p_A} = \mathop {p_A^o{\chi _A}}\limits_{Mole\,fraction} = p_A^o \times {{{n_A}} \over {{n_A} + {n_B}}} = p_A^o \times {{{n \over A}} \over {2{n \over A}}} = {{p_A^o} \over 2}$

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)

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

Putting in above equation (i),

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

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