JEE MAIN - Chemistry (2021 - 20th July Morning Shift - No. 21)
At 20$$^\circ$$ C, the vapour pressure of benzene is 70 torr and that of methyl benzene is 20 torr. The mole fraction of benzene in the vapour phase at 20$$^\circ$$ above an equimolar mixture of benzene and methyl benzene is _____________ $$\times$$ 10$$-$$2. (Nearest integer)
Answer
78
Explanation
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,
nA = number of moles of methyl benzene, nB
Mole fraction of benzene in vapour phase, yA = $${{{p_A}} \over {{p_T}}}$$ .... (i)
where pA is pressure of benzene in mixture.
$${p_A} = \mathop {p_A^o{\chi _A}}\limits_{Mole\,fraction} = p_A^o \times {{{n_A}} \over {{n_A} + {n_B}}} = p_A^o \times {{{n \over A}} \over {2{n \over A}}} = {{p_A^o} \over 2}$$
pT 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),
$${y_A} = {{{{70} \over 2}} \over {{{(70 + 20)} \over 2}}} = {{70} \over {90}}$$
yA = 0.78 = 78 $$\times$$ 10$$-$$2
Vapour pressure of pure methyl benzene, $$p_B^o$$ = 20 Torr
This mixture is equimolar, so number of moles of benzene,
nA = number of moles of methyl benzene, nB
Mole fraction of benzene in vapour phase, yA = $${{{p_A}} \over {{p_T}}}$$ .... (i)
where pA is pressure of benzene in mixture.
$${p_A} = \mathop {p_A^o{\chi _A}}\limits_{Mole\,fraction} = p_A^o \times {{{n_A}} \over {{n_A} + {n_B}}} = p_A^o \times {{{n \over A}} \over {2{n \over A}}} = {{p_A^o} \over 2}$$
pT 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),
$${y_A} = {{{{70} \over 2}} \over {{{(70 + 20)} \over 2}}} = {{70} \over {90}}$$
yA = 0.78 = 78 $$\times$$ 10$$-$$2
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