At 20° 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° C above an equimolar mixture of benzene and methyl benzene is × 10-2. (Nearest integer)

Question

At 20° 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° C above an equimolar mixture of benzene and methyl benzene is × 10-2. (Nearest integer) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

PB°=40 PT°=20 KB=0.5=KT Now yB=(KB PB°/KB P° B+KT P°) =(70 × 0.5/70 × 0.5+20 × 0.5)

Q. At $2 0^{\circ} C$ , 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 $2 0^{\circ} C$ above an equimolar mixture of benzene and methyl benzene is ________ $\times 1 0^{- 2}$ . (Nearest integer)

$P_{B}^{\circ} = 40$
$P_{T}^{\circ} = 20$
$K_{B} = 0.5 = K_{T}$
Now $y_{B} = \frac{K _{B} P _{B}^{\circ}}{K _{B} P ^{\circ} _{B} + K _{T} P ^{\circ}}$
$= \frac{70 \times 0.5}{70 \times 0.5 + 20 \times 0.5}$

Q. At 20∘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∘C above an equimolar mixture of benzene and methyl benzene is ________ ×10−2. (Nearest integer)

PB∘​=40 PT∘​=20 KB​=0.5=KT​ Now yB​=KB​P∘B​+KT​P∘KB​PB∘​​ =70×0.5+20×0.570×0.5​

Q. At $2 0^{\circ} C$ , 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 $2 0^{\circ} C$ above an equimolar mixture of benzene and methyl benzene is ________ $\times 1 0^{- 2}$ . (Nearest integer)

$P_{B}^{\circ} = 40$
$P_{T}^{\circ} = 20$
$K_{B} = 0.5 = K_{T}$
Now $y_{B} = \frac{K _{B} P _{B}^{\circ}}{K _{B} P ^{\circ} _{B} + K _{T} P ^{\circ}}$
$= \frac{70 \times 0.5}{70 \times 0.5 + 20 \times 0.5}$