At 20^{circ} C,the vapour pressure of benzene | vedclass

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(D) Given: $P_{B}^{\circ} = 70 \, torr$,$P_{M}^{\circ} = 20 \, torr$. Since the mixture is equimolar,the mole fractions in the liquid phase are $X_{B}

Vedclass · Accepted Answer

$78$

Vedclass · Answer

(D) Given: $P_{B}^{\circ} = 70 \, torr$,$P_{M}^{\circ} = 20 \, torr$. Since the mixture is equimolar,the mole fractions in the liquid phase are $X_{B} = 0.5$ and $X_{M} = 0.5$. The partial pressure of benzene is $P_{B} = X_{B} 	imes P_{B}^{\circ} = 0.5 	imes 70 = 35 \, torr$. The partial pressure of methyl benzene is $P_{M} = X_{M} 	imes P_{M}^{\circ} = 0.5 	imes 20 = 10 \, torr$. The total pressure is $P_{total} = P_{B} + P_{M} = 35 + 10 = 45 \, torr$. The mole fraction of benzene in the vapor phase $(y_{B})$ is given by $y_{B} = \frac{P_{B}}{P_{total}} = \frac{35}{45} = 0.777...$. Rounding to the nearest integer for $y_{B} 	imes 10^{-2}$,we get $78 	imes 10^{-2}$.

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 vapor phase at $20^{\circ} C$ above an equimolar mixture of benzene and methyl benzene is $..... \times 10^{-2}$ (Nearest integer).

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