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(C) According to Raoult's Law,the partial pressure of a component in the vapour phase is given by $P_i = P^o_i X_i$. Since the mixture is $1:1$ molar,

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The vapour will contain a higher percentage of benzene.

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(C) According to Raoult's Law,the partial pressure of a component in the vapour phase is given by $P_i = P^o_i X_i$. Since the mixture is $1:1$ molar,the mole fractions in the liquid phase are equal $(X_{	ext{benzene}} = X_{	ext{toluene}} = 0.5)$. The total pressure $P = P^o_{	ext{benzene}} X_{	ext{benzene}} + P^o_{	ext{toluene}} X_{	ext{toluene}}$. The mole fraction in the vapour phase is $Y_i = \frac{P_i}{P} = \frac{P^o_i X_i}{P}$. Since $X_{	ext{benzene}} = X_{	ext{toluene}}$,the component with the higher pure vapour pressure $(P^o)$ will have a higher mole fraction in the vapour phase. Given $P^o_{	ext{benzene}} = 12.8\ kPa$ and $P^o_{	ext{toluene}} = 3.85\ kPa$,benzene is more volatile. Therefore,the vapour will contain a higher percentage of benzene.

Which of the following statements about the composition of the vapour over an ideal $1:1$ molar mixture of benzene and toluene is correct? Assume that the temperature is constant at $25\ ^oC$. (Given,vapour pressure data at $25\ ^oC$,benzene $= 12.8\ kPa,$ toluene $= 3.85\ kPa$)

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Heptane and octane form an ideal solution. At $373 \, K$,the vapour pressures of the two liquid components are $105.2 \, kPa$ and $46.8 \, kPa$ respectively. What will be the vapour pressure of a mixture of $26.0 \, g$ of heptane and $35 \, g$ of octane (in $, kPa$)?

An aqueous solution of hydrochloric acid:

Which of the following form an ideal solution?
$I$. Chloroethane and bromoethane
$II$. Benzene and toluene
$III$. $n$-hexane and $n$-heptane
$IV$. Phenol and aniline

What is vapour pressure? And what is an ideal solution?

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