Q

# Confused! kindly explain, Which of the following statement 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°C. (Given: Vapour Pressur Data at 25

Which of the following statement 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°C. (Given: Vapour Pressur Data at 25°C, benzene=12.8 kPa, toluene=3.85 kPa)

• Option 1)

The vapour will contain a higher percentage of benzene

• Option 2)

The vapour will contain a higher percentage of toluene

• Option 3)

The vapour will contain equal amounts of benezene and toluene

• Option 4)

Not enough information is given to make a predication

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As learnt

Rault's Law -

The total vapour pressure of binary mixture of miscible liquids be having ideally is given by

$P_{T}= P_{A}^{0}x_{A}+P_{B}^{0}x_{B}$

Where $x_{A}$  and  $x_{B}$ are mole fraction of A and B in liquid phase.

- wherein

$P_{A}^{0}$  and $P_{B}^{0}$ are vapour pressures of pure liquids.

Vapour pressure due to benzene=$P_{A}x_{A}$

$x_{A}=$ mole fraction of benzene in solution

$P_{A}=$ vapour pressure of benzene= 12.8 kPa

Vapour pressure due to toluene=$P_{B}x_{B}$

$x_{B}=$ mole fraction of toluene

$P_{B}=$ vapour pressure of toluene = 3.85 kPa

Since, $x_{A}= x_{B}$

$\\ \therefore P_{A}x_{A} \: > \: P_{B}x_{B} \\ \\ as \: P_{A} \: > \: P_{B}$

Therefore, vapour will contain more benzene due to it's higher pressure.

Option 1)

The vapour will contain a higher percentage of benzene

This option is correct

Option 2)

The vapour will contain a higher percentage of toluene

This option is incorrect

Option 3)

The vapour will contain equal amounts of benezene and toluene

This option is incorrect

Option 4)

Not enough information is given to make a predication

This option is incorrect

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