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Standard XII

Chemistry

Non-Ideal Solutions

A mixture of ...

Question

A mixture of toluene and benzene boils at $100^{\circ} C$ . Assuming ideal behaviour, the mole fraction of toluene in the mixture is closest to:
[Vapour pressures of pure toluene and pure benzene at $100^{\circ} C$ are $0.742$ and $1.800$ bar, respectively. $1 a t m = 1.013 b a r]$ .

0.824

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0.744

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0.544

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0.624

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Solution

The correct option is A $0.744$
For ideal solution:

$P = p_{A}^{o} \times x_{A} + p_{B}^{o} \times x_{B}$

$1.013 = 0.742 \times x_{T o l u e n e} + 1.8 \times x_{B e n z e n e}$

$1.013 = 0.742 \times x_{T o l u e n e} + 1.8 \times (1 - x_{T o l u e n e})$

Solving we get,

$x_{T o l u e n e} = 0.744$

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A mixture of toluene and benzene boils at 100∘C. Assuming ideal behaviour, the mole fraction of toluene in the mixture is closest to:[Vapour pressures of pure toluene and pure benzene at 100∘C are 0.742 and 1.800 bar, respectively. 1 atm=1.013 bar].

The correct option is A 0.744For ideal solution: P=poA×xA+poB×xB1.013=0.742×xToluene+1.8×xBenzene1.013=0.742×xToluene+1.8×(1−xToluene)Solving we get,xToluene=0.744

A mixture of toluene and benzene boils at $100^{\circ} C$ . Assuming ideal behaviour, the mole fraction of toluene in the mixture is closest to:
[Vapour pressures of pure toluene and pure benzene at $100^{\circ} C$ are $0.742$ and $1.800$ bar, respectively. $1 a t m = 1.013 b a r]$ .