Question

Equal moles of benzene and toluene are mixed the V.P. of benzene and toluene in pure state are 700 and 600 mm Hg respectively. The mole fraction of benzene in vapour state is

Answer 1

Explanation:

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Answer 2

Answer:

The mole fraction of benzene equals to 0.54 in an equimolar solution of benzene and toluene in vapor phase.

Explanation:

Consider x are the number of moles of benzene and toluene .

So the mole fraction of benzene or toluene  $=\frac{x}{x+x}=\frac{x}{2x}=\frac{1}{2}$

In pure state, vapor pressure: $P^{o}_{b}=700mmHg$  and $P^{o}_{t}=600mmHg$

Total vapor pressure  $P_{total}=P^{o}_{b}*x_{b}+P^{o}_{t}*x_{t}$

$P_{total}=(700)(0.5)+(600)(0.5)\\P_{total}=350+300\\P_{total}=650mmHg$

Then mole fraction of benzene in vapor phase:

$Y(b)=\frac{(P^{o}_{b})(x_{b})}{P_{total}}$

      $=\frac{(700)(0.5)}{650}$

      $=0.538$ ≈ $0.54$

Therefore mole fraction of benzene comes out to be 0.54 in vapor state.