A mixture of toluene and benzene boils at 100 | vedclass

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(B) According to Raoult's law,the total pressure $P_{total}$ is given by $P_{total} = P_{toluene}^0 \chi_{toluene} + P_{benzene}^0 \chi_{benzene}$.Sin

Vedclass · Accepted Answer

$0.744$

Vedclass · Answer

(B) According to Raoult's law,the total pressure $P_{total}$ is given by $P_{total} = P_{toluene}^0 \chi_{toluene} + P_{benzene}^0 \chi_{benzene}$.Since the mixture boils at $100^{\circ} C$ at $1 \ atm$ pressure,$P_{total} = 1.013 \ bar$.Let $\chi_{toluene} = x$,then $\chi_{benzene} = 1 - x$.Substituting the values: $1.013 = 0.742x + 1.800(1 - x)$.$1.013 = 0.742x + 1.800 - 1.800x$.$1.013 - 1.800 = (0.742 - 1.800)x$.$-0.787 = -1.058x$.$x = \frac{0.787}{1.058} \approx 0.744$.

$A$ 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 $0.742 \ bar$ and $1.800 \ bar$,respectively. $1 \ atm = 1.013 \ bar$)

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