Benzene and toluene form an ideal solution ov | vedclass

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(B) Given: $P_{B}^{\circ} = 50 \ mmHg$,$P_{T}^{\circ} = 40 \ mmHg$ $M_{B} = 78 \ g \ mol^{-1}$,$M_{T} = 92 \ g \ mol^{-1}$ Moles of benzene $(n_{B})$

Vedclass · Accepted Answer

$0.21$

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(B) Given: $P_{B}^{\circ} = 50 \ mmHg$,$P_{T}^{\circ} = 40 \ mmHg$ $M_{B} = 78 \ g \ mol^{-1}$,$M_{T} = 92 \ g \ mol^{-1}$ Moles of benzene $(n_{B})$ $= \frac{117 \ g}{78 \ g \ mol^{-1}} = 1.5 \ mol$ Moles of toluene $(n_{T})$ $= \frac{46 \ g}{92 \ g \ mol^{-1}} = 0.5 \ mol$ Mole fraction of benzene $(X_{B})$ $= \frac{1.5}{1.5 + 0.5} = 0.75$ Mole fraction of toluene $(X_{T})$ $= \frac{0.5}{1.5 + 0.5} = 0.25$ Partial pressure of benzene $(P_{B})$ $= P_{B}^{\circ} 	imes X_{B} = 50 	imes 0.75 = 37.5 \ mmHg$ Partial pressure of toluene $(P_{T})$ $= P_{T}^{\circ} 	imes X_{T} = 40 	imes 0.25 = 10 \ mmHg$ Total vapour pressure $(P_{total})$ $= P_{B} + P_{T} = 37.5 + 10 = 47.5 \ mmHg$ Mole fraction of toluene in vapour phase $(Y_{T})$ $= \frac{P_{T}}{P_{total}} = \frac{10}{47.5} \approx 0.21$

Similar Questions

Which of the following does not show positive deviation from Raoult's law?

$1 \ mol$ of liquid $A$ and $2 \ mol$ of liquid $B$ make a solution having an observed vapour pressure of $42 \ torr$. The vapour pressures of pure $A$ and pure $B$ are $45 \ torr$ and $36 \ torr$ respectively. The described solution:

$A$ 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. $1 \ atm = 1.013 \ bar$)

Which of the following is not correct?

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