Benzene and toluene form nearly ideal solutio | vedclass

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(A) According to Raoult's law,the partial vapour pressure of a component is given by $P_i = P_i^{\circ} 	imes X_i$. First,calculate the number of mol

Vedclass · Accepted Answer

$50$

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(A) According to Raoult's law,the partial vapour pressure of a component is given by $P_i = P_i^{\circ} 	imes X_i$. First,calculate the number of moles of benzene $(C_6H_6)$ and toluene $(C_7H_8)$: $n_{	ext{benzene}} = \frac{78\,g}{78\,g/mol} = 1\,mol$. $n_{	ext{toluene}} = \frac{46\,g}{92\,g/mol} = 0.5\,mol$. Next,calculate the mole fraction of benzene $(X_B)$: $X_B = \frac{n_{	ext{benzene}}}{n_{	ext{benzene}} + n_{	ext{toluene}}} = \frac{1}{1 + 0.5} = \frac{1}{1.5} = \frac{2}{3}$. Now,calculate the partial vapour pressure of benzene $(P_B)$: $P_B = P_B^{\circ} 	imes X_B = 75\,torr 	imes \frac{2}{3} = 50\,torr$.

Benzene and toluene form nearly ideal solutions. At $20\,^{\circ}C$,the vapour pressure of benzene is $75\,torr$ and that of toluene is $22\,torr$. The partial vapour pressure of benzene at $20\,^{\circ}C$ for a solution containing $78\,g$ of benzene and $46\,g$ of toluene in $torr$ is

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