Benzene and toluene form an ideal solution ov | vedclass

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(B) Given: $p^{\circ}_{benzene} = 50 \ mm \ Hg$,$p^{\circ}_{toluene} = 40 \ mm \ Hg$. Number of moles of benzene $(n_b) = \frac{117 \ g}{78 \ g \ mol^

Vedclass · Accepted Answer

$0.21$

Vedclass · Answer

(B) Given: $p^{\circ}_{benzene} = 50 \ mm \ Hg$,$p^{\circ}_{toluene} = 40 \ mm \ Hg$. Number of moles of benzene $(n_b) = \frac{117 \ g}{78 \ g \ mol^{-1}} = 1.5 \ mol$. Number of moles of toluene $(n_t) = \frac{46 \ g}{92 \ g \ mol^{-1}} = 0.5 \ mol$. Total moles $= 1.5 + 0.5 = 2.0 \ mol$. Mole fraction of benzene $(\chi_b) = \frac{1.5}{2.0} = 0.75$. Mole fraction of toluene $(\chi_t) = \frac{0.5}{2.0} = 0.25$. Partial pressure of benzene $(p_b) = p^{\circ}_b 	imes \chi_b = 50 	imes 0.75 = 37.5 \ mm \ Hg$. Partial pressure of toluene $(p_t) = p^{\circ}_t 	imes \chi_t = 40 	imes 0.25 = 10 \ mm \ Hg$. Total vapour pressure $(p_T) = p_b + p_t = 37.5 + 10 = 47.5 \ mm \ Hg$. Mole fraction of toluene in vapour phase $(y_t) = \frac{p_t}{p_T} = \frac{10}{47.5} \approx 0.21$.

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