Vapour pressure of C 6 H 6 and C 7 H 8 mixture at 500 C is given by P ( mm Hg ) = 180 X B + 90 , where X B is the mole fraction of C 6 H 6 . A solution is prepared by mixing 936g benze

Question

Vapour pressure of C 6 ​ H 6 ​ and C 7 ​ H 8 ​ mixture at 500 C is given by P ( mm Hg ) = 180 X B ​ + 90 , where X B ​ is the mole fraction of C 6 ​ H 6 ​ . A solution is prepared by mixing 936g benzene and 736g toluene and if the vapours over this solution are removed and condensed into liquid and again brought to the temperature of 50°C, what would be mole fraction of C 6 ​ H 6 ​ in the vapour state?

TestHub · Accepted Answer

Initial Mole Fractions: Moles of benzene ( $n_{B} = \frac{936}{78} = 12$ ) and toluene ( $n_{T} = \frac{736}{92} = 8$ ) give an initial liquid mole fraction $X_{B} = \frac{12}{20} = 0.6$ .

Vapor Pressures: Comparing $P = 180 X_{B} + 90$ with Raoult's law gives pure vapor pressures $P_{T}^{\circ} = 90 mm H g$ and $P_{B}^{\circ} = 270 mm H g$ .

First Vapor Phase: At $X_{B} = 0.6$ , the total pressure $P_{1} = 180 (0.6) + 90 = 198 mm H g$ , yielding a first vapor mole fraction $Y_{B} = \frac{270 \times 0.6}{198} = \frac{9}{11}$ .

Condensed Liquid Phase: The condensed liquid has new mole fractions $X_{B}^{'} = \frac{9}{11}$ and $X_{T}^{'} = \frac{2}{11}$ .

Final Vapor Phase: The second vapor mole fraction is $Y_{B}^{'} = \frac{P _{B}^{\circ} X _{B}^{'}}{P _{B}^{\circ} X _{B}^{'} + P _{T}^{\circ} X _{T}^{'}} = \frac{270 \times ( 9/11 )}{270 \times ( 9/11 ) + 90 \times ( 2/11 )} = \frac{2430}{2610} \approx 0.93$ .

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