Question

A solution has a 1:4 mole ratio of pentane to hexane. the vapour pressures of the pure hydrocarbon

Answer 1

your question is incomplete. A complete question is -----> A solution has 1:4 mole ratio of pentane to hexane the vappour pressure of pure hydrocarbon at 20degree are 400mm hg for pentane and 120 mm hg for hexane the mole fraction of pentane in the vapour pressure would be

solution , mole fraction of pentane , $X_p$ = 1/(1 + 4) = 1/5 = 0.2

mole fraction of hexane , $X_h$ = 4/(1 + 4) = 4/5 = 0.8

vapor pressure of pure pentane at 20°C , $P^{\circ}_p$ = 400mm Hg

vapor pressure of pure hexane at 20°C , $P^{\circ}_h$ = 120mm Hg

now, total pressure of solution, $P_T=X_pP^{\circ}_p+X_hP^{\circ}_h$

= 0.2 × 400 + 0.8 × 120

= 80 + 96

= 176 mm Hg

now mole fraction of pentane in vapor phase, $Y_p=X_p\frac{P^{\circ}_p}{P_T}$

= 0.2 × 400/176 = 80/176 =0.45

Answer 2

Answer:

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