A solution has 1:4 mole ratio of pentane to bexane. The vapour pressure of the pure hydrocarbhonse at `20^(@)C` and 440 mm Hg is for pentane and 120 m

Question

A solution has 1:4 mole ratio of pentane to bexane. The vapour pressure of the pure hydrocarbhonse at `20^(@)C` and 440 mm Hg is for pentane and 120 m

A solution has 1:4 mole ratio of pentane to bexane. The vapour pressure of the pure hydrocarbhonse at `20^(@)C` and 440 mm Hg is for pentane and 120 mm Hg for bexane: The mole fraction of pentane in the vapour phase will be :
A. 0.2
B. 0.549
C. 0.786
D. 0.478

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - c
`(n_(C_(5)H_(12)))/(n_(C_(6)H_(14)))=1/4`
`X_(C_(5)H_(12))=1/(1+4)=1/5`
`X_(C_(6)H_(14))=4/(1+4)=4/5`
`P_(C_(5)H_(12))^(@)=440mmHg,P_(C_(6)H_(14))^(@)=120 mm Hg`
`P_(("total"))=P_(C_(5)H_(12))^(@)xxX_(C_(5)H_(12))+P_(C_(6)H_(14))^(@)xxX_(C_(6)H_(14))`
`=440xx1//5+120xx4//5`
`=88+96=184 mm Hg`
Mole fraction `C_(5)H_(12)` in vapour phase
`X_(C_(5)H_(12))=("Partoa V.P. of "C_(5)H_(12))/("Partoa V.P. of "C_(6)H_(14))=(88mm)/(184mm)=0.478`