1.

A solutions 16 g of methanol and 90 g of water. The mole fraction of methanol in the solution is :

Answer»

`0.200`
`0.549`
`0.786`
`0.478`

Solution :p (pentane) `=440` mm HG
p (HEXANE)`=120` mm Hg
`X` (pentane)`=(1)/(1+4)=0.2`
`x` (hexane)`=(4)/(1+4)=0.8`
p (pentane)`=0.2xx440=88`mm
p (hexane)`=0.8xx120=96` mm
total vapour pressure `=88+96=184` mm
MOLE fraction of pentane in vapour PHASE
y (pentane) `=(88)/(184)=0.478`


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