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The vapour pressure of benzene is 75 mm and that of toluene is 22 m at 20^(@)C temperature. The solution prepared by mixing 78 gram benzene and 46 gram toulene then, what will be the partial pressure of benzene in the mixture ?

Answer»

25
50
100
75

Solution :`p_(1)^(0)= 75 mm, p_(2)^(0)=22mm`
`w_(1)=78 g,w_(2)=56 g`
`n_(1)=1` mole , `n_(2)=0.5` mole
`X_(1)=(1)/(1.5), X_(2)=(0.5)/(1.5)`
PARTIAL PRESSURE of benzene,
`P=P_(1)^(0)X_(1)=(75)/(1.5)=50` mm.


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