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 ?

Options

A. 25
B. 50
C. 100
D. 75

Answer

B

Explanation

`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.