A stream of air is bubbled slowly through liquid benzene in a flask at `20.0^(@)C` against an ambient pressure of `100.56kPa`. After the passage of `4.80L` of air, measured at `20.0^(@)C` and `100.56kPa` before it contains benzonb vapor, it is found that `1.705g` of benzene have been evaporated. Assuming that the air saturated benzene vapor when it leaves the flask. Calculate the equilibrium vapor pressure of the benzene at `20.0^(@)C`.

Answer

`11.08kPa`

Explanation

Volume of Benzene `=4.80 lt`
no of moles of Benzene `=(1.705)/(78)`
Temperature `=20^(@)C=293K`
`(nRT)/(V)=0.109` atm
`=11.08 kPa`.