Relative humidity of air is `60^(@)C` and the saturation vapour pressure of water vapour in air is `3.6kPa`. The amount of water vapours present in `2 L` air at `300 K` is

Options

A. `52 g`
B. `31.2 g`
C. `26 g`
D. `5.2 g`

Explanation

`PV=(w)/(m)RT` (for vapours of `H_(2)O`)
`P=3.6xx10^(3)Pa`, `V=2xx10^(-3)m^(3)`, `T=300 K`
`:. W_(H_(2)O)=(3.6xx10^(3)xx18xx2xx10^(-3))/(8.314xx300=0.052)`
`w_(H_(2)O)=52 g`
Since, realtive humidity `=60%`, therefore, amount of `H_(2)O=52xx0.6=31.2 g`