If `10^(-4) dm^(3)` of water is introduced into a `1.0 dm^(3)` flask to `300K` how many moles of water are in the vapour phase when equilibrium is established ? (Given vapour pressure of `H_(2)O` at `300K` is `3170Pa R =8.314 JK^(-1) mol^(-1))` .

Options

A. `1.27 xx 10^(-3) mol`
B. `5.56 xx 10^(-3) mol`
C. `1.53 xx 10^(-2) mol`
D. `4.46 xx 10^(-2) mol`

Answer

A

Explanation

`n = (PV)/(RT) = (3170 xx 10^(-3))/(8.314 xx 300) = 1.27 xx 10^(-3)` mol
Here `V = 1 dm^(-3) = 10^(-3) m^(3)`