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 11^(-3) " mol"`
C. `1.53 xx 10^(-2) " mol"`
D. `4.46 xx 10^(-2) " mol"`

Answer

A

Explanation

volume occupied by water molecules in the vapour phase
`( 1- 10^(-4)) ~~ 1 dm^(3)`
`PV=nRT`
`.^(n)H_(2)O = (PV)/(RT) =((3170 Nm^(-2))xx (10^(-3) m^(3)))/((8.314 NmK^(-1) " mol"^(-1))xx(300 K))`
`= 1.27 xx 10^(-3) "mol"`