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

Answer

B

Explanation

Vapour pressure does not depend upon the amount of water of size of the vessel. Hence, `P=3170 Pa. "Also",V=1 "dm"^(3)=10^(-3) m^(3) , T = 300 K. PV = nRT,`
`n = (PV)/(RT)=(3170 xx 10^(-3))/(8.314 xx 300)= 1.27 xx 10^(-3) "moles"`