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If 10^(-4) dm^(3) of water is introduced...

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

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`

Text Solution

Verified by Experts

The correct Answer is:
A
`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)`
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