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

If `10^(-4) dm^(3)` of water is introduced into a `1.0 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))` .

A

`5.56xx10^(-3)mol`

B

`1.56xx10^(-2) mol`

C

`4.46xx10^(-2) mol`

D

`1.27xx10^(-3) mol`

Text Solution

Verified by Experts

The correct Answer is:
D
`PV=nRT rArr V=1dm^(3)=10^(-3)m^(3) rArr P=3170Pa`
`R=8.314JK^(-1)mol^(-1) rArr T=300K rArr 3170xx10^(-3)=nxx8.314xx300`
`n=(3170xx10^(-3))/(8.314xx300)=1.27xx10^(-3)mol`.
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