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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

`1.27xx10^(-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:
C
PV = nRT
`3170 xx 10^(-3) = n xx 8.314 xx 300`
`n = (3170 xx 10^(-3))/(8.314xx300), V = 1 Lt =1 "dm"^(2)= 10^(-3) m^(3)`
`n = 1.27 xx 10^(-3)` mol
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