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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)"mole"`

B

`5.56xx10^(-3)"mole"`

C

`1.53xx10^(-2)"mole"`

D

`4.46xx10^(-2)"mole"`

Text Solution

Verified by Experts

The correct Answer is:
a
(a) `n=(PV)/(RT)=(3170xx10^(-3))/(8.314xx300)=1.27xx10^(-3)"mole"`
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