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The volume of 0.0168 mol of O2 obtained ...

The volume of 0.0168 mol of `O_2` obtained by decomposition of `KClO_3` and collected by displacement of water is 428 mL at a pressure of 754 mm Hg at `25^(@)C`. The pressure of water vapour at `25^(@)C` is

A

18 mm Hg

B

20 mm Hg

C

22 mm Hg

D

24 mm Hg.

Text Solution

Verified by Experts

The correct Answer is:
D
Applying , PV=nRT for dry gas
`P=? , V=428 mL = 0.428 L , n =0.0168 `mol
`R= 0.0821 L " atm " K^(-1) " mol"^(-1) , T = 25^@ C = (273 + 25 ) K = 298 K`
`P=(0.0168 xx 0.0821 xx 298)/(0.428) = 0.96 atm `
`=(0.96 xx 760) mm Hg ~~ Hg `
` P_("moist gas ") = P_("dry gas ") + P_(" water vapours")`
`P_("water vapour") =754-730=24 mm Hg`
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