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Helium gas collected over water measures...

Helium gas collected over water measures 350 mL at `20^(@)`C. If atmospheric pressure is 752.5 mm of Hg and vapour pressure of water at `20^(@)C` be 17.5 mm of Hg, what is the percentage weight of water vapour in moist helium gas?

A

`98.76 %`

B

`9.67 %`

C

`5.32 %`

D

`13.83 %`

Text Solution

Verified by Experts

The correct Answer is:
B
Partial pressure of dry He(g) = 752.2 - 17.5 = 735 mm of Hg
Moles of He = `(pV)/(RT) = (735)/(760) xx (0.35)/(0.082 xx 293) = 0.0141` mol `implies ` Mass of He = 0.01414 `xx 4 = 0.05635 g`
Similarly , moles of `H_(2) O (v) = (17.5)/(760) xx (0.35)/(0.082 xx 293) = 3.35 xx 10^(-4)` mol
`implies` Mass of `H_(2) O(v) = 6.03 xx 10^(-3)` g
`implies` Total mass = 0.05635 + 6.037 `xx 10^(-3) = 62.387 xx 10^(-3)` g
Mass % of `H_(2) O (v) = (6.03 xx 10^(-3))/(62.387 xx 10^(-3)) xx 100 = 9.67 %`
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