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Question

The time taken for a certain volume of gas to diffuse through a small hole was 2 min. Under similar conditions an equal volume of oxygen took 4 min to pass. The molecular mass of the gas is:

A
32.0 g
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B
16.0 g
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C
8.0 g
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D
4.0 g
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Solution

The correct option is C 8.0 g
Graham's law of effusion,
r1M
Where r= rate of effusion
M= molar mass of gas
Also,
r=Volume of gas effused(V)Time taken(t)
r=Vt
rArB=MBMA
VAtAVBtB=MBMA
Where VA and VB are the volume of gas A and gas B.
tA and tB are time taken by gas for effusion.
According to question,
VA=VB=V
tA=2 min
tB=4 min
MB=32 g (Molar mass of O2)
V2V4=32MA
42=32MA
MA=8 g/mol

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