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Question : 127 of 160
Marks:
+1,
-0
Solution:
Let, the volume of methane effused in
25 s be
xmL.
Now, from the Graham's law of diffusion,
Rate of effusion
∝...(i) Also, rate of effusion
=| Volumediffused |
| Timetaken |
...(ii) From
Eq.(i) and
(ii) we get,
⇒=√ ⇒=√⇒=√ ⇒x=300√2=424.26mL≃424mL
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