Question

a vessel contains equal volume of so2 and CH4 through a small hole the gases effused into vaccum. after 200second the total volume is reduced to half. What is the ratio of SO2 and CH4 remaining in the vessel ? a) 1 : 2 b)2 : 1 c)1 :1 d) 1: 4

Answer 1

c) this is dancer because the vessel contains equal amount of s O2 and ch4 then the ratio will be 1:1

Answer 2

Answer:

2:1

Explanation:

rate of diffusion $r = 1/\sqrt{M}$( i.e., rate of diffusion is inversely proportional to molar mass )

$\frac{r_{CH4} }{r_{SO2} } = \frac{\sqrt{M_{SO2} } }{\sqrt{M_{CH4} } } }$

= $\frac{\sqrt{64} }{\sqrt{16} }$  = $\frac{8}{4}$  = $\frac{2}{1}$

therefore the ratio of so2 and ch4 = 2:1