A given volume of methane diffuses in 20 seconds. How long will it take the same volume of sulphur (IV) oxide to diffuse under the same conditions? [CH4 = 16; SO2 = 64]

A. 5 seconds

B. 20 seconds

C. 40 seconds

D. 60 seconds

E. 80 seconds

Correct Answer: Option C

**C. 40 seconds**

##### Explanation

Applying the Graham’s law defination; i.e,

R(SO2)R(CH4)=M(CH4)M(SO2)−−−−−−√=t(CH4)t(SO2)R(SO2)R(CH4)=M(CH4)M(SO2)=t(CH4)t(SO2)

Where t(SO22) = ?, M(SO22) = 64

t(Ch44) = 20 sec, M(Ch44) = 16

therefore, t(SO2)20t(SO2)20 = 6416−−√6416

t(SO2)20t(SO2)20 = 8484

t(SO22) = 8 x 204204

= 40 sec.

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