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  • Solve it, Sulphur dioxide and oxygen were allowed to diffuse through a porous partition.20 dm3 of SO2 diffuses through the porous partition in 60 seconds. The volume of O2 in dm3 which diffuses under the similar condition in 30 seconds will be (atomi

 Sulphur dioxide and oxygen were allowed to diffuse through a porous partition.20 dm3 of SO2 diffuses through the porous partition in 60 seconds. The volume of O2 in dm3 which diffuses under the similar condition in 30 seconds will be (atomic mass of sulphur = 32 u) :

  • Option 1)

     7.09

  • Option 2)

     14.1

  • Option 3)

     10.0

  • Option 4)

    28.2

 

Answers (1)

best_answer

As we have learned

Graham’s law of diffusion -

\dpi{100} r_{1}/r_{2}=\sqrt{M_{1}/M_{2}}  where r is rate of diffusion of gas , M is molar mass.

-

 

 Rate of diffusion \propto \frac{1}{M^{1/2}}

\frac{r_1}{r_2}= \frac{\sqrt {M_2}}{\sqrt {M_1}}

\frac{20/60}{V/30}= \frac{\sqrt {32}}{\sqrt {64}}

\Rightarrow V = 10 \sqrt 2 = 14.14

 

 

 


Option 1)

 7.09

Option 2)

 14.1

Option 3)

 10.0

Option 4)

28.2

Posted by

SudhirSol

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