Question #120532
The rate of effusion of NH3 gas through a porous barrier is observed to be 1.15*10^-4 mol/h.

Under the same conditions, the rate of effusion of N2 gas would be _________
mol/h.
1
Expert's answer
2020-06-08T15:30:43-0400

As per Graham's law,

Rate of diffusion /effusion is inversely proportional to the molar mass of the gas.

And MNH3=17 g;MN2=28 gM_{NH_3}=17\ g; M_{N_2}=28\ g

rN2rNH3=MNH3MN2\frac{r_{N_2}}{r_{NH_3}}=\sqrt{\frac{M_{NH_3}}{M_{N_2}}}     \implies rN21.15×104=1728\frac{r_{N_2}}{1.15\times10^{-4}}=\sqrt{\frac{17}{28}}     rN2=8.97×105mol/h\implies r_{N_2}=8.97\times 10^{-5} mol/h


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