Answer:
[tex]1.167\times 10^{-3}\ \text{mol/h}[/tex]
Explanation:
[tex]R_1[/tex] = Rate of diffusion of [tex]CO_2[/tex] = [tex]9.31\times 10^{-4}\ \text{mol/h}[/tex]
[tex]R_2[/tex] = Rate of diffusion of [tex]N_2[/tex]
[tex]M_1[/tex] = Molar mass of [tex]CO_2[/tex] = 44.01 g/mol
[tex]M_2[/tex] = Molar mass of [tex]N_2[/tex] = 28.0134 g/mol
From Graham's law we have the relation
[tex]\dfrac{R_1}{R_2}=\sqrt{\dfrac{M_2}{M_1}}\\\Rightarrow R_2=\dfrac{R_1}{\sqrt{\dfrac{M_2}{M_1}}}\\\Rightarrow R_2=\dfrac{9.31\times 10^{-4}}{\sqrt{\dfrac{28.0134}{44.01}}}\\\Rightarrow R_2=1.167\times 10^{-3}\ \text{mol/h}[/tex]
The rate of effusion of the [tex]N_2[/tex] gas would be [tex]1.167\times 10^{-3}\ \text{mol/h}[/tex].
