Question

F2(g) (M=38.00 g/mol) effuses through a pinhole at a rate of 0.01287 mol/s. At what rate does N2O4(g) (M=92.02 g/mol) effuse at the same pinhole?

Answer 1

$\frac{v_A}{v_B}=\frac{\sqrt{M_B}}{\sqrt{M_A}}$

$v_{N_2O_4}=v_{F_2}\sqrt{M_{F_2}M_{N_2O_4}}$

$v_{N_2O_4}=0.01287 \sqrt{38x92.02}=0.761 mole/s$