Question

The two gases, H2 and O2, were allowed to effuse under the same conditions through a pinhole. If the rate of effusion of O2 was 4.80 x 10-2 ms-1, what is the rate of effusion of H2?​

Answer 1

The rate of effusion of H₂ : 7.2 x 10⁻² m/s

Further explanation  

Graham's law: the rate of effusion of a gas is inversely proportional to the square root of its molar masses or  

the effusion rates of two gases = the square root of the inverse of their molar masses:  

$\rm \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }$

or  

$\rm M_1\times r_1^2=M_2\times r_2^2$

MW₁ O₂ = 32 g/mol

MW₂ H₂ = 2 g/mol

$\tt \dfrac{1.8\times 10^{-2}}{r_2}=\sqrt{\dfrac{2}{32} }\\\\\dfrac{(1.8\times 10^{-2})^2}{r_2^2}=\dfrac{2}{32}\\\\r_2=0.072=7.2\times 10^{-2}~m/s$