Question

The rate of diffusion of a gas having molecular weight just double of nitrogen gas is $56mL{s}^{-1}$. The rate of diffusion of nitrogen will be :

• A. $79.19mL{s}^{-1}$
• B. $112.0mL{s}^{-1}$
• C. $56mL{s}^{-1}$
• D. $90.0mL{s}^{-1}$

Answer 1

The correct option is A $79.19mL{s}^{-1}$

The rate of diffusion of a gas is inversely proportional to the square root of its molar mass.
$\frac{r_X}{r_{N_2}}=\sqrt{\frac{M_{N_2}}{M_X}}$
The molecular weight of the gas is double of nitrogen gas :

$M_X=2M_{N_2}$
Hence,

$\frac{56mL/s}{r_{N_2}}=\sqrt{\frac{M_{N_2}}{2M_{N_2}}}$

$r_{N_2}=79.19mL/s$

Therefore, the correct option is A.