The rate of diffusion of a gas having molecular weight just double of nitrogen gas is 56 mL s. The rate of diffusion of nitrogen will be

Options

A. `79.19 mL s^(-1)`
B. `112.0 mL s^(-1)`
C. `56 mL s^(-1) `
D. `90.0 mL s^(-1)`

Answer

A

Explanation

`(r_(N_2))/(r_("gas")) = sqrt((M_("gas"))/(M_(N_2))) = sqrt(2)implies r_(N_2) = 56 sqrt(2) = 79.19 mL s^(-1)`