Ammonia gas diffuses through a fine hole at the rate `0.5 lit min^-1`. Under the same conditions find the rate of diffusion of chlorine gas.

Explanation

Rate of diffusion of ammonia `(r_1) = 0.5 lit mi n^(-1)`
Molecular weight of ammonia`(M_1) = 17`
Rate of diffusion of chlorine ` (r_2) = ?`
Molecular weight of chlorine ` (M_2) = 71`
According to Graham.s law of diffusion
` ( r_1)/(r_2) = sqrt((M_2)/(M_1))`
` i.e., ( 0.5 )/(r_2) = sqrt((71 )/( 17 )) = 2.04`
` therefore ` r_2 = ( 0.5 ) /(2.04 ) = 0.245 lit mi n^(-1)`
` therefore ` Rate of diffusion of `Cl_2` = 0.245 lit/min