`80` mL of ammonia gas effuse out form a vessel in `8` mixture at `5^(@)C` and `500 mm` Hg pressure. How long will `40` mL of a certain gas with molecular mass 68 take to effuse out form the same vessel at the same temperature and pressure ?

Explanation

Volume of ammounia `(NH_(3)) = 80 mL`
volume of gas `X = 40 mL`
Rate of emission of `NH_(3)(r_(NH_(3))) = (80)/(8) = 10 mL "min"^(-1)`
Rate of diffusion of gas `X (r_(X)) = (40)/(t) mL sin^(-1)`
Molar mass of `NH_(3) (M_(NH_(3))) = 17`
According to Graham's Law of diffusion:
`(r_(NH_(3)))/(r_(X)) = sqrt((M_(X))/(M_(NH_(3)))) = sqrt((68)/(17)) = 2`
`(10 xx t)/(40) = 2` or `t = 8 "min"`