The reaction between gaseous `NH_(3)` and `HBr` produces a white solide `NH_(4)Br`. Suppose a small quantity of gaseous `NH_(3)` and gaseous `HBr` are introduced simultaneously into opposite ends of an open tube which is one metre long. Calculate the distance of white solid formed from the end which was used to introduce `NH_(3)`

Explanation

Let the distance of white solid from `NH_(3) " end" x = cm`
The distance of which solid from `HBr " end " = (100 - x) cm`.
Rates of diffusion shall be proportional to these distances.
`(r_(1))/(r_(2)) = (x)/((100 - x)) = sqrt((M_(HBr))/(M_(NH_(3))))`
Mol. mass of `HBr = 1 + 80 = 81`
Mol. mass of `NH_(3) = 14 + 3 = 17`
So, `(x)/((100 - x)) = sqrt((81)/(17))`
or, `(x)/((100 - x)) = 2.18`
So, `x = 100 xx 2.18 - 2.18 x`
So, `x = (100 xx 2.18)/(3.18) = 68.55 cm`