A small quantity of gaseous `NH_(3)" and "HBr` are introduced simultaneously into the opposite ends of an open tube that is 1 m long. Calculate the distance of the white solid `NH_(4)Br` formed from the end that was used to introduce `NH_(3)`.

Options

A. `68.55 cm`
B. `63.5 cm`
C. `60.09 cm`
D. `65.24 cm`

Answer

A

Explanation

By Graham.s law of diffusion `(r_(NH_3))/(r_(HBr))=sqrt((M_(HBr))/(M_(NH_3)))= sqrt((81)/(17))= 2.18`
Thus, `NH_(3)` travels `2.18` times faster than HBr. In other words, `NH_(3)` will travel `2.18 cm` in the same time in which HBr travel 1 cm. Given that the length of the tube = 100 cm.
Therefore, distance travelled in the tube by `NH_(3)= (2.18)/(2.18 +1)xx 100= 68.55 cm`.