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 travels 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