The reaction between gaseous $N H_{3}$ and $H B r$ produces a white solid $N H_{4} B r$ . Suppose a small quantity of gaseous $N H_{3}$ and gaseous $H B r$ 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 $N H_{3}$ .

Open in App

Solution

Let the distance of white solid formation from $N H_{3}$ end $= x cm$ .
The distance of white solid formation from $H B r$ end $= (100 - x) cm$ .
Rate of diffusion shall be proportional to these distances.
$\frac{r_{N H_{3}}}{r_{H B r}} = \frac{x}{(100 - x)} = \sqrt{\frac{M_{H B r}}{M_{N H_{3}}}}$
Mol. mass of $H B r = 1 + 80 = 81$
Mol. mass of $N H_{3} = 14 + 3 = 17$
So, $\frac{x}{(100 - x)} = \sqrt{\frac{81}{17}}$
or, $\frac{x}{(100 - x)} = 2.18$
So, $x = 100 \times 2.18 - 2.18 x$
or, $3.18 x = 100 \times 2.18$
So, $x = \frac{100 \times 2.18}{3.18}$
$= 68.55 cm$

Suggest Corrections

Similar questions

Q. The reaction between gaseous

{N H}_{3}

and

H B r

produces a white solid

{N H}_{4} B r

. Suppose that

{N H}_{3}

and

H B r

are introduced simultaneously into the opposite ends of an open tube of

1

metre long. Where would you expect the white solid to form?

A small quantity of gaseous $N H_{3}$ and HBr are introduced simultaneously into the opposite ends of an open tube that is 1m long. Calculate the distance of the white solid $N H_{4} B r$ formed from the end that was used to introduce $N H_{3}$ .

Q. Two gases NH₃ and HBr are introduced simultaneously into opposite ends of an open tube 1m long. The distance at which a ring of NH₄Br is formed from the NH₃ end is

N H_{3}

and

H C l

gas are introduced simultaneously from the two ends of a long tube. A white ring of

N H_{4} C l

appears first :

Q. When HCl fumes and

N H_{3}

fumes are introduced into opposite ends of a long, dry glass tube, a white ring forms in the tube. Which of the following statement explains this phenomenon best?

Join BYJU'S Learning Program

Explore more

Graham's Law of Effusion

Standard XII Chemistry

Join BYJU'S Learning Program