Question

A bottle of dry Ammonia$\left({\mathrm{NH}}_3\right)$and a bottle of dry Hydrogen chloride $\left(\mathrm{HCl}\right)$connected through a long tube are opened simultaneously at both ends. The white ammonium chloride (${\mathrm{NH}}_4\mathrm{Cl}$) ring first formed will be:

• A. At the center of the tube
• B. Near the hydrogen chloride bottle
• C. Near the ammonia bottle
• D. Throughout the length of the tube

Answer 1

The correct option is B

Near the hydrogen chloride bottle

Explanation for correct option

Option (B) Near the hydrogen chloride bottle

• ${\mathrm{NH}}_3$ is lighter than $\mathrm{HCl}$ thus it diffuses faster.
• Hence, a white ring of ${\mathrm{NH}}_4\mathrm{Cl}$ is formed in the tube near the $\mathrm{HCl}$.

Explanation for incorrect option

Option a) At the centre of the tube.

• Since the rate of diffusion of both hydrochloric acid and ammonia is different
• Hence the ring will not form at the centre .This option is incorrect

Option c) Near the Ammonia bottle :

• we know that the rate of diffusion of ammonia is faster so it will be easily diffused through another end of the tube
• Hence the ring will not form near the ammonia bottle.Option is incorrect

Option d) Throughout the length of the tube.

• This option is incorrect since ammonia is diffusing at a faster rate
• Thus it will reach the other bottle soon and react to give the product.

Hence The white ammonium chloride ${\mathrm{NH}}_4\mathrm{Cl}$ ring will be first formed Near the Hydrogen chloride bottle.