Q17 / 180 · COMEDK 2013 · Chemistry

$\mathrm{NH}_{3}$ and $\mathrm{HCl}$ gas are introduced simultaneously from the two ends of a long tube. A white ring of $\mathrm{NH}_{4} \mathrm{Cl}$ appears first

  • Correct answer
    A
    nearer to the $\mathrm{HCl}$ end
  • B
    at the centre of the tube
  • C
    throughout the tube
  • D
    nearer to the $\mathrm{NH}_{3}$ end

Solution

Distance travelled by gas is directly proportional to rate of diffusion $$ \begin{aligned} &\therefore \quad r_{\mathrm{NH}_{3}}\left(l_{1}\right) \propto \frac{1}{\sqrt{M_{\mathrm{NH}_{3}}}} \propto \frac{1}{\sqrt{17}} \\ &\quad r_{\mathrm{HCl}}\left(l_{2}\right) \propto \frac{1}{\sqrt{M_{\mathrm{HCl}}}} \propto \frac{1}{\sqrt{365}} \\ &\because \quad r_{\mathrm{NH}_{3}}>r_{\mathrm{HCl}} \end{aligned} $$ $\therefore$ Length travelled by $\mathrm{NH}_{3}$ is greater than $\mathrm{HCl}$. Hence, white ring of $\mathrm{NH}_{4} \mathrm{Cl}$ is formed nearer to $\mathrm{HCl}$ end.

Hence, the correct answer is option: 'A'.

Relevant Questions