Small amounts of NH3 and HCl(g) are released simultaneously at the opposite ends of a 2 metres long tube. At what point the formation of NH4Cl would start? (Either end of the tube can be used to tell the answer.)

May 9, 2017

The formation of $\text{NH"_4"Cl}$ would start at 0.8 m from the $\text{HCl}$ end of the tube.

Explanation:

Graham's law of diffusion states that the rate of diffusion of a gas is inversely proportional to the square root of its molecular mass.

We can write the law as

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} {r}_{1} / {r}_{2} = \sqrt{{M}_{2} / {M}_{1}} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

where:

${r}_{1}$ and ${r}_{2}$ are the rates of effusion of two gases and
${M}_{1}$ and ${M}_{2}$ are their molecular masses.

If Gas 1 is ${\text{NH}}_{3}$ and gas 2 is $\text{HCl}$, the formula becomes

${r}_{1} / {r}_{2} = \sqrt{\left(36.46 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{u"))))/(17.03 color(red)(cancel(color(black)("u}}}}\right)} = \sqrt{2.141} = 1.463$

Thus, the ${\text{NH}}_{3}$ travels 1.463 times at fast as $\text{HCl}$.

A white ring of $\text{NH"_4"Cl}$ will form when the two gases meet.

Let $x =$ the distance travelled by the $\text{HCl}$.

Then $1.463 x =$ the distance travelled by the ${\text{NH}}_{3}$.

$x + 1.463 x = \text{2 m}$

$2.463 x = \text{2 m}$

$x = \text{2 m"/2.463 = "0.8 m}$

So, the white ring will be formed at a distance of 0.8 m from the $\text{HCl}$ end.
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