Question

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.)

Answer 1

The formation of `"NH"_4"Cl"` would start at 0.8 m from the `"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

`color(blue)(bar(ul(|color(white)(a/a)r_1/r_2 =sqrt(M_2/M_1)color(white)(a/a)|)))" "`

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 `"NH"_3` and gas 2 is `"HCl"`, the formula becomes

`r_1/r_2 = sqrt((36.46 color(red)(cancel(color(black)("u"))))/(17.03 color(red)(cancel(color(black)("u"))))) = sqrt(2.141) = 1.463`

Thus, the `"NH"_3` travels 1.463 times at fast as `"HCl"`.

A white ring of `"NH"_4"Cl"` will form when the two gases meet.

Let `x =` the distance travelled by the `"HCl"`.

Then `1.463x =` the distance travelled by the `"NH"_3`.

`x + 1.463x = "2 m"`

`2.463x = "2 m"`

`x = "2 m"/2.463 = "0.8 m"`

So, the white ring will be formed at a distance of 0.8 m from the `"HCl"` end.
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