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.)

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Answer 1 · 2017-05-09T01:04:58

The formation of ${NH}_{4} Cl$ $"NH"_4"Cl"$ would start at 0.8 m from the $HCl$ $"HCl"$ end of the tube.

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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