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.)

1 Answer
May 9, 2017

Answer:

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.
.