Question

The vapour density of a gaseous mixture of non-reactive gases 'A' and 'B' is 40. If the molar mass of gas
'A' is 20 gm/mol and the mixture contains the gases in 2 : 3 volume ratio, then Calculate molar mass of
gas 'B' :

Answer 1

The molar mass of gas B is 120grams.

Given:

The vapour density of a gaseous mixture of non-reactive gases 'A' and 'B' is 40. If the molar mass of gas 'A' is 20 gm/mol and the mixture contains the gases in a 2 : 3 volume ratio.

To Find:

The molar mass of gas B.

Solution:

To find the molar mass of gas B we will follow the following steps:

As we know,

Molecular weight is directly proportional to twice vapour density.

Also, molar volume is the volume of 1 mole of a substance and is directly proportional to molar mass.

So,

The molecular weight of the mixture = 2 × 40 = 80 grams.

Also,

The average molecular mass of mixture = mass of A in mixture + mass of B in the mixture. ------> equation 1

The mass of A in the mixture =

$\frac{2}{3} \times molecular \: weight \: of \: A = \frac{2}{5} \times 20 = 8 \: grams$

Similarly,

The mass of B in the mixture =

$\frac{3}{5} \times \: molecular \: mass \: of \: B$

Now,

Putting values in the equation first we get,

$80 = 8 + \frac{3}{5} \times molecular \: mass \: of \: B$

Now,

$80 - 8 = \frac{3}{5} \times molecular \: mass \: of \: \: B$

$molecular \: mass \: of \: \: B = \frac{5}{3} \times 72 = 24 \times 5 = 120grams$

Henceforth, the molar mass of gas B is 120grams.

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