The rate of diffusion of methane at a given temperature is twice that of X

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Concept:

• Diffusion is the intermixing of two or more gases irrespective of density relationship and without the help of an external agency.
• Gases when provided in an open space, escape from their vessel and mixes with the other gases present in the surroundings.
• This occurs due to diffusion, it is seen that gases with low molecular masses move and intermix more quickly than those of heavier ones.
• So, it can be said that the lighter gases diffuse more quickly than the heavier ones.
• The rate of diffusion is seen to be inversely proportional to the square roots of their densities under similar conditions of temperature and pressure. This is known as Graham's law of diffusion.
• Mathematically, we can say:

\(r = \sqrt{1 \over d}\), where r = rate and d = density

Calculation:

Given:

Gas is represented as 'X'.

The rate of diffusion of methane is twice that of 'X'.

• We know, that the Vapour density of gas and Molecular weight is related as:

Molecular mass = 2 × Vapour Density

Hence, rate =

\(r = \sqrt{1 \over m}\)

Molecular mass = CH4 = 12 + 4 = 16

Rate of diffusion of methane = 

\(r_{methane} = \sqrt{1 \over 16}\)

Rate of diffusion of gas 'x' with mass M =

\(r_{x} = \sqrt{1 \over M} \)

The ratio of rates of diffusion of Methane to 'X' is:

\({r_{methane}\over r_{x}} = \sqrt{M \over 16} = 2\)

or, \({M \over 16} = 4 ;M = 16\times 4=64g\)

Hence, the molecular weight of X is 64g.

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