Question

Helium diffuses twice as fast as another gas B. If the vapour density of helium is 2, the molecular weight of B is

(a) 4

(b) 8

(c) 16

(d) 24

Answer 1

Correct option is (c) 16

According to Graham's diffusion law,

At constant pressure and temperature, the rate of diffusion is inversely proportional to the square root of the molar mass of the gas.

\(r \propto \frac 1{\sqrt M}\)

Given, r_B = r, r_He = 2r

Vapour density of He = 2

And we know that Vapour density = Realtive molecular mass / 2

So, the relative molecular mass of He = Vapour density × 2 = 2 × 2 = 4

And for two gases, the ratio of rates of diffusions are

\(\frac{r_{He}}{r_B} = \sqrt{\frac{M_B}{M_{He}}}\)

Then, \(\frac{2r}r = \sqrt{\frac{M_B}4} = 16\)

From the above calculation, it is clear that the molecular weight of B is 16.