Question

A vessel contains an equal number of moles of $He$ and $C{H}_4$ gases. If half of the gaseous mixture effused out then ratio of number of moles of He and $C{H}_4$ remaining in the vessel is:

• A. 2:1
• B. 1:2
• C. 4:1
• D. 1:4

Answer 1

Answer: (B)

1:2

According to the Graham's law of diffusion, the ratio of the rates of the gases is equal to the reciprocal of the ratio of the square roots of their molecular weights.

$\frac{r_1}{r_2}=\sqrt{\frac{m_2}{m_1}}$

This is also equal to the ratio of the number of moles of two gases that diffuses in a given time.

The molecular wight of methane (16 g/mol) is four times the molecular weight of helium (4 g/mole).

Hence, in a given time, the number of the moles of helium that diffuse in a given time is one half the number of moles of methane that diffuse in the same time.

The ratio of the number of mole of Helium and methane remaining in the vessel is 1:2.