Question

A gaseous mixture of ${{\text{O}}_{\text{2}}}$ and X containing 20% of X, diffuses through a small hole in 234 s while pure ${{\text{O}}_{\text{2}}}$ take 224 s to diffuse through the same hole. The molecular mass of the mixture is:
(a) 34.9
(b) 46.6
(c) 32
(d) 44

Answer 1

Hint: In order to solve the question, we need to know the relationship between rate of diffusion and molecular mass. Using the formula, we can solve the question.

Complete Complete step-by-step answer:
We know that the rate of gas diffuse is inversely proportional to the density of the gas.
From the properties of gases, we are also aware that the rate of diffusion is inversely proportional to square root of molecular mass. When same quantities of two gases diffuses,
$\dfrac{{{\text{t}}_{\text{mixture}}}}{{{\text{t}}_{O_2}}}\text{=}\sqrt{\dfrac{{{\text{M}}_{\text{mixture}}}}{{{\text{M}}_{O_2}}}}$
$\dfrac{\text{234}}{\text{224}}\text{=}\sqrt{\dfrac{{{\text{M}}_{\text{mixture}}}}{\text{32}}}$
$\text{1}\text{.0446=}\sqrt{\dfrac{{{\text{M}}_{\text{mixture}}}}{\text{32}}}$
$\text{1}\text{.091=}\dfrac{{{\text{M}}_{\text{mixture}}}}{\text{32}}$
${{\text{M}}_{\text{mixture}}}\text{=34}\text{.9 g/mol}$
Therefore, when a correct option is Option A. \[\text{34}\text{.9 g/mol}\]
gaseous mixture of ${{\text{O}}_{\text{2}}}$ and X containing 20% of X diffuses through a small hole in 234 seconds, while ${{\text{O}}_{\text{2}}}$ takes 224 seconds to diffuse through that same hole, the molecular mass of the mixture is \[\text{34}\text{.9 g/mol}\].

Note: The rate of diffusion depends upon the following factors:
1. The concentration gradient: The greater will be the difference in concentration, the rapid will be the diffusion.
2. Surface Area: Increase in surface area results in the increase in rate of diffusion.
3. Distance the gas particles must travel.
4. Membrane permeability: Diffusion rate increases with the increase in permeability of the membrane.
5. Temperature also plays a vital role upon the rate of diffusion.
6. Pressure: Gasses diffuse from a region of higher partial pressure to a region of lower partial pressure.