Question

The ratio of rates of diffusion of two gases X and Y is $4:\sqrt{11}$ .If the molecular mass of Y is double to the molecular mass of oxygen, then X is

• A. CO
• B. SO₃
• C. CO₂
• D. NO

Answer 1

The correct option is C

CO₂

Explanation for the correct option:

Graham's law of diffusion: It states that the rate of diffusion of different gases are inversely proportional to the square root of their densities at constant temperature and pressure. This law was given by Thomas Graham.

• Diffusion: The movement of gas particles from a region of their higher concentration to a region of their comparatively lower concentration.

Mathematically, Graham's law is:

• $\frac{r_X}{r_Y}=\sqrt{\frac{d_y}{d_X}}=\sqrt{\frac{M_Y}{M_X}}$where r= rate of diffusion, d= density and M= Molecular masses of X and Y gas.

Step-1: Calculate the value of M_X

$\frac{r_X}{r_Y}=\frac{4}{\sqrt{11}}$

$\frac{r_X}{r_Y}=\sqrt{\frac{d_y}{d_X}}=\sqrt{\frac{M_Y}{M_X}}$

Given, My=2 (Molecular mass of Oxygen)

Atomic weight of Oxygen (O)= 16 u

Molecular mass of Oxygen (O₂)= 2× 16= 32 u

Therefore, My=2 (32)= 64 u

$\frac{r_X}{r_Y}=\sqrt{\frac{M_Y}{M_X}}=\sqrt{\frac{64}{M_X}}$

$\frac{4}{\sqrt{11}}=\sqrt{\frac{64}{M_X}}$

On squaring both sides

$\frac{16}{11}=\frac{64}{M_X}$

$M_X=\frac{64\times 11}{16}=44$ u

Step-2: Find the gas with same molecular mass as M_X

Atomic weight of Carbon (C)= 12 u

Atomic weight of Oxygen (O)= 16 u

Molecular mass of Carbon dioxide (CO₂)= 12+2(16)= 44 u

Explanation for the incorrect options:

Atomic weight of Sulphur (S)= 32 u

Atomic weight of Nitrogen (N)= 14 u

(i) Molecular mass of Carbon monoxide (CO)= 12+16= 28 u

(ii) Molecular mass of Sulphur trioxide (SO₃)= 32 +3(16)= 80 u

(iii) Molecular mass of Nitric oxide (NO)= 14+16= 30 u

As M_X comes out to be 44 u, then X is Carbon dioxide.

Hence, option (c) is correct