Question

The time taken by for a certain volume of gas to diffuse through a small hole was 2min. Under similar conditions an equal volume of oxygen took 5.65 minute to pass. The molecular mass of the gas is

Answer 1

Given: Time taken for a certain volume of some gas to diffuse, T_x = 2 min.
Time taken for equal volume of oxygen to diffuse = 5.65 min.

To find: Molecular mass of gas, x

Solve: Rate of diffusion of gas is inversely proportional to time taken by gas to diffuse.
Let the rate of diffusion and molar mass of gas x is R_x and M_x and rate of diffusion and molar mass of oxygen is R_O2 and M_O2 respectively.

$\frac{R_x}{R_{O_2}}=\frac{t_{O_2}}{t_x}$.........................(1)

According to Graham's law of diffusion, rate of diffusion is inversely proportional to square root of molar mass.

$\frac{R_x}{R_{O_2}}=\sqrt{\frac{M_{O_2}}{M_x}}$..................(2)

Molar mass of O₂, = 32 g mol^-1

From (1) and (2), we can write

$$\begin{gathered} \frac{t_{O_2}}{t_x}=\sqrt{\frac{M_{O_2}}{M_x}} \\ \text{By Putting the values we get} \\ \frac{2}{5.65}=\sqrt{\frac{32}{M_x}} \\ 0.35=\sqrt{\frac{32}{M_x}} \\ \text{Squaring on both sides we get} \\ 0.1225=\frac{32}{M_x} \\ {M}_x=\frac{32}{0.1225} \\ {M}_x=261.22 \end{gathered}$$

Molar mass of gas, x = 261.22 g mol^-1