Question

A gaseous mixture containing $He,C{H_4},$ and $S{O_2}$ was allowed to effuse through a fine hole, then find what molar ratio of gases coming out initially? (Given mixture contains $He,C{H_4},$ and $S{O_2}$ in $1:2:3$ mole ratio).

Answer 1

Given that there are three gases and allowed to effuse through a fine hole. According to graham’s law of effusion, the rate of effusion of gases is inversely proportional to the square root of molar mass of particles. Thus, the molar mass of the gases in the mixture should be determined.

Complete answer:
Given that a gaseous mixture containing helium gas, methane and Sulphur dioxide. These gases are allowed to effuse through a fine hole. This mixture contains the three gases in the mole ratio of $1:2:3$ .
Scientist Graham discovered a law known as graham’s law of effusion. According to graham’s law of effusion, the rate of effusion of gases is inversely proportional to the square root of molar mass of particles.
The molar ratio of $He$ and $C{H_4}$ can be determined from the mole ratio and molar mass, the molar mass of $He$ is $4$ and molar mass of $C{H_4}$ is $16$ . Thus, the mole ratio will be $\dfrac{1}{2}\sqrt {\dfrac{{16}}{4}} = 1:1$
The mole ratio of $He$ and $S{O_2}$ can be determined from the mole ratio and molar mass, the molar mass of $He$ is $4$ and molar mass of $S{O_2}$ is $64$ . Thus, the mole ratio will be $\dfrac{1}{3}\sqrt {\dfrac{{64}}{4}} = 4:3$
The molar ratio of $He$ and $C{H_4}$ is $1:1$ . The molar ratio of $He$ and $S{O_2}$ $4:3$
The molar ratio of $He,C{H_4},$ and $S{O_2}$ is $4:4:3$ .

Note:
The molar ratio of the gases coming out are the product of mole ratio and inversely proportional to the square root of the molar mass of the gas. Finally, the molar ratio of helium gas, methane should be multiplied by $4$ as the ratio obtained is $1:1$ .