Question

The molecular weight of a gas which diffuses through a porous plug of 1/6 th of the speed of hydrogen under identical condition is

Answer 1

HERE IS UR ANSWER

Molecular mass of a gas which diffuse through a Porus plug at 1/6th of the speed of hydrogen under identical condition

rg=16rH2rg=16rH2

Mg=MH2.[rH2rg]2Mg=MH2.[rH2rg]2

2×62=2×36=72

Answer 2

Answer: The molar mas of unknown gas is 72 g/mol.

Explanation:

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows:

$\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$

$\frac{Rate_{X}}{Rate_{H_2}}=\sqrt{\frac{M_{H_2}}{M_{X}}$

$\frac{{\frac{1}{6}}}{1}=\sqrt{\frac{2}{M_{X}}$

Squaring both sides and solving for $M_{X}$

$M_{X}=72g/mol$

Hence, the molar mas of unknown gas is 72 g/mol.