Answer:
125.84 g/mol is the molar mass of the unknown gas.
Explanation:
Let the volume of the gases effusing out be V.
Effusion rate of the unknown gas = [tex]R=\frac{V}{371 s}[/tex]
Effusion rate of the nitrogen gas = [tex]r=\frac{V}{175 s}[/tex]
Molar mass of unknown gas = m
Mass of nitrogen gas = 28 g/mol
Graham's 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 the equation:
[tex]\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}[/tex]
[tex]\frac{R}{r}=\sqrt{\frac{28 g/mol}{M}}[/tex]
[tex]\frac{\frac{V}{371 s}}{\frac{V}{175s}}=\sqrt{\frac{28 g/mol}{M}}[/tex]
[tex]\frac{175 s}{371 s}=\sqrt{\frac{28 g/mol}{M}}[/tex]
[tex]M=\frac{28 g/mol\times 371\times 371}{175\times 175}=125.84 g/mol[/tex]
125.84 g/mol is the molar mass of the unknown gas.
