Respuesta :
From Graham's law of diffusion the effusion rate of a gas is inversely proportional to the square root of its molar mass.
Therefore for helium and radon; Helium has a mass of 4 g while randon has 222, hence;
= √222/√4
= 14.90/2
= 7.45
Thus the ratio of effusion rate between helium and randon is 7.45
Therefore for helium and radon; Helium has a mass of 4 g while randon has 222, hence;
= √222/√4
= 14.90/2
= 7.45
Thus the ratio of effusion rate between helium and randon is 7.45
Answer: The ratio of rate of effusion of helium and radon is 7.45
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 the equation:
[tex]\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}[/tex]
We are given:
Molar mass of Helium = 4.00 g/mol
Molar mass of Radon = 222.1 g/mol
Taking their ratios, we get:
[tex]\frac{Rate_{He}}{Rate_{Rn}}=\sqrt{\frac{M_{Rn}}{M_{He}}}[/tex]
[tex]\frac{Rate_{He}}{Rate_{Rn}}=\sqrt{\frac{222.1}{4.00}}\\\\\frac{Rate_{He}}{Rate_{Rn}}=7.45[/tex]
Hence, the ratio of rate of effusion of helium and radon is 7.45
