Answer: The fluorine gas will effuse 2.05 times faster than bromine gas.
Explanation:
To calculate the rate of diffusion of gas, we use Graham's Law.
This law states that the rate of effusion of gas is inversely proportional to the square root of the molar mass of that gas. The equation given by this law follows:
[tex]\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}[/tex]
We are given:
Molar mass of Fluorine = 38 g/mol
Molar mass of Bromine = 160 g/mol
By taking their ratio, we get:
[tex]\frac{Rate_{F_2}}{Rate_{Br_2}}=\sqrt{\frac{M_{Br_2}}{M_{F_2}}}[/tex]
[tex]\frac{Rate_{F_2}}{Rate_{Br_2}}=\sqrt{\frac{160}{38}}\\\\\frac{Rate_{F_2}}{Rate_{Br_2}}=2.05\\\\Rate_{F_2}=2.05\times Rate_{Br_2} [/tex]
Hence, the fluorine gas will effuse 2.05 times faster than bromine gas.
