Question

How many times greater is the rate of effusion of molecular fluorine than that of molecular bromine at the same temperature and pressure

Answer 1

According to Graham's Law of Diffusion,"the rates of diffusion of two gases are inversely proportional to the square root of their Molar masses or Densities at the same pressure and temperature".

                                           r₁ / r₂  =  $\sqrt{M2 / M1}$
Where,
            r₁  =  Rate of Fluorine

            r₂  =  Rate of Bromine

            M₂  =  Molar mass of Bromine  =  159.8 g/mol

            M₁  =  Molar mass of Fluorine  =  37.98 g/mol

Putting values,

                                           r₁ / r₂  =  $\sqrt{159.8 / 37.98}$

                                           r₁ / r₂  =  $\sqrt{4.20}$

                                           r₁ / r₂  =  2.04

Result:
          Fluorine effuses 2 times faster than Bromine gas.