Answer:

The molar mass of the gas which diffuses 0.25 times as fast as helium is 64.

The steps to be taken are in line with the calculation techniques of the mathematical representation of Graham's law of diffusion.

According to Graham's law of diffusion;

At constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass.

The molar mass of helium, m2= 4.

Therefore, r1/r2 = 1/4 since the gas diffuses 0.25 times as fast as helium.

Therefore;

(0.25)² = 4/m1

0.0625 = 4/m1

m1 = 4/0.0625

m1 = 64.

Therefore, the molar mass of the gas is 64.

The steps to be taken are in line with the calculation techniques of the mathematical representation of Graham's law of diffusion.