Heavy water, D2O (molar mass 20.03) can be separated from ordinary water by the difference in the relative rates of diffusion of the molecules in the gas phase. Calculate the relative rate of diffusion of H2O to D2O. (Atomic weights: O

Question

Chemistry

Karlee Wilkerson

25 February, 08:24

Heavy water, D2O (molar mass 20.03) can be separated from ordinary water by the difference in the relative rates of diffusion of the molecules in the gas phase. Calculate the relative rate of diffusion of H2O to D2O. (Atomic weights: O

+5

Answers (1)

Know the Answer?

Lilianna Zuniga · Answer 1

1.58

Explanation:

This problem can be solved by using a law that was proposes in the year 1848, the law is known as the Graham's law of effusion. The law shows the relationship between square root of the density of a particular gas and the rates of effusion. The Graham' s law can be represented mathematically by using the formula below;

(NOTE: the question asked us to Calculate the relative rate of diffusion of H2O to D2O).

==> Rate of H2O effusion / rate of D2O effusion = √ (M{D2O} / M{H2O}).

Also, we should know that Graham's law shows the relationship between the square root of density and the rate of effusion it can also show the relationship between the square root of the molar mass and the rate of effusion.

So, by using the Graham's law formula we will have;

Rate of H2O effusion / rate of D2O effusion = √ (M{D2O} / M{H2O}). Where the molar mass of H2O = 18 gmol^ - and D2O molar mass = 20 grams per mole.

Rate of H2O effusion / rate of D2O effusion = √ (20 / 18).

Rate of H2O effusion / rate of D2O effusion = √ 2.5.

Rate of H2O effusion / rate of D2O effusion = 1.58.

.