Answer to Question #110935 in Chemistry for Joy

Question

Equal volumes of oxygen and nitrogen were allowed to diffuse under the similar conditions of temperature and pressure.If oxygen took 18s to diffuse , determine the time that would be taken by nitrogen

Answer 1

According to Graham's law of diffusion, the ratio of the rates of diffusion of two gases ($r_{O_2}$,$r_{N_2}$) is the square root of the inverse ratio of their molar masses ($M_{O_2}$ , $M_{N_2}$ ):

$\frac{r_{O_2}}{r_{N_2}} = \sqrt\frac{M_{N_2}}{M_{O_2}}$ .

The rate of diffusion is the amount of gas passed through an area per unit time. Therefore, it is inversely proportional to the time of diffusion $t$ :

$\frac{r_{O_2}}{r_{N_2}} =\frac{t_{N_2}}{t_{O_2}}$ .

Combine the equations above:

$\sqrt\frac{M_{N_2}}{M_{O_2}}=\frac{t_{N_2}}{t_{O_2}}$ .

The molar masses of O₂ and N₂ are 32.00 g/mol and 28.01 g/mol, respectively. The time taken by nitrogen is:

$t_{N_2}=t_{O_2}\sqrt\frac{M_{N_2}}{M_{O_2}} = 18(\text{s})\cdot \sqrt\frac{28.01\text{ (g/mol)}}{32.00\text{ (g/mol)}} = 16.8 (\text{s})$

Answer: if oxygen took 18s to diffuse , the time that would be taken by nitrogen is 16.8s.