Answer:- [tex]N_2[/tex] gas effuses at a rate that is 1.73 times that of Kr gas under the same conditions.
Solution:- Rate of effusion of gases is inversely proportional to the square root of the molar mass of the gas. It is known as Graham's law. The formula is written as:
[tex]\frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}}[/tex]
Since we have nitrogen and krypton gases, so let's write the formula as:
[tex]\frac{r_N_2}{r_K_r}=\sqrt{\frac{M_K_r}{M_N_2}}[/tex]
where r stands for rate of effusion and M stands for molar mass.
Molar mass of nitrogen gas is 28.01 gram per mol and the molar mass of krypton gas is 83.8 gram per mol.
let's plug in the values in the formula:
[tex]\frac{r_N_2}{r_K_r}=\sqrt{\frac{83.8}{28.01}}[/tex]
[tex]\frac{r_N_2}{r_K_r}[/tex] = 1.73
[tex]r_N_2=1.73(r_K_r)[/tex]
So, the rate of effusion of nitrogen gas is 1.73 times that of Kr.
