Question

"n2(g effuses at a rate that is ______ times that of br2(g under the same conditions."

Answer 1

Graham's law of effusion states that the rate of diffusion of a certain gas iis inversely proportional to the sqrt of the molar mass. 
Molar mass of N2 is 28 g/mol and that of Br2 is 158.81 g/mol.

The sqrt of the ratio of their molar mass is equal to 5.67, the sqrt of this number is 2.38. Thus, N2 diffuses 2.38 times faster than Br2. 

Answer 2

Answer:

$N_2$ effuses at a rate that is 2.38 times that of $Br_2$

Explanation:

We have two components: $N_2$ and $Br_2$. We need to compare the rate of effusion under the same conditions.

According to the Graham effusion law: if we have two components Component 1 (C1) and Component 2 (C2), with Molecular Weights M1 and M2 respectively. They will have a rate of effusion V1 and V2 respectively as follows:

$\frac{V1}{V2} =\sqrt{\frac{M2}{M1} }$

Now, we will replace for our components:

Component 1 - Nitrogen $N_2$  

Component 2 - Bromine $Br_2$

Molecular weight Nitrogen M1 = M($N_2$) = 28 g/mol

Molecular weight Bromine M2 =M( Bromine $Br_2$) = 159 g/mol

After that, we should replace the values:  

$\frac{V1}{V2} =\sqrt{\frac{159}{28} } \\\frac{V1}{V2} =\sqrt{5.67}\\\\ \frac{V1}{V2} =2.38$

As V1 is the effusion rate of Nitrogen and V2 is the effusion rate of Bromine. We can reorganize the equation:

$V1 = 2.38 V2$

The conclusion is that V1 is 2.38 times V2.

It means $N_2$ effuses at a rate that is 2.38 times that of $Br_2$ .