Question

A certain gas effuses four times as fast as oxygen gas. What is the molar mass of the gas? 

Answer 1

The equation for the effusion rate is:

$\frac{ER_{1}}{ER_{2}}=(\frac{M_{2}}{M_{1}})^{1/2}$

,where ER is the effusion rate.

Here we have $\frac{ER(X)}{ER(O_{2})}=4$ and the molar mass of the oxygen gas is equal to 32 g/mol.

Thus, $\frac{ER(X)}{ER(O_{2})}=(\frac{M(O_{2})}{M(X)})^{1/2}$

Using this formula, taking its square and slightly rearrange it, we could get the following expression $M(X)=M(O_{2})*(\frac{ER(O_{2})}{ER(X)})^2$

and finally $M(X)=32g/mol*(\frac{1}{4})^2=2g/mol$

It is obviously, that the unknown gas is hydrogen gas or H₂.