Answer:
[tex]V_2=19.23L[/tex]
Explanation:
Hello,
In this case, by using the Avogadro's law which allows us to understand the volume-moles behavior as a directly proportional relationship:
[tex]\frac{V_2}{n_2} =\frac{V_1}{n_1}[/tex]
We can compute the volume of 34.3 g of argon by representing it in mole as shown below:
[tex]n_1=1 mol\\\\n_2=34.3g*\frac{1mol}{39.95g} =0.859mol[/tex]
Thus, we find:
[tex]V_2=\frac{V_1*n_2}{n_1}=\frac{22.4L*0.859mol}{1mol} \\\\V_2=19.23L[/tex]
Best regards.
