Question

Given: There are 39.95 grams of Argon (39.95 g/1 mole) and one mole has a volume of 22.4 Liters (1 mole/22.4 L). What is the volume, in Liters, of 34.3 grams of Argon?

Answer 1

Answer:

$V_2=19.23L$

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:

$\frac{V_2}{n_2} =\frac{V_1}{n_1}$

We can compute the volume of 34.3 g of argon by representing it in mole as shown below:

$n_1=1 mol\\\\n_2=34.3g*\frac{1mol}{39.95g} =0.859mol$

Thus, we find:

$V_2=\frac{V_1*n_2}{n_1}=\frac{22.4L*0.859mol}{1mol} \\\\V_2=19.23L$

Best regards.