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

`V_2=19.23L`

Step-by-step 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:

`(V_2)/(n_2) =(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*(1mol)/(39.95g) =0.859mol`

Thus, we find:

`V_2=(V_1*n_2)/(n_1)=(22.4L*0.859mol)/(1mol) \\\\V_2=19.23L`

Best regards.