Question

A sample of argon gas at STP occupies 14.0 L. What mass of argon is present in the container?

Answer 1

At STP ( temperature273^oK and pressure 1 atm) argon gas occupies 14.0 L (V₁) in a container.

Let us assume , volume of this gas at NTP (( temperature, 293.15^oC and pressure 1 atm) is V₂ ml

Then, we know V₁/T₁ = V₂/T₂ at constant pressure (here P = 1 atm)

So, V₂ = V₁T₂ / T₁= (14 x 293.15 ) / 273 = 15.03 L

At NTP, 22.4 L is volme of 1 mole of any gas

So, at NTP, 15.03 L is volume of 15.03 / 22.4 = 0.6709 mole of Argon gas

So, 0.6709 mole of Argon is present in the container.

1 mole of Argon = 39.95 g of Argon

Then, 0.6709 mole of Argon = 39.95 x 0.6709 g = 26.8 g of Argon is present in the container (ANSWER)