The volume of a gas at STP is 7.30 L. What is the quantity of gas in this sample?

0.326 moles

Explanation:

We are given;

Volume of a gas at STP as 7.30 L

We are supposed to determine the quantity of the gas sample;

We need to know that 1 mole of a gas occupies a volume of 22.4 L at standard temperature and pressure. On the other hand, at room temperature and pressure, 1 mole of a gas will occupy a volume of 24 L

In this case;

1 mole = 22.4 L

Therefore;

Moles = Volume : Molar gas volume at STP

Therefore,

Moles of the gas sample = 7.30 L : 22.4 L

= 0.326 moles

Therefore, the quantity of the gas sample is 0.326 moles

The quantity of gas is 0.33 mol.

Explanation:

Given dа ta:

Volume of gas = 7.30 L

Pressure of gas = 1 atm

Temperature of gas = 273 K

Amount of gas = ?

Solution:

PV = nRT

n = PV / RT

n = 1 atm. 7.30 L / 0.0821 atm. L. mol⁻¹. K⁻¹ * 273 K

n = 7.30 atm. L / 22.4 atm. L. mol⁻¹

n = 0.33 mol