Question

A mixture of 0.224 g of H2, 1.06 g of N2, and 0.834 g of Ar is stored in a closed container at STP. Find the volume (in L) of the container, assuming that the gases exhibit ideal behavior.

Answer 1

Answer: The volume of given container is 3.83 L.

Step-by-step explanation:

Given: Mass of
`H_(2)` = 0.224 g

Mass of
`N_(2)` = 1.06 g

Mass of Ar = 0.834 g

Since, moles is the mass of a substance divided by its molar mass. Therefore, moles of given substances present in the mixture are as follows.

Moles of
`H_(2)` are:

`Moles = (mass)/(molar mass)\\= (0.224 g)/(2 g/mol)\\= 0.112 mol`

Moles of
`N_(2)` are:

`Moles = (mass)/(molar mass)\\= (1.06 g)/(28 g/mol)\\= 0.038 mol`

Moles of Ar are:

`Moles = (mass)/(molar mass)\\= (0.834 g)/(40 g/mol)\\= 0.021 mol`

Total moles = (0.112 + 0.038 + 0.021) mol = 0.171 mol

Now, using ideal gas equation the volume is calculated as follows.

PV = nRT

where,

P = pressure

V = volume

n = no. of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

`V = (nRT)/(P)\\= (0.171 mol * 0.0821 L atm/mol K * 273 K)/(1 atm)\\= 3.83 L`

Thus, we can conclude that the volume of given container is 3.83 L.