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.

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:

$Moles = \frac{mass}{molar mass}\\= \frac{0.224 g}{2 g/mol}\\= 0.112 mol$

Moles of $N_{2}$ are:

$Moles = \frac{mass}{molar mass}\\= \frac{1.06 g}{28 g/mol}\\= 0.038 mol$

Moles of Ar are:

$Moles = \frac{mass}{molar mass}\\= \frac{0.834 g}{40 g/mol}\\= 0.021 mol$

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.

$V = \frac{nRT}{P}\\= \frac{0.171 mol \times 0.0821 L atm/mol K \times 273 K}{1 atm}\\= 3.83 L$

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