Question

What is the pressure of a mixture of 0.200 g of H2, 1.00 g of N2 , and 0.820 g of Ar in a container with a volume of 2.00 L at 20 °C?

Answer 1

P(mixture) = 1.92 atm

Step-by-step explanation:

Given data:

Mass of H₂ = 0.200 g

Mass of N₂ = 1.00 g

Mass of Ar = 0.820 g

Volume = 2 L

Temperature = 20°C

Pressure of mixture = ?

Solution:

Pressure of hydrogen:

Number of moles of hydrogen = mass / molar mass

Number of moles of hydrogen = 0.200 g / 2 g/mol

Number of moles of hydrogen = 0.1 mol

P = nRT / V

P = 0.1 mol× 0.0821 atm. L.mol⁻¹ .k⁻¹ × 293 K / 2L

p = 2.41 atm. L /2 L

P = 1.2 atm

Pressure of nitrogen:

Number of moles of nitrogen = mass / molar mass

Number of moles of nitrogen = 1 g / 28 g/mol

Number of moles of nitrogen = 0.04 mol

P = nRT / V

P = 0.04 mol× 0.0821 atm. L.mol⁻¹ .k⁻¹ × 293 K / 2L

p = 0.96 atm. L /2 L

P = 0.48 atm

Pressure of argon:

Number of moles of argon = mass / molar mass

Number of moles of argon = 0.820 g / 40 g/mol

Number of moles of argon = 0.02 mol

P = nRT / V

P = 0.02 mol× 0.0821 atm. L.mol⁻¹ .k⁻¹ × 293 K / 2L

p = 0.48 atm. L /2 L

P = 0.24 atm

Total pressure of mixture:

P(mixture) = pressure of hydrogen + pressure of nitrogen + pressure of argon

P(mixture) = 1.2 atm + 0.48 atm + 0.24 atm

P(mixture) = 1.92 atm