what is the pressure in mm of Hg, of a gas mixture that contains 1g of H2, and 8.0g of Ar in a 3.0 L container at 27 degree C

4367.72 mmHg

Explanation:

Given dа ta:

Mass of hydrogen = 1 g

Mass of argon = 8 g

Volume of container = 3.0 L

Temperature = 27 °C = 27 + 273 = 300 K

Pressure of mixture = ?

Solution:

Moles of hydrogen:

Number of moles = mass / molar mass

Number of moles = 1 g / 2 g/mol

Number of moles = 0.5 mol

Moles of argon:

Number of moles = mass / molar mass

Number of moles = 8 g / 40 g/mol

Number of moles = 0.2 mol

Pressure of hydrogen:

PV = nRT

P = nRT/V

P = 0.5 mol * 0.0821 atm. L. mol⁻¹. K⁻¹ * 300 K / 3.0 L

P = 12.315 atm / 3.0

P = 4.105 atm

Pressure of argon:

PV = nRT

P = nRT/V

P = 0.2 mol * 0.0821 atm. L. mol⁻¹. K⁻¹ * 300 K / 3.0 L

P = 4.926 atm / 3.0

P = 1.642 atm

Total pressure:

T (total) = P (H₂) + P (Ar)

T (total) = 4.105 atm + 1.642 atm

T (total) = 5.747 atm

atm to mmHg:

5.747 * 760 = 4367.72 mmHg