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 data:

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

the answer to your question is c

6.4 cm (max) - 25 cm (min)