5 mole H₂O₂(l) is placed in a container of volume 490 mL at 300 K, where it is completely decomposed into H₂O(l) and O₂(g). find total exact pressure exerted by all gases.

(Given: H₂O = 1g/mL, Aqueous tension H₂O(l) at 300 K = 38 mm of Hg)

( Take R = 0.08 atm^-L/mol^-K)

Explanation

Given reaction:

2H₂O₂(l) → 2H₂O(l) + O₂(g)

For every 2 moles of H₂O₂, 1 mole of O₂ gas is produced.

Moles of O₂ produced:

Since 5 moles of H₂O₂ are decomposed:

Moles of O₂ = 5 / 2 = 2.5 moles.

Pressure due to O₂:

Using the ideal gas law: P = nRT / V

Substituting values:

n = 2.5 moles

R = 0.08 atm-L/mol-K

T = 300 K

V = 490 mL = 0.490 L

P_O2 = (2.5 × 0.08 × 300) / 0.490 = 122.45 atm.

Pressure due to aqueous tension of H₂O:

Aqueous tension at 300 K is given as 38 mmHg, which is converted to atm:

P_H2O = 38 / 760 = 0.05 atm.

Total pressure exerted by all gases:

P_total = P_O2 + P_H2O = 122.45 + 0.05 = 150.05 atm.

Final Answer

The total pressure exerted by all gases is 150.05 atm.