Answer :
The percentage of sulfuric acid in the chamber is 50.5%
Step 1: Calculate the number of moles of sodium hydroxide (NaOH) used to neutralize the sulfuric acid.
First, calculate the number of moles of NaOH:
Molarity of NaOH = 1 M (given)
Volume of NaOH = 50 cc = 0.05 L (converted to liters)
Moles of NaOH = Molarity * Volume (in liters) = 1 M * 0.05 L = 0.05 mol
Step 2: Calculate the number of moles of sulfuric acid (H₂SO₄) neutralized.
The reaction equation is: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
From the equation, 1 mole of H₂SO₄ reacts with 2 moles of NaOH.
Moles of H₂SO₄ = Moles of NaOH / 2 = 0.05 mol / 2 = 0.025 mol
Step 3: Calculate the mass of sulfuric acid.
Molar mass of H₂SO₄ = 98.08 g/mol
Mass of H₂SO₄ = Moles * Molar mass = 0.025 mol * 98.08 g/mol = 2.452 g
Step 4: Calculate the volume of sulfuric acid solution.
Given: 2.84 cc of chamber solution contains 2.452 g of H₂SO₄.
Specific gravity of H₂SO₄ solution = 1.71
Step 5: Calculate the percentage of sulfuric acid.
Mass percentage = (Mass of H₂SO₄ / Mass of solution) * 100
To find the mass of the solution, use the specific gravity:
Mass of solution = Volume * Specific gravity = 2.84 cc * 1.71 = 4.8544 g
Percentage = (2.452 g / 4.8544 g) * 100 ≈ 50.5%
