Question

The 25mL of a 0.15M solution of lead nitrate, Pb(NO₃)₂, reacts with all the aluminium sulphate, Al₂(SO₄)₃, present in 20mL of solution. What is the molar concentration of the solution of Al₂(SO₄)₃? 3Pb(NO₃)₂ + Al₂(SO₄)₃ → 3PbSO₄ + 2Al(NO₃)₃

• A. \(6.25\times10^{-2}\,\text{M}\)
• B. \(2.421\times10^{-2}\,\text{M}\)
• C. \(0.1875\,\text{M}\)
• D. None of these

Hint:

Always use balanced equations to relate reacting moles.

Answer 1

The Correct Option is A

Step 1: Moles of Pb(NO₃)₂: n = 0.025 × 0.15 = 0.00375
Step 2: From stoichiometry, required moles of Al₂(SO₄)₃: n = (1)/(3)×0.00375 = 0.00125
Step 3: Molarity: M = (0.00125)/(0.020) = 0.0625M