The molarity of 1 L orthophosphoric acid (H₃ PO₄) having 70 % purity by weight (specific gravity 1.54 g cm⁻³) is __________ M.
(Molar mass of H₃ PO₄=98 g mol⁻¹)

Solution

Chemistry ma’am · 3 steps

Take the 1 L of solution, find its mass from the density, then keep only the 70% that is acid.

1. Mass of solution. Specific gravity 1.54 g cm⁻³ means 1 mL weighs 1.54 g. So 1000 mL of solution weighs 1000 × 1.54 = 1540 g.
2. Mass of acid. 70% by weight means 70 g of H₃PO₄ in every 100 g of solution. So the acid in 1540 g is 0.70 × 1540 = 1078 g.

   M = (moles of solute)/(volume in L) = (1078 / 98)/1 = 11 mol L⁻¹

3. So. The 1 L of solution holds 11 mol of H₃PO₄, so the molarity is 11 M.

The common slip: 7.14. Treating 70% by weight as 70 g per 100 mL ignores the density and gives 700/98 ≈ 7.14, not 11.

The idea to keep. For % w/w with a density, use M = (10 × d × w%)/(molar mass), with d in g cm⁻³. Here it is (10 × 1.54 × 70)/98 = 11.

Answer from the NTA final answer key. Solution by PrepWiser.