The molarity of HNO₃in a sample which has density 1.4 g/mL and mass percentage of 63% is _____.
(Molecular Weight of HNO₃ = 63)

Solution

Chemistry ma’am · 4 steps

Take one litre of the solution, use the density to get its mass, then pick out the 63% that is acid.

1. Mass of one litre. One litre is 1000 mL. With density 1.4 g mL⁻¹, its mass is 1000 × 1.4 = 1400 g.
2. Mass of acid. 63% of that mass is HNO₃: 0.63 × 1400 = 882 g.

   M = (882 / 63)/(1 L) = (% w/w × d × 10)/(Molar mass) = (63 × 1.4 × 10)/63

3. Moles per litre. 882 / 63 = 14 mol of HNO₃ sit in that one litre.
4. So. The molarity of the acid is 14 mol L⁻¹.

The common slip: 10. Treating 100 g of solution as 100 mL ignores the density and gives 10. The 100 g actually fills only 100/1.4 = 71.4 mL.

The idea to keep. Mass percent and density together give molarity: M = (% × d × 10)/(molar mass), with d in g/mL. Always turn the solution's mass into its volume through the density.

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