Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL^–1?

Given:

Concentration of Nitric Acid, HNO₃ = 68%

Density of solution, d = 1.504 g/ml

To find: Molarity, M_o

Formula:

Density, d

Molarity, M_o

Solution:

68% of Nitric acid by mass in aqueous solution means that 68g [[68 × 100]/100] of Nitric acid present in 100g of solution.

⇒ Molecular mass of Nitric Acid, HNO₃ = [1 × 1] + [1 × 14] + [16 × 3]

= 63g

⇒ Number of moles of Nitric Acid = [68/63]

= 1.079 moles

⇒ Given Density, d = 1.504 g/ml

⇒ Volume, v = [100/1.504]

= 66.489 ml

⇒ Molarity, M_o = [1.079/66.489] × 1000

= 16.23 M

Therefore the molarity of the sample is 16.24 M.