Question

Concentrated nitric acid used in laboratory 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?

Answer 1

68% by mass = 68 / 100 => 0.68

Density = 1.504 g/mL⁻¹

Molar mass HNO₃ =63.01 g/mol

M = 1000 * T* D / mm

M = 1000 * 0.68 * 1.504 / 63.01

M = 1022.72 / 63.01

M = 16.23 mol/L

hope this helps!