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?
The molarity of the concentrated nitric acid sample is 16.23 M.
Step 1: Calculate mass of solution Let us consider 1 L (1000 mL) of the solution. Mass of solution = Volume × Density = 1000 mL × 1.504 g mL⁻¹ = 1504 g
Step 2: Calculate mass of nitric acid (HNO₃) Since the solution is 68% nitric acid by mass, Mass of HNO₃ = (68/100) × 1504 g = 1022.72 g
Step 3: Calculate moles of HNO₃ Molar mass of HNO₃ = 1 + 14 + () = 63 g mol⁻¹ Moles of HNO₃ = 1022.72 g / 63 g mol⁻¹ = 16.23 mol
Step 4: Calculate Molarity Molarity = Moles of solute / Volume of solution in litre Molarity = 16.23 mol / 1 L = 16.23 mol L⁻¹
Therefore, the molarity of the concentrated nitric acid is 16.23 M.
Explanation
This question tests the student's understanding of molarity calculations from mass percentage and density. The key steps involve: (1) finding the mass of a known volume of solution using density, (2) calculating the mass of solute from the percentage composition, (3) converting mass to moles using molar mass, and (4) applying the molarity formula. The molar mass of HNO₃ (63 g mol⁻¹) is calculated by adding atomic masses of H(1), N(14), and O(=48).
Solution Steps
Step 1: Calculate mass of 1 L solution using density = 1000 mL × 1.504 g mL⁻¹ = 1504 g
Step 2: Calculate mass of HNO₃ using percentage = 68% × 1504 g = 1022.72 g
Step 3: Calculate moles of HNO₃ = 1022.72 g ÷ 63 g mol⁻¹ = 16.23 mol
Step 4: Calculate Molarity = 16.23 mol ÷ 1 L = 16.23 M