Problem 55 - Chemistry: The Central Science

Question

Commercial aqueous nitric acid has a density of \(1.42 \mathrm{~g} / \mathrm{mL}\) and is \(16 \mathrm{M}\). Calculate the percent \(\mathrm{HNO}_{3}\) by mass in the solution.

Step-by-Step Solution

Answer

The percent of HNO₃ by mass in the commercial aqueous nitric acid solution is approximately 71.0%.

Step 1: Calculate the mass of 1 liter of solution

Given that the density of the solution is 1.42 g/mL, we can easily convert this value to grams per liter (1 L = 1000 mL): Density = 1.42 g/mL × 1000 mL/L = 1420 g/L So, the mass of 1 liter of the solution is 1420 g.

Step 2: Calculate the moles of HNO₃ in 1 liter of solution

The molarity of the solution is given as 16 M. This means that there are 16 moles of HNO₃ in 1 liter of the solution.

Step 3: Calculate the mass of HNO₃ in 1 liter of solution

To calculate the mass of HNO₃ in 1 liter of solution, we need to know its molar mass. The molar mass of HNO₃ = 1.01 (for H) + 14.01 (for N) + 3 × 16.00 (for O) = 63.02 g/mol. Now, we can calculate the mass of 16 moles of HNO₃: Mass of HNO₃ = moles × molar mass = 16 mol × 63.02 g/mol = 1008.32 g

Step 4: Calculate the mass percent of HNO₃ in the solution

Finally, we can use the mass of HNO₃ and the mass of the solution to find the mass percent of HNO₃ in the solution: Mass percent of HNO₃ = (mass of HNO₃ / mass of solution) × 100% Mass percent of HNO₃ = (1008.32 g / 1420 g) × 100% ≈ 71.0% The percent of HNO₃ by mass in the commercial aqueous nitric acid solution is approximately 71.0%.