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.

Short answer

The percent of $\mathrm{HNO}_{3}$ by mass in the commercial aqueous nitric acid solution is approximately 71%.

Step-by-step solution

Find the mass of HNO3 in the given volume

To find the mass of HNO3, we need to use the given molarity and volume. From the given exercise, we have a molarity (M) of 16 M (16 moles of HNO3 per liter). Let's start by assuming 1L of the solution. So, in 1L of solution, there are 16 moles of HNO3. We need to determine the mass of these 16 moles using the molar mass of HNO3. The molar mass of HNO3 is 63.01 g/mol, and we can calculate the mass as follows: Mass of HNO3 = Molarity × Volume × Molar mass of HNO3 Mass of HNO3 = 16 mol/L × 1 L × 63.01 g/mol = 1008.16 g

Calculate the mass of the 1L solution

We are given the density of the commercial aqueous nitric acid solution, which is 1.42 g/mL. To find the mass of 1L of the solution, we will convert the density to g/L and then use: Mass of the solution = Volume × Density Density = 1.42 g/mL × \(1000 \frac{mL}{L}\) = 1420 g/L Mass of the solution = 1L × 1420 g/L = 1420 g

Calculate the percentage of HNO3 by mass

Now that we have the mass of HNO3 and the mass of the entire solution, we can find the percent by mass of HNO3: Percent by mass of HNO3 = \(\frac{\text{Mass of HNO3}}{\text{Mass of the solution}}\) × 100% Percent by mass of HNO3 = \(\frac{1008.16 \mathrm{~g}}{1420 \mathrm{~g}}\) × 100% Percent by mass of HNO3 ≈ 71% Therefore, the percent of HNO3 by mass in the commercial aqueous nitric acid solution is approximately 71%.