You can purchase nitric acid in a concentrated form that is 70.3% HNO3 by mass and has a density of 1.41 g>mL. Describe exactly how you would prepare 1.15 L of 0.100 M HNO3 from the concentrated solution.

Verified step by step guidance

1. Calculate the number of moles of HNO3 needed using the formula: \( \text{moles} = \text{molarity} \times \text{volume} \). Convert the volume from liters to milliliters if necessary.
2. Determine the mass of HNO3 required using the molar mass of HNO3 (63.01 g/mol).
3. Calculate the volume of the concentrated HNO3 solution needed using its density and the percentage by mass. Use the formula: \( \text{mass} = \text{density} \times \text{volume} \) and \( \text{mass of HNO3} = \text{mass of solution} \times \text{percentage by mass} \).
4. Convert the mass of HNO3 to volume using the density of the concentrated solution.
5. Dilute the calculated volume of concentrated HNO3 to a final volume of 1.15 L with distilled water, ensuring thorough mixing.

Key Concepts

Molarity (M)

Molarity is a measure of concentration defined as the number of moles of solute per liter of solution. It is expressed in moles per liter (mol/L). To prepare a specific molarity, one must calculate the amount of solute needed based on the desired volume and concentration, which is crucial for diluting concentrated solutions.

Dilution

Dilution is the process of reducing the concentration of a solute in a solution, typically by adding more solvent. The dilution equation, M1V1 = M2V2, relates the initial and final molarities and volumes, allowing one to determine how much concentrated solution is needed to achieve a desired concentration in a larger volume.

Density and Mass Percent

Density is the mass of a substance per unit volume, and in this context, it helps convert the mass percent of nitric acid into a usable form. The mass percent indicates how many grams of solute are present in 100 grams of solution. By using the density of the concentrated solution, one can calculate the mass of HNO3 in a given volume, which is essential for preparing the desired molarity.