What is the molality of aqueous nitric acid HNO3 (63 g/mol) that has a density of 1.42 g/mL and is 16.7 M? Report your answer to three significant figures. Do not include units.

[HNO₃] = 45.4 m

Explanation:

We need to be organized to solve this:

Solute: HNO₃

Solvent: Water

Solution: Mass of solute + Mass of solvent

Density → For the solution → Solution mass / Solution Volume

M → moles of solute / 1L of solution

Let's use density to determine solution mass

1.42 g/mL = Solution mass / 1000mL → Solution mass = 1420 g

Let's determine the moles of the solute (HNO₃)

Moles . molar mass → 16.7 mol . 63 g/mol = 1052.1 g

Now we have solution mass and solute mass; let's find out solvent mass.

Solvent mass = Solution mass - Solute mass → 1420 g - 1052.1 g = 367.9 g

Let's convert the solvent mass to kg → 367.9 g . 1kg / 1000 g = 0.6379 kg

Molality → Moles of solute / 1kg of solvent → 16.7 mol / 0.6379 kg = 45.4 m

h let me think

just did the assignment and the answer was #1 +2 n