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17 February, 03:17

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.

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  1. 17 February, 03:40
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    [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
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