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2 August, 00:22

An aqueous solution is 4.44 M nitric acid and the density of the solution is 1.42 g/mL. Calculate the mole fraction of this solution.

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  1. 2 August, 00:24
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    The mole fraction of HNO3 is 0.225

    Explanation:

    1. Given data

    Density = 1.429 / ml

    Mass% = 63.01 g HNO3 / 100g of solution

    The mass of 63.01 g is in 100 / 1.142 / ml of solution

    Or 63.01 g in 55.7 mL

    Molarity = 15.39 moles / L

    Mass of water in 100g = 100 - 63.01=36.99 g

    So 63.01 grams in 36.99 grams of water

    So mass of HNO3 in 1000grams of water = 63.01 * x 1000 / 36.99 = 1703

    Moles of HNO3 in 1000g = 1703 / 63.01 = 27.03 moles

    Molality = 27.03 molal (mole / Kg)

    Mole fraction = Mole of HN03 / Moles of water + mole of HNO3

    Mole of water = 62 / 18 = 3.44

    Moles of HNO3 = 63.01 / 63.01 = 1.000

    Mole fraction = 1.000 / 3.44 + 1.000 = 0.225

    The mole fraction of HNO3 is 0.225
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