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.

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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