A 2.44 mass % aqueous solution of nitric acid (HNO3) has a density of 1.05 g/mL. Calculate the molality of the solution.

[HNO₃] = 0.39 m

Explanation:

Let's analyse the information given.

2.44 % mass is the grams of solute in 100 g of solution. In this case, the solution has 2.44 g of HNO₃

In this case, we do not have to use density, because we have solution mass and solute mass.

Total mass of solution = Solute mass + Solvent mass

100 g = 2.44 g + Solvent mass

100 g - 2.44 g = 97.56 g → Solvent mass.

To find molality we have to convert solvent mass in g to kg

97.56 g. 1kg/1000 g = 0.09756 kg

And afterwards, the mass of solute into moles (mass / molar mass)

These steps can be done in an indistinct order. The important thing is that the molality implies the ratio of solute moles between 1 kg of solvent.

2.44 g / 63 g/mol = 0.0387 moles.

Molality = mol / kg → 0.0387 mol / 0.09756 kg = 0.39 m