Answer to Question #308887 in Chemistry for Junard

Question

What is the molality of 60.5% by mass of nitric acid (HNO3) solution?

Answer 1

Solution:

solute = nitric acid (HNO₃)

solvent = water (H₂O)

Suppose we are given 100 g of an aqueous HNO₃ solution

Therefore,

Mass of HNO₃ = Mass of solution × w(HNO₃) / 100% = (100 g × 60.5%) / (100%) = 60.5 g

Mass of H₂O = Mass of solution − Mass of HNO₃ = 100 g − 60.5 g = 39.5 g

The molar mass of HNO₃ is 63.01 g/mol

Therefore,

Moles of HNO₃ = (60.5 g HNO₃) × (1 mol HNO₃ / 63.01 g HNO₃) = 0.960 mol HNO₃

Kilograms of H₂O = (39.5 g H₂O) × (1 kg / 1000 g) = 0.0395 kg H₂O

Molality = Moles of solute / Kilograms of solvent

Therefore,

Molality of HNO₃ solution = Moles of HNO₃ / Kilograms of H₂O

Molality of HNO₃ solution = (0.960 mol) / (0.0395 kg) = 24.3 mol/kg = 24.3 m

Molality of HNO₃ solution = 24.3 m

Answer: The molality of nitric acid (HNO₃) solution is 24.3 m