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An aqueous solution of nitric acid contains 70. 0 % hno3 by mass. The density of hno3 is 1. 42 (g/cm3). Express the mass percent of hno3 in molarity.

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The mass percent of HNO₃ in molarity is 15.76 M.

Assumption

Let the mass of the solution be 100 g.

Therefore, the mass of 70% HNO₃ in the solution is 70 g

Determination of the mole of HNO₃

  • Mass = 70 g
  • Molar mass of HNO₃ = 1 + 14 + (3×16) = 63 g/mol
  • Mole of HNO₃ =?

Mole = mass / molar mass

Mole of HNO₃ = 70 / 63

Mole of HNO₃ = 1.11 mole

Determination of the volume of the solution

  • Mass of solution = 100 g
  • Density of solution = 1.42 g/cm³
  • Volume of solution =?

Volume = mass / density

Volume of solution = 100 / 1.42

Volume of solution = 70.42 cm³

Divide by 1000 to express in L

Volume of solution = 70.42 / 1000

Volume of solution = 0.07042 L

Determination of the molarity

  • Mole of HNO₃ = 1.11 mole
  • Volume of solution = 0.07042 L
  • Molarity =?

Molarity = mole / Volume

Molarity = 1.11 / 0.07042

Molarity = 15.76 M

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