Question

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.

Answer 1

Answer:

[HNO₃] = 15.7 M

Explanation:

70% HNO₃ by mass, means that in 100 g of solution, we have 70 g of solute(in this case our solute is the HNO₃)

Let's determine the volume of solution, by density.

Solution density = Solution mass / Solution volume

1.42 g/mL = 100 g / Solution volume

100 g / 1.42 g/mL = Solution volume  →  70.4 mL

This is the volume occupied by the 70 grams of solute.

Molarity is: moles of solute / L of solution

Let's convert the mass to moles like this:

70 g / 63 g/m = 1.11 moles

Let's convert the volume from mL to L

70.4 mL / 1000 = 0.0704L

M = mol/L → 1.11 mol / 0.0704L = 15.7 M

Answer 2

The mass percent of a solution which contains 70% of HNO₃ in molarity is 15.78 M.

Let the mass of the solution be 100 g. Thus, 70% of HNO₃ in the solution will be 70 g.

• Next we shall determine the number of mole in 70 g of HNO₃.

Mass of HNO₃ =  70 g

Molar mass of HNO₃ = 1 + 14 + (16×3) = 63 g/mol

Mole of HNO₃ =?

Mole = mass / molar mass

Mole of HNO₃ = 70 / 63

Mole of HNO₃ = 1.111 mole

• Next, we shall determine 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

• Finally, we shall determine the molarity of HNO₃

Mole of HNO₃ = 1.111 mole

Volume of solution = 70.42 cm³ = 70.42 / 1000 = 0.07042 L

Molarity of HNO₃ =?

Molarity = mole / Volume

Molarity of HNO₃ = 1.111 / 0.07042

Molarity of HNO₃ = 15.78 M

Therefore, the mass percent of HNO₃ in molarity is 15.78 M.

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