Question

Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL⁻¹?

Answer 1

Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in an aqueous solution. This means that 68 g of nitric acid are dissolved in 100 g of the solution.

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

Then, the number of moles of HNO₃ = `68/63` mol

= 1.079 mol

Given,

Density of solution = 1.504 g mL⁻¹

∴ Volume of solution = `(100  g)/(1.504  g  mL^(-1))`

= 66.5 mL

= 0.0665 L

Molarity of Solution = `"Number of moles of the solute"/"Volume of solution in L"`

= `1.079/0.0665`

= 16.23 M

Answer 2

Since nitric acid is 68% by mass, 100 g of solution will contain 68 g of the acid.

Volume of 100 g solution = `"Mass"/"Density"`

= `100/1.504`

= 66.49 mL

∵ `w = (M xx M' xx v)/1000`

We have,

`M = (w xx 1000)/(M' xx v)`

= `(68 xx 1000)/(63 xx 66.49)`    ...[Molar mass of HNO₃ = 63]

= 16.23

Hence, the molarity of the sample is 16.23 M.