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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?
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Solution
Let the mass of the solution = 100 g
Then the mass of nitric acid = 68 g
Molar mass of nitric acid (HNO3) = 1 × 1 + 1 × 14 + 3 × 16 = 63 g mol−1
∴ Number of moles of HNO3 = `68/63` mol
= 1.079 mol
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
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