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?
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 is dissolved in 100 g of the solution.
Molar mass of nitric acid (HNO3) = 1 × 1 + 1 × 14 + 3 × 16 = 63 g mol−1
Then, number of moles of HNO3 = `68/63` mol
Density of solution = 1.504 g mL−1
∴ Volume of 100 g solution = `100/1.504 mL`
Molarity o Solution = `(1.079 "mol")/(66.49xx10^(-3)L)`
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