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

Answer:

68% by mass implies that 68 g of HNO₃are present in 100 g of solution.
Volume of solution =

\frac{Mass of solution}{Density of solution}

= \frac{100 g}{1.504 g {mL}^{- 1}} = 66.5 {cm}^{3} = 0.665 L .

Molar mass,

M_{B} of {NHO}_{3} = 1 + 14 + 48 = 63 .

Molarity = \frac{ω_{A}}{M_{B} \times V} = \frac{68 g}{63 g {mol}^{- 1} \times 0.0665} = 16.23 M .

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