Concentrated nitric acid used for laboratory works 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 solution is `1.504 gmL^(-1)` ?

Explanation

68% `HNO_(3)`, means that 68 g `HNO_(3)` is present in 100 g of solution.
Moles of `HNO_(3) = (68)/(63)= 1.08`
Density of solution ` = 1.504 g mL^(-1)`
Volume of solution ` = (1000)/(1.504) = 66.46 mL`
Molarity ` = (1.08)/(66.49) xx 1000 = 16.24 M`