Concentrated nitric acid used in the laboratory work is 68% nitric acid by mass in aquequs solution. What should be molarity of such sample of the acid if the density of solution is 1.504 g mL`""^(-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 )=66.49`
Molarity `= (1.08)/(66.49)xx1000`