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 1.504 gm `L^(-1)` is ?

Explanation

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 `(HNO_(3))`
`= 1xx1+1xx14+3xx16`
`= 63 g mol^(-1)`
Then, number of moles `HNO_(3)=(68)/(63)mol`
= 1.079 mol
Given,
Density of solution `= 1.504 g mL^(-1)`
Therefore, Volume of 100 g solution
`= (100)/(1.504)mL`
`= 66.49 mL`
`= 66.49xx10^(-3)L`
Molarity of solution `= (1.079 mol)/(66.49xx10^(-3)L)-16.23 M`