Concentrated nitric acid used in the laboratory work is `68%` nitric acid by mass in aqueous solution. What should be the molaritiy of such a sample of the acid if the density of solution is `1.504 g mL^(-1)`?

Answer

`16.23M`

Explanation

`68%` by mass that `68g` of `HNO_(3)` is present in `100g` of solution.
Volume of solution `=(Mass of solution )/(Density of solution )`
`=(100g)/(1.50gmL^(-1))=66.5cm^(3)=0.65L`
Molar mass, `Mw_(B)` of `HNO_(3)=1+14+48=63`
Molarity`=(W_(A))/(Mw_(B)xxV)`
`=(68g)/(63g mol^(-1)xx0.0665)=16.23M`