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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 is 1.504g `mL^(-1)`?

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The correct Answer is:
`16.23M`

`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`
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