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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)`? |
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Answer» Correct Answer - `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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