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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 molarity of such a sample of the acid if the denisty of the solution is `1.504" mL"^(-1)` ? |
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Answer» Given `68%" HNO"_(3)` by mass that means 68gms mass of `"HNO"_(3)` present in 100 gms of solution. Molecular weight of `"HNO"_(3)=63` Number of moles of `"HNO"_(3)=("weight")/("GMW")=(68)/(63)=1.079` Given density of the solution = 1.504 gm/ml Volume of solution `= ("Mass of solution")/("density")` `=(100)/(1.504)=66.5" ml"` Molarity `=nxx(1000)/("V"(ml))` `=(1.079xx1000)/(66.5)=16.23" M"` |
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