Concentrated nitric acid used in laboratory work is $ 68 \% $ nitric acid by aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is $ 1.504 \mathrm{~g} \mathrm{~mL}^{-1} $ ?

Question

Concentrated nitric acid used in laboratory work is $ 68 \% $ nitric acid by aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is $ 1.504 \mathrm{~g} \mathrm{~mL}^{-1} $ ?,

MARKS App · Accepted Answer

68% HNO 3 by mass means Mass of nitric acid = 68 g Mass of solution - 100 g Molar mass of HNO 3 = (1) + (14) + (16 x 3) = 63 g mol -1 Number of moles of HNO 3 n HNO 3 = W M = 68 g 63 g / mol = 1.079 mol Density of solution = 1.504 g mL-1 Volume of solution = Mass Density = 100 g 1.504 g mL - 1 = 66.5 mL or 0.0665 L Molarity = Moles of solute Volume of solution in litres = 1.079 mol 0.0665 L = 16.23 mol L -1 or 16.23 M

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