concentrated nitric used in laboratory work is 68%. Nitric acid by mass in aqueous solutions. What should be the molarity of such a sample of the acid. If the density of the solution is 1.50 4 gram per milliliter ?
Asked by sumit satpaty | 30th May, 2011, 10:19: PM
68% means 68g oh HNO3 is present in 100g of solution.
Volume of solution = Mass / Density
= 100 / 1.504 = 66.49ml
Moles of HNO3 = 68 / 63 = 1.08
Molarity = 1.08 x 1000 / 66.49 = 16.24 M
Answered by | 31st May, 2011, 09:13: AM
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