concentration HNO3 used in a laboratory work 68% nitric acid by mass in aqueous solution what be the molarity of such a sample of the acid if the density of solution is 1.504g/ml
Answers
Answer:
Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in an aqueous solution. From statement, its clear that 68 g of nitric acid is mixed in 100 g of the solution.
Given, density = 1.504g/mL-1
Density = mass / volume
1.504g/mL-1 = 1000 / volume
Volume = 1000/1.504 = 66.49 mL
From the formula of nitric acid (HNO3), the molar mass would be = 1 × 1 + 1 × 14 + 3 × 16 = 63 g mol - 1
Therefore, We can calculate number of moles of HNO3 = 68 / 63 mol
= 1.08 mol
Molarity of sample = (Number of moles / Volume) x mass
= (1.08/66.49) x 1000 = 16.24 M
The above answer is correct if you don't use numeric 5 in the question otherwise question is incorrect according to me.
Explanation:
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