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Question
Concentrated nitric acid used in 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 density of the solution is 1.504 g mL−1?
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Solution 1
Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in an aqueous solution. This means that 68 g of nitric acid are dissolved in 100 g of the solution.
Molar mass of nitric acid (HNO3) = 1 × 1 + 1 × 14 + 3 × 16 = 63 g mol−1
Then, the number of moles of HNO3 = `68/63` mol
= 1.079 mol
Given,
Density of solution = 1.504 g mL−1
∴ Volume of solution = `(100 g)/(1.504 g mL^(-1))`
= 66.5 mL
= 0.0665 L
Molarity of Solution = `"Number of moles of the solute"/"Volume of solution in L"`
= `1.079/0.0665`
= 16.23 M
Solution 2
Since nitric acid is 68% by mass, 100 g of solution will contain 68 g of the acid.
Volume of 100 g solution = `"Mass"/"Density"`
= `100/1.504`
= 66.49 mL
∵ `w = (M xx M' xx v)/1000`
We have,
`M = (w xx 1000)/(M' xx v)`
= `(68 xx 1000)/(63 xx 66.49)` ...[Molar mass of HNO3 = 63]
= 16.23
Hence, the molarity of the sample is 16.23 M.
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