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Karnataka Board PUCPUC Science 2nd PUC Class 12

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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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?

Numerical
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Solution

Let the mass of the solution = 100 g

Then the mass of nitric acid = 68 g

Molar mass of nitric acid (HNO3) = 1 × 1 + 1 × 14 + 3 × 16 = 63 g mol−1

∴ Number of moles of HNO3 = `68/63` mol

= 1.079 mol

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

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Chapter 1: Solutions - Exercises [Page 27]

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NCERT Chemistry Part 1 and 2 [English] Class 12
Chapter 1 Solutions
Exercises | Q 1.4 | Page 27
Nootan Chemistry [English] Class 12 ISC
Chapter 1 Solutions
'NCERT TEXT-BOOK' Exercises | Q 2.4 | Page 123

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