Question

Concentrated nitric acid used in the laboratory 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 $1.504\, g\, mL^{-1}$?

• A. $20\, M$
• B. $12.5\, M$
• C. $12.23\, M$
• D. $40.5\, M$

Answer 1

Answer: (C)

Let mass of solution $= 100\, g$
Then mass of nitric acid $= 68\, g$
Molar mass of $HNO_{3} = 63\, g\,$ mol $^{-1}$
Number of moles of $HNO_{3} = \frac{68}{63} = 1.079$ mol
Density of solution $= 1.504\, g\, mL^{-1}$
$\therefore $ Volume of solution $= \frac{100\,g}{1.504 \,g\, mL^{-1}}$
$= 66.5 \,mL = 0.06665\, L$
Molarity of the solution $= \frac{\text{Number of moles of the solute}}{\text{Volume of solution in L}}$
$ = \frac{1.079}{0.0665}\, M = 16.23\,M$