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 $1.504gm{L}^{-1}$?

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

Answer 1

Answer: (C)

Let mass of solution $=100g$
Then mass of nitric acid $=68g$
Molar mass of $HN{O}_3=63g$ mol ${}^{-1}$
Number of moles of $HN{O}_3=\frac{68}{63}=1.079$ mol
Density of solution $=1.504gm{L}^{-1}$
$\therefore$ Volume of solution $=\frac{100g}{1.504gm{L}^{-1}}$
$=66.5mL=0.06665L$
Molarity of the solution $=\frac{\text{Number of moles of the solute}}{\text{Volume of solution in L}}$
$=\frac{1.079}{0.0665}M=16.23M$