Question

Commercial grade fuming nitric acid contains about 90.0% HNO3 by mass with a density of 1.50 g/mL, calculate the molarity of the HNO3 solution.

Answer 1

Answer:

Since molarity is defined as moles of solute per liter of solution, we need to find the number of moles of nitric acid, and the volume of solution.

molar mass of nitric acid (HNO3) = 1 + 14 + (3x16) = 15 + 48 = 63 g/mole

1.50 g/ml x 1000 ml = 1500 g/liter

1500 g/liter x 0.90 = 1350 g/liter of pure HNO3 (the 0.9 is to correct for the fact that it is 90% pure)

1350 g/liter x 1 mole/63 g = 21.43 moles/liter = 21 Molar HNO3

= 21 Molar of HNO3

Answer 2

Answer:

21.4 M.

Explanation:

Mass of HNO3 = 90%

= 90 g of HNO3 in 100 g of solution.

Density = 1.5 g/ml

Molar mass of HNO3 = 1 + 14 + (3*16)

= 63 g/mol

Number of moles = mass/molar mass

= 90/63

= 1.429 mol.

Volume of solution = mass/density

= 100/1.5

= 66.67 ml of solution.

= 0.0667 l.

Molarity is defined as the number of moles of a substance per unit volume of solution.

Molarity, M = number of moles/volume

= 1.429/0.0667

= 21.4 M.