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.

The molarity of 90% HNO3 with a density of 1.50 g/mL would be 21.34 M

Recall that: Density of a solution = mass/volume

For 100 g HNO3 solution,

      volume = 100/1.50

                     = 66.67 mL

mole of 90% HNO3 = 90/63

                                 = 1.43 moles

Molarity = mole/volume

               = 1.43/0.067

                  = 21.34 M

More on density and molarity can be found here: link

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

i believe it is 40–40–20%

explanation:

i'm assuming that the correct answer is indeed the first one/ "a"..