alculate the molarity of a solution of Nitric Acid (HNO3) if it is 71% pure HNO3 and has a specific gravity of 1.43? (Atomic weight of H=1, N=14, O=16).

Question

answer fast lol , I have an exam rn

I hope I don't fail thanksssss

Answer 1

Molar mass HNO3 = 1 + 14 + 3x16 = 63 g/mol

Let's say you have 1000 ml (1 L) of this solution. How many grams do you have?

1000 ml x 1.43 g/ml = 1430 g

Since it is only 71% HNO3, that means you have 1430 g x 0.71 = 1015.3 g HNO3 in 1 liter

How many moles is this?

1015.3 g x 1 mol / 63 g = 16 moles

16 mol/L = 16 M (answer)