How many mL of 50.0% by mass HNO3 solution, with a density of 2.00 g/mL is required to make a 500 mL of a 2.00 M HNO3?
initial Molarity X initial Volume = final Molarity X final Volume
The Attempt at a Solution
(50 g HNO3)(1 mL/2.00 g)(0.001 L/1 mL) = 0.025 L
(50 g HNO3)(1 mol HNO3/63.012 g HNO3) = 0.79 mol
0.79 mol/0.025 L = 31.6 M
(500 mL)(2 M) = (31.6 M)V
V = 31.6 mL
I don't know if the process that I used is right. Which leads me to doubt the answer. Any help is much appreciated.