Question

Prepare 300mL of 0.5M HNO3 from concetrated HNO3 which is 55% w/w, density is 1.34Kg/L (molar mass HNO3=63.012g/mol)

Answer 1

First find the moles of HNO3 necessary to make the solution. Then find the mass that makes those moles, and find the volume with the density of the stock solution.

Answer 2

Take a hypothetical sample of 1.000 L of the concentrated acid: (1.000 L) x (1.34 kg/L) x (1000 g/kg) x (0.55) / (63.01296 g HNO3/mol) = 11.696 mol
So the concentration of the concentrated acid is 11.696 mol/L.
(0.300 L) x (0.5 mol/L HNO3) / (11.696 mol/L) = 12.8 mL
So take 12.8 mL of the concentrated acid and dilute it to exactly 300 mL.