What volume of pure nitric acid would you need to measure out in order to prepare 250cm3 of 1.0 M nitric acid?

cap'n kane -

This answer was posted in the Science forum by methos5000:

"First, please check my math before you accept the result; I don't have a calculator handy.
I'd be impressed if you could find pure nitric acid, but "concentrated" nitric acid is >98.5%, so it's close enough that we'll treat it as pure unless you need very exact quantities. You need two more pieces of information to calculate this, the density of concentrated nitric (1.5 g/ml), and the molecular weight of nitric acid (HNO3 -> 1+14+16*3 = 63 g/mol).
take the volume (250 cm3 = 0.250 L) and multiply it by the concentration (0.1 M) to get the moles of nitric in the final solution (0.250L*0.1mol/L = 0.025 mol). multiply this by the molecular weight to find the mass of nitric in the final solution (0.025 mol * 63 g/mol = 1.575 g). divide this by the density to find the volume of nitric needed (1.575 g / 1.5 g/ml = 1.05 ml)."