posted by Sarah on .
A commercial sample of concentrated aqueous nitric acid (HNO3) is 18.2% nitric acid base mass and has a density of 1.12 g/mL.
(a) What is the molarity (M) of the HNO3 solution?
(b) What is the molality (m) of the HNO3 solution?
I know that....
(a) molarity = moles of solute/liters of solution
(b) molality = moles of solute/kg of solvent
But, I don't know how to find the moles of HNO3 (63 amu) with the information I have been given.
assume some volume of the acid (it wont matter, it divides out).
Assume a volume of 2liters.
massnitric=.182*density*volume and you know density, and volume.
Now compute the moles of nitric given mass and its mole mass.
So then the mass would equal .407 g?
And that would be .401/63 = .0065 mol?
Is that correct so far?
I don't really understand how I am able to just pic a random value for volume. The volume effects the mass, which effects the moles, right?
You are not correct so far. I think you must have multiplied 2 L x 0.182 x 1.12. Note that the density is in g/mL so the units don't come out right.