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.

Question

(a) What is the molarity (M) of the HNO3 solution?

(b) What is the molality (m) of the HNO3 solution?

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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.

Answer 1

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.

Answer 2

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?

Answer 3

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.