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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.

  • chemistry -

    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.

  • chemistry -

    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?

  • chemistry -

    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.

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