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Calculation of molarity - please help

  1. Jul 27, 2008 #1

    I have to make up a 0.5N solution of HNO3 (69%). I have attempted to do the calculation but am not confident if it is correct. Please see below:

    no. moles = cv/1000

    c = 0.5N v = 100ml

    so no.moles = 0.5 x 100/ 1000 = 0.05 moles

    mass = 0.05moles/63 = 3.15g

    HN03 (69%) = 2.17g

    Therefore 2.17g of HNO3 (69%) in 100ml water = 0.5N

    Is this correct? (As someone else seems to have got 4.56g as the answer and I dont know where they obtained this from)

    Thank you very much for any help.
  2. jcsd
  3. Jul 27, 2008 #2


    User Avatar

    Staff: Mentor

    0.5N or 0.5M? N stands for normality, and in the case of nitric acid normality may but may not equal molarity.

    You are rihht that you need 3.15g of nitiric acid, however, your conversion to mass of 69% solution is wrong. That's obvious at the first sight - you need 3.15 g of acid, yet you plan to use 2.17 g of solution that is not pure acid - in no way 2.17 g will contain 3.15 g.

    Try to start with percentage definition and solve it for the mass of the solution. You will know where 4.56 came from.
  4. Jul 28, 2008 #3
    ok thanks for you help, i understand where I've gone wrong. I should have divided 3.15g and not multiplied.
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