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C asked in Science & MathematicsChemistry · 5 years ago

Chemistry Question... Due date is quickly approaching & I have been trying to figure this out with no such luck... PLEASE HELP?

How many total ions are present in 20.0 mL of a 1.50 M solution of Ca(NO3)2?

I remember that M= moles/Liters & 1 L= 1000 mL so 20.0 mL=0.02 L but I don't know where to go from there...

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  • KennyB
    Lv 7
    5 years ago

    First, note that this molecule breaks into three ions - one Ca+2 and two NO3-2

    Second, volume times molarity gives moles - 20.0 mL x 1.50 M is 30 millimoles.

    3 x 10^-2 moles x 6.02 x 10^23 molecules/mole x 3 ions/molecule is about 5.4 x 10^22 ions.

    Check my math (I just SWAGed it) and report the answer to three significant figures.

  • 5 years ago

    Solution: Find the number of moles and multiply by Avagadro's Number.

    M= moles/Liters

    moles= M*Liters

    moles= (1.5M)(.02L)= .03 moles of Ca(NO3)2

    Now convert from moles to molecules (ions in this case):

    .03 x (6.02x10^23) = 1.806x10^22 molecules of Calcium Nitrate

    Now when dissolved it will be Ca+2 and NO3- ***NOTE We have 2 nitrate ions for every 1 Calcium*

    Multiply the number of molecules by the number of ions:

    (1.806x10^22)x3 = 5.418 x 10^22 ions in the solution

    Source(s): Chemistry Major
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