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If one assumes the volumes are additive, what is the concentration, in mol/L, of NO3− in a solution obtained by mixing 263 mL of 0.167 M KNO3, 311 mL of 0.404 M Mg(NO3)2, and 713 mL of H2O?

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3 answers
  1. mols NO3^- in KNO3 = M x L = ?
    mols NO3^- in Mg(NO3)2 = 2*M x L = ?
    M NO3^- = total mols/total L solution

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  2. In this problem, you need to find two things: the moles of solute (NO₃⁻) and the volume of the total solution.

    First, find the moles of NO₃⁻ we have in each solution.

    n₁ = (0.263 L)(0.167 M KNO₃) = 0.043921 mol KNO₃
    Now, since there is one mole of NO₃⁻ per one mole of KNO₃, we can say easily say that the number of moles of NO₃⁻ is also 0.043921 mol.

    n₂ = (0.311 L)(0.404 M Mg(NO₃)₂) = 0.125644 mol Mg(NO₃)₂
    IMPORTANT: since there are TWO moles per one mole of Mg(NO₃)₂, we have to say that the number of moles of NO₃⁻ is DOUBLED the number of moles in Mg(NO₃)₂. Thus, it is 0.251288 mol NO₃⁻

    Now we need to find the TOTAL number of moles of NO₃⁻.

    n = 0.043921 mol + 0.251288 mol = 0.295209 mol NO₃⁻

    Next, find the total volume. (pretty easy...)

    V = 0.263 L + 0.311 L + 0.713 L = 1.287 L

    Finally, use all the info we calculated to find the concentration of NO₃⁻ in the resulting solution.

    M = (0.295209 mol) / (1.287 L) = 0.229 M NO₃⁻

    Note: Notice how the final answer has only 3 sig figs. You do not round until the end. Remember that :)
    Thus, the NO₃⁻ concentration of the resulting solution is 0.229 M NO₃⁻

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  3. Cheers, thanks man

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