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If 92.75 mg of Na2CO3 is dissolved in 0.5mL of water, what would be the normality of the solution after the addition of 0.67 mL of water.

  • chemistry -

    0.09275g/molar mass = mols Na2CO3
    (We assume, I suppose, that the volumes of water are additive--they aren't always additive) so add 0.67 mL + 0.50 mL and convert to L.
    M = mols/L.
    For N = M x 2.

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