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25 July, 21:00

A solution is prepared by dissolving 0.7236 g oxalic acid (H2C2O4) in enough water to make 100.0 mL of solution. A 10.00-mL aliquot (portion) of this solution is then diluted to a final volume of 250.0 mL. What is the final molarity of the diluted oxalic acid solution?

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  1. 26 July, 00:07
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    C H2C2O4 = 3.216 E-3 M

    Explanation:

    m H2C2O4 = 0.7236 g

    ∴ Mw H2C2O4 = 90.03 g/mol

    ⇒ n H2C2O4 = (0.7236 g) (mol/90.03 g) = 8.04 E-3 mol

    ⇒ C H2C2O4 = (8.04 E-3 mol / 0.1 L) = 0.0804 mol/L

    dilution factor (f):

    ∴ f = 10 mL/250 mL = 0.04

    final molarity:

    ⇒ C H2C2O4 = (0.0804mol/L) (0.04) = 3.216 E-3 M
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