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18. Calculate the molar mass, by titration calculations, of oxalic acid (H2C204) if 25.00 mL of a 0.750
M KOH solution were required to titrate 0.846 grams of oxalic acid.

Sagot :

Explanation:

[tex]1000 \: ml \: contain \: 0.750 \: moles \\ 25.00 \: ml \: will \: contain \: ( \frac{25 \times 0.75}{1000} ) \: moles \\ = 0.01875 \: moles \\ from \: reaction \: eqn \\ H _{2} C _{2} 0 _{4} + 2KOH - - > K _{2}C _{2} 0 _{4} +2H _{2} 0 \\ 1 \: mole \: of \: oxalic \: reacts \: with \: 2 \: moles \: of \: KOH \\ 0.01875 \: moles \: react \: with \: ( \frac{0.01875 \times 2}{1} ) \\ = 0.0375 \: moles \\ 0.0375 \: moles \: weigh \: 0.846g \\ 1 \: mole \: weighs \: 56g[/tex]

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