Question

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.

Answer 1

Step-by-step explanation:

`1000 \: ml \: contain \: 0.750 \: moles \\ 25.00 \: ml \: will \: contain \: ( (25 * 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 \: ( (0.01875 * 2)/(1) ) \\ = 0.0375 \: moles \\ 0.0375 \: moles \: weigh \: 0.846g \\ 1 \: mole \: weighs \: 56g`