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Question

$$18.5 \,cm^3$$ of oxalic acid was completely neutralized by $$20.0 \,cm^3, 0.125 \,N$$ base. Calculate the (a) normality (b) molarity and (c) mass of oxalic acid crystals in $$1 \,dm^3$$ of solution.

Solution
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(a) Normality of acid $$= \dfrac{(V \times N)_{base}}{Volume \,of \,acid} = \dfrac{20.0 \,cm^3 \times 0.125 \,N}{18.5 \,cm^3} = 0.1351 \,N.$$

Oxalic acid is a dibasic acid.

(b ) $$\therefore$$ Molarity of acid $$= \dfrac{Normality \,of \,acid}{2} = 0.06757 \,M.$$

(c) Mass of oxalic acid crystals in $$1 \,dm^3$$ of solution = normality $$\times$$ equivalent mass $$= 0.1351 \,N \times 63.0$$ (equivalent mass of oxalic acid crystals $$= 63.0$$) = $$8.513 \,g.$$

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