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Question

Calculate the normality of oxalic acid solution containing $$0.895 \,g$$ crystals in $$250 \,cm^3$$ of its solution.

Solution
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Molecular formula of oxalic acid crystals $$= H_2C_2O_4.2H_2O = 126$$

Equivalent mass $$= \dfrac{molecular \,mass}{basicity} = \dfrac{126}{2} = 63$$

$$Mass / lit = N \times g. \,eq. \,mass$$

$$\dfrac{Mass}{250 \,ml} = \dfrac{N \times g. \,eq. \,mass}{4} \Rightarrow N \ times g. \,eq. \,mass = \dfrac{4 \times Mass}{250 \,ml}$$

$$N = \dfrac{\dfrac{4 \times Mass}{250 \,ml}}{g. eq. mass} = \dfrac{4 \times 0.895}{63} = 0.0568.$$

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