Redox and Equivalent ConceptHard

Question

A quantity of 0.10 g of anhydrous organic acid requires 25 ml of 0.10 N-NaOH for neutralization. A quantity of 0.245 g of the hydrated acid requires 50 ml of the same alkali. The number of moles of water of crystallization per equivalent of the anhydrous acid is

Options

A.1.0
B.2.0
C.0.5
D.4.0

Solution

neq anhydrous acid = neq NaOH

Or $\frac{0.10}{E} = \frac{25 \times 0.1}{1000} \Rightarrow E = 40$

and neq hydrated acid = neqNaOH

Or $\frac{0.245}{E + 18x} = \frac{50 \times 0.1}{1000} \Rightarrow x = 0.5$

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