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2.505 g of hydrated dibasic acid requires 35 ml of 1 N NaOH solution for complete neutralization. When 1.01 g of the hydrated acid is heated to constant weight 0.72 g of the anhydrous acid is obtained. The degree of hydration (number of water molecules) of the hydrated dibasic acid is approximately _________

Correct answer

2

Step-by-step solution

Let formula of hydrated dibasic acid be H ₂ ~A xH ₂ O Equivalent of dibasic acid = Equivalent of NaOH aligned & 2.505 2 M = 35 1000 1 & M= 2.505 2 1000 35 1 =143.14 & H ₂ ~A xH ₂ O H ₂ ~A + xH ₂ O aligned Mole of H ₂ O formed =x mole of hydrated dibasic acid aligned & (1.01-0.72) 18 =x 1.01 143.14 & x=2.28 aligned degree of hydration is 2 .

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