2.505 g of hydrated dibasic acid 35 ml of 1N 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 _____________.
(Round off to the nearest integer)

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answer is 2.

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Detailed Solution

Let formula of hydrated dibasic acid be $H_{2} {A.xH}_{2} O$
Equivalent of diabasic acid = Equivalent of NaOH
$\frac{2.505×2}{M} = \frac{35}{1000} ×1 M= \frac{2.505 \times 2 \times 1000}{35 \times 1} = 143.14 H_{2} {A.xH}_{2} O \overset{Δ}{\to} H_{2} {A+xH}_{2} O$

Mole of $H_{2} O$ formed $= x \times \frac{1.01}{143.14}$
X = 2.28
Degree of hydration is $\approx 2$ .