1.245 g of a sample of `CuSO_4.xH_2O` was dissolved in water and `H_2S` passed till CuS was complete precipitated. The filtrate contained liberated `H_2SO_4`, which required 20 " mL of " `(N)/(2)` NaOH for complete neutralisation. Calculate x, the number of molecules of water associated with `CuSO_4(Cu=63.6)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the equivalents of NaOH used
Given that 20 mL of \( \frac{N}{2} \) NaOH is used for neutralization, we can calculate the equivalents of NaOH.

\[ \text{Equivalents of NaOH} = \text{Volume (L)} \times \text{Normality} = 0.020 \, \text{L} \times \frac{N}{2} = 0.020 \times 0.5 = 0.010 \, \text{equivalents} \]
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