1 g of a complex `[Cr(H_(2)O)_(5)Cl]Cl_(2).H_(2O` (mol. Wt. 266.5) was passed through a cation exchanger to produce HCl. The acid liberated was diluted to 1 litre. The normality of this acid solution is

Options

A. `5xx10^(-3)N`
B. `7.5xx10^(-3)N`
C. `7.5xx10^(-2)N`
D. `7.5xx10^(-1)N`

Answer

To solve the problem step by step, we will follow the outlined approach to determine the normality of the HCl solution produced from the complex.

### Step 1: Identify the Given Information
- Weight of the complex: 1 g
- Molecular weight of the complex: 266.5 g/mol

### Step 2: Calculate the Moles of the Complex
To find the number of moles of the complex, we use the formula:
\[ \text{Moles of complex} = \frac{\text{Weight of complex}}{\text{Molecular weight of complex}} \]
Substituting the values:
\[ \text{Moles of complex} = \frac{1 \text{ g}}{266.5 \text{ g/mol}} = \frac{1}{266.5} \text{ mol} \]

### Step 3: Determine the Moles of HCl Produced
From the dissociation of the complex \([Cr(H_2O)_5Cl]Cl_2 \cdot H_2O\), we know that:
- 1 mole of the complex produces 2 moles of Cl\(^-\).
- Each Cl\(^-\) corresponds to 1 mole of HCl.

Thus, the moles of HCl produced will be:
\[ \text{Moles of HCl} = 2 \times \text{Moles of complex} = 2 \times \frac{1}{266.5} = \frac{2}{266.5} \text{ mol} \]

### Step 4: Calculate the Molarity of the HCl Solution
The total volume of the diluted solution is given as 1 liter. Molarity (M) is defined as:
\[ \text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}} \]
Substituting the values:
\[ \text{Molarity of HCl} = \frac{\frac{2}{266.5}}{1} = \frac{2}{266.5} \text{ M} \]

### Step 5: Calculate the Normality of the HCl Solution
Normality (N) is related to molarity by the number of equivalents. For HCl, which provides 1 H\(^+\) ion per molecule, the normality is equal to the molarity:
\[ \text{Normality} = \text{Molarity} \times \text{Number of H}^+ \]
Here, the number of H\(^+\) ions from HCl is 1:
\[ \text{Normality} = \frac{2}{266.5} \times 1 = \frac{2}{266.5} \text{ N} \]

### Step 6: Calculate the Final Value
Calculating the numerical value:
\[ \text{Normality} = \frac{2}{266.5} \approx 0.00749 \text{ N} \]
This can be approximated as:
\[ \text{Normality} \approx 7.5 \times 10^{-3} \text{ N} \]

### Conclusion
The normality of the acid solution is approximately \(7.5 \times 10^{-3} \text{ N}\).

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