STATEMENT-1 : The equivalent mass of Mohr's salt is `(M)/(4)` [If M is molecular mass of Mohr's salt]
and
STATEMENT-2: The normality of Mohr's salt is higher than molarity for same amount and volume at constant temperature.

Options

A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-7
B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-7
C. Statement-1 is True, Statement-2 is False
D. Statement-1 is False, Statement-2 is True

Answer

To solve the problem, we need to analyze both statements regarding Mohr's salt and determine their validity.

### Step 1: Understanding Mohr's Salt
Mohr's salt is represented as \( \text{(NH}_4)_2\text{SO}_4 \cdot \text{FeSO}_4 \cdot 6\text{H}_2\text{O} \). It contains ammonium ions and ferrous ions.

### Step 2: Finding the Equivalent Mass
The equivalent mass of a compound is calculated using the formula:
\[ \text{Equivalent mass} = \frac{\text{Molecular mass}}{n \text{ factor}} \]
Where \( n \text{ factor} \) is the total change in oxidation state per formula unit of the compound.

### Step 3: Determining the Oxidation States
1. The oxidation state of ammonium \( \text{NH}_4^+ \) is +1. Since there are 2 ammonium ions:
\[ \text{Total from ammonium} = 2 \times (+1) = +2 \]
2. The oxidation state of iron in \( \text{FeSO}_4 \) is +2.

### Step 4: Total Oxidation State
Adding these together gives:
\[ \text{Total oxidation state} = +2 (\text{from NH}_4) + +2 (\text{from Fe}) = +4 \]
Thus, the \( n \text{ factor} \) for Mohr's salt is 4.

### Step 5: Calculating Equivalent Mass
Now, substituting into the equivalent mass formula:
\[ \text{Equivalent mass} = \frac{M}{4} \]
This confirms that **Statement 1** is true.

### Step 6: Understanding Normality and Molarity
- **Molarity (M)** is defined as:
\[ \text{Molarity} = \frac{\text{Number of moles of solute}}{\text{Volume of solution in liters}} \]
- **Normality (N)** is defined as:
\[ \text{Normality} = \text{Molarity} \times n \text{ factor} \]

### Step 7: Comparing Normality and Molarity
For Mohr's salt:
\[ \text{Normality} = \text{Molarity} \times 4 \]
This means that the normality of Mohr's salt is indeed higher than its molarity.

### Conclusion
Both statements are true:
- **Statement 1**: The equivalent mass of Mohr's salt is \( \frac{M}{4} \) (True).
- **Statement 2**: The normality of Mohr's salt is higher than the molarity (True).

However, **Statement 2** does not explain **Statement 1**; they are independently true.

### Final Answer
Both statements are true, but Statement 2 is not a correct explanation for Statement 1.

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