What weight of ferrous ammonium sulphate is needed to prepare 100 ml of 0.1 normal solution (mol. wt. 392)

Options

A. 39.2 gm
B. 3.92 gm
C. 1.96 gm
D. 19.6 gm

Explanation

To find the weight of ferrous ammonium sulfate needed to prepare 100 ml of a 0.1 normal solution, we can follow these steps:

### Step 1: Understand Normality
Normality (N) is defined as the number of equivalents of solute per liter of solution. For ferrous ammonium sulfate (Fe(NH₄)₂(SO₄)₂·6H₂O), we need to determine the number of equivalents.

### Step 2: Calculate the Equivalent Weight
The equivalent weight is calculated as:
\[ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{n} \]
where \( n \) is the number of equivalents per mole. For ferrous ammonium sulfate, it typically provides one ferrous ion (Fe²⁺) per formula unit, thus \( n = 1 \).

Given:
- Molecular Weight = 392 g/mol
- Therefore, Equivalent Weight = 392 g/mol.

### Step 3: Use the Normality Formula
The formula for normality is given by:
\[ N = \frac{\text{Number of Gram Equivalents}}{\text{Volume in Liters}} \]
We can rearrange this to find the number of gram equivalents:
\[ \text{Number of Gram Equivalents} = N \times \text{Volume in Liters} \]
For our case:
- \( N = 0.1 \)
- Volume = 100 ml = 0.1 L

Calculating the number of gram equivalents:
\[ \text{Number of Gram Equivalents} = 0.1 \, \text{N} \times 0.1 \, \text{L} = 0.01 \, \text{equivalents} \]

### Step 4: Calculate the Weight of Ferrous Ammonium Sulfate
Now, we can find the weight of ferrous ammonium sulfate required using:
\[ \text{Weight} = \text{Number of Gram Equivalents} \times \text{Equivalent Weight} \]
Substituting the values:
\[ \text{Weight} = 0.01 \, \text{equivalents} \times 392 \, \text{g/equivalent} = 3.92 \, \text{g} \]

### Final Answer
The weight of ferrous ammonium sulfate needed to prepare 100 ml of a 0.1 normal solution is **3.92 grams**.

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