The amount of `FeSO_(4).7H_(2)O` to be weighted to prepare a solution of 100 ml of 0.1 M concentration is (Mol.wt.of `FeSO_(4)`. `7H_(2)0=278)`

Options

A. 27.8 g
B. 90 g
C. 2.78 g
D. 5.5 g

Explanation

To solve the problem of how much `FeSO4.7H2O` is needed to prepare a 100 mL solution of 0.1 M concentration, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for Molarity**:
Molarity (M) is defined as the number of moles of solute divided by the volume of the solution in liters.
\[ M = \frac{\text{Number of moles of solute}}{\text{Volume of solution in liters}} \]

2. **Convert Volume from mL to L**:
The volume of the solution is given as 100 mL. We need to convert this to liters.
\[ \text{Volume in L} = \frac{100 \text{ mL}}{1000} = 0.1 \text{ L} \]

3. **Calculate the Number of Moles of Solute**:
Using the molarity formula, we can find the number of moles of `FeSO4.7H2O`.
\[ \text{Number of moles} = M \times \text{Volume in L} = 0.1 \text{ M} \times 0.1 \text{ L} = 0.01 \text{ moles} \]

4. **Use the Relation Between Moles, Weight, and Molecular Weight**:
The relationship between the number of moles, weight (mass), and molecular weight (MW) is given by:
\[ \text{Number of moles} = \frac{\text{Weight}}{\text{Molecular Weight}} \]
Rearranging this gives:
\[ \text{Weight} = \text{Number of moles} \times \text{Molecular Weight} \]

5. **Substitute the Values**:
We know the number of moles (0.01 moles) and the molecular weight of `FeSO4.7H2O` (278 g/mol).
\[ \text{Weight} = 0.01 \text{ moles} \times 278 \text{ g/mol} = 2.78 \text{ grams} \]

6. **Final Answer**:
The amount of `FeSO4.7H2O` to be weighed to prepare the solution is **2.78 grams**.