The amount of anhydrous `Na_2 CO_3` present in 250 ml of 0.25 M solution is

Options

A. 6.225 g
B. 66.25 g
C. 6.0 g
D. 6.625 g

Answer

To find the amount of anhydrous Na₂CO₃ present in 250 ml of a 0.25 M solution, 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 solution in liters.
\[ M = \frac{\text{Number of moles of solute}}{\text{Volume of solution in liters}} \]

2. **Convert the volume from ml to liters**:
Since the volume is given in milliliters (ml), we need to convert it to liters.
\[ 250 \text{ ml} = \frac{250}{1000} \text{ L} = 0.250 \text{ L} \]

3. **Calculate the number of moles of Na₂CO₃**:
Rearranging the molarity formula gives us:
\[ \text{Number of moles of solute} = M \times \text{Volume in liters} \]
Substituting the values:
\[ \text{Number of moles of Na₂CO₃} = 0.25 \text{ M} \times 0.250 \text{ L} = 0.0625 \text{ moles} \]

4. **Calculate the molecular weight of Na₂CO₃**:
The molecular weight (molar mass) of Na₂CO₃ can be calculated as follows:
- Sodium (Na) = 23 g/mol, and there are 2 sodium atoms: \( 23 \times 2 = 46 \text{ g/mol} \)
- Carbon (C) = 12 g/mol, and there is 1 carbon atom: \( 12 \times 1 = 12 \text{ g/mol} \)
- Oxygen (O) = 16 g/mol, and there are 3 oxygen atoms: \( 16 \times 3 = 48 \text{ g/mol} \)
- Adding these together gives:
\[ \text{Molecular weight of Na₂CO₃} = 46 + 12 + 48 = 106 \text{ g/mol} \]

5. **Calculate the weight of Na₂CO₃**:
Now that we have the number of moles and the molecular weight, we can find the weight of Na₂CO₃ using the formula:
\[ \text{Weight} = \text{Number of moles} \times \text{Molecular weight} \]
Substituting the values:
\[ \text{Weight} = 0.0625 \text{ moles} \times 106 \text{ g/mol} = 6.625 \text{ g} \]

### Final Answer:
The amount of anhydrous Na₂CO₃ present in 250 ml of 0.25 M solution is **6.625 grams**.