The number of gram of anhydrous Na2CO3 present in 250 ml of 0.25 M solution is :
(A) 6.625 g
(B) 0.625 g
(C) 0.567 g
(D) 7.125 g

To find the number of grams of anhydrous Na₂CO₃ (sodium carbonate) present in 250 ml of a 0.25 M solution, we can follow these steps:

### Step 1: Calculate the number of moles of Na₂CO₃

The molarity (M) of a solution is defined as the number of moles of solute (in this case, Na₂CO₃) per liter of solution. The formula to calculate the number of moles (\( n \)) is:

\[
n = M \times V
\]

where:
- \( n \) is the number of moles,
- \( M \) is the molarity of the solution (0.25 M),
- \( V \) is the volume of the solution in liters.

Since we have 250 ml of solution, we need to convert this volume into liters:

\[
V = \frac{250 \, \text{ml}}{1000} = 0.250 \, \text{L}
\]

Now, we can calculate the number of moles:

\[
n = 0.25 \, \text{M} \times 0.250 \, \text{L} = 0.0625 \, \text{moles}
\]

### Step 2: Calculate the mass of Na₂CO₃

To find the mass of Na₂CO₃, we use the formula:

\[
\text{mass} = n \times \text{molar mass}
\]

First, we need to calculate the molar mass of Na₂CO₃. The molar masses of the elements are approximately:
- Sodium (Na): 23 g/mol
- Carbon (C): 12 g/mol
- Oxygen (O): 16 g/mol

Now, calculate the molar mass of Na₂CO₃:

\[
\text{Molar mass of Na}_2\text{CO}_3 = 2(23) + 12 + 3(16) = 46 + 12 + 48 = 106 \, \text{g/mol}
\]

Now, substitute the number of moles and the molar mass into the mass formula:

\[
\text{mass} = 0.0625 \, \text{moles} \times 106 \, \text{g/mol} = 6.625 \, \text{g}
\]

### Final Answer

Thus, the number of grams of anhydrous Na₂CO₃ present in 250 ml of a 0.25 M solution is:

**(A) 6.625 g**