Calculate the mass of anhydrous `Na_2CO_3` required to prepare 250 ml 0.25 M solution .

Options

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

Answer

To calculate the mass of anhydrous Na₂CO₃ required to prepare a 250 ml solution with a molarity of 0.25 M, follow these steps:

### Step 1: Calculate the number of moles of Na₂CO₃
The formula for calculating the number of moles is:
\[ \text{Number of moles} = \text{Molarity} \times \text{Volume (in liters)} \]
Given:
- Molarity (M) = 0.25 M
- Volume (V) = 250 ml = 0.250 L (since 1 L = 1000 ml)

Now, substituting the values:
\[ \text{Number of moles} = 0.25 \, \text{M} \times 0.250 \, \text{L} = 0.0625 \, \text{moles} \]

### Step 2: Calculate the mass of Na₂CO₃
To find the mass, use the formula:
\[ \text{Mass} = \text{Number of moles} \times \text{Molar mass} \]
The molar mass of Na₂CO₃ can be calculated as follows:
- Sodium (Na) = 23 g/mol, and there are 2 Na atoms
- Carbon (C) = 12 g/mol, and there is 1 C atom
- Oxygen (O) = 16 g/mol, and there are 3 O atoms

Calculating the molar mass:
\[ \text{Molar mass of Na₂CO₃} = (2 \times 23) + (1 \times 12) + (3 \times 16) = 46 + 12 + 48 = 106 \, \text{g/mol} \]

Now, substituting the number of moles and the molar mass into the mass formula:
\[ \text{Mass} = 0.0625 \, \text{moles} \times 106 \, \text{g/mol} = 6.625 \, \text{grams} \]

### Final Answer
The mass of anhydrous Na₂CO₃ required to prepare the solution is **6.625 grams**.

---