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 sodium carbonate (Na₂CO₃) required to prepare a 250 ml solution with a molarity of 0.25 M, follow these steps:

### Step 1: Convert the volume of the solution from milliliters to liters.
- Given volume = 250 ml
- To convert ml to liters, use the conversion factor: 1 L = 1000 ml.

\[ \text{Volume in liters} = \frac{250 \, \text{ml}}{1000} = 0.250 \, \text{L} \]

### Step 2: Calculate the number of moles of Na₂CO₃ needed.
- Molarity (M) is defined as moles of solute per liter of solution.
- Given molarity = 0.25 M

Using the formula:

\[ \text{Moles of Na}_2\text{CO}_3 = \text{Molarity} \times \text{Volume in liters} \]

\[ \text{Moles of Na}_2\text{CO}_3 = 0.25 \, \text{M} \times 0.250 \, \text{L} = 0.0625 \, \text{moles} \]

### Step 3: Calculate the molar mass of Na₂CO₃.
- The molar mass of Na₂CO₃ can be calculated as follows:
- Sodium (Na) = 23 g/mol (2 atoms)
- Carbon (C) = 12 g/mol (1 atom)
- Oxygen (O) = 16 g/mol (3 atoms)

\[ \text{Molar mass of Na}_2\text{CO}_3 = (2 \times 23) + (1 \times 12) + (3 \times 16) \]

\[ = 46 + 12 + 48 = 106 \, \text{g/mol} \]

### Step 4: Calculate the mass of Na₂CO₃ required.
- Use the formula:

\[ \text{Mass} = \text{Moles} \times \text{Molar mass} \]

\[ \text{Mass of Na}_2\text{CO}_3 = 0.0625 \, \text{moles} \times 106 \, \text{g/mol} = 6.625 \, \text{g} \]

### Final Answer:
The mass of anhydrous Na₂CO₃ required to prepare 250 ml of a 0.25 M solution is **6.625 g**.

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