Calculate the weight aof anhydrous sodium carbonate present in 200 mL of 0.2 M solution.

Answer

To calculate the weight of anhydrous sodium carbonate (Na2CO3) present in 200 mL of a 0.2 M solution, we can follow these steps:

### Step 1: Determine the Molar Mass of Anhydrous Sodium Carbonate
The molar mass of Na2CO3 can be calculated as follows:
- Sodium (Na): 23 g/mol (2 sodium atoms contribute: 2 × 23 = 46 g/mol)
- Carbon (C): 12 g/mol (1 carbon atom contributes: 1 × 12 = 12 g/mol)
- Oxygen (O): 16 g/mol (3 oxygen atoms contribute: 3 × 16 = 48 g/mol)

Adding these together:
\[ \text{Molar Mass of Na2CO3} = 46 + 12 + 48 = 106 \, \text{g/mol} \]

### Step 2: Use the Molarity Formula
Molarity (M) is defined as the number of moles of solute per liter of solution. The formula can be expressed as:
\[ \text{Molarity} = \frac{\text{Number of moles of solute}}{\text{Volume of solution in liters}} \]

Given that the molarity is 0.2 M and the volume of the solution is 200 mL (which is 0.2 L), we can set up the equation:
\[ 0.2 = \frac{\text{Number of moles}}{0.2} \]

### Step 3: Calculate the Number of Moles
From the equation above, we can rearrange to find the number of moles:
\[ \text{Number of moles} = 0.2 \times 0.2 = 0.04 \, \text{moles} \]

### Step 4: Calculate the Weight of Anhydrous Sodium Carbonate
Using the number of moles and the molar mass, we can find the weight (W) of Na2CO3:
\[ W = \text{Number of moles} \times \text{Molar Mass} \]
\[ W = 0.04 \, \text{moles} \times 106 \, \text{g/mol} \]
\[ W = 4.24 \, \text{grams} \]

### Final Answer
The weight of anhydrous sodium carbonate present in 200 mL of a 0.2 M solution is **4.24 grams**.

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