The mass of `Na_(2)CO_(3)` required to prepare 500ml of `0.1M` solution is

Options

A. `10.6g`
B. `5.3g`
C. `2.65g`
D. `7.95g`

Explanation

To find the mass of Na₂CO₃ required to prepare 500 ml of a 0.1 M solution, we can follow these steps:

### Step 1: Understand the formula for molarity
Molarity (M) is defined as the number of moles of solute per liter of solution. The formula for molarity is:
\[ M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]

### Step 2: Convert volume from ml to liters
Since the volume of the solution is given in milliliters, we need to convert it to liters:
\[ 500 \, \text{ml} = 0.5 \, \text{liters} \]

### Step 3: Calculate the number of moles of Na₂CO₃
Using the molarity formula, we can rearrange it to find the number of moles of Na₂CO₃:
\[ \text{moles of Na₂CO₃} = M \times \text{volume in liters} \]
\[ \text{moles of Na₂CO₃} = 0.1 \, \text{M} \times 0.5 \, \text{L} = 0.05 \, \text{moles} \]

### Step 4: Calculate the molar mass of Na₂CO₃
To find the mass of Na₂CO₃ needed, we first need to calculate its molar mass:
- Sodium (Na) = 23 g/mol, and there are 2 sodium atoms: \( 2 \times 23 = 46 \, \text{g/mol} \)
- Carbon (C) = 12 g/mol: \( 1 \times 12 = 12 \, \text{g/mol} \)
- Oxygen (O) = 16 g/mol, and there are 3 oxygen atoms: \( 3 \times 16 = 48 \, \text{g/mol} \)

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

### Step 5: Calculate the mass of Na₂CO₃ required
Now, we can calculate the mass of Na₂CO₃ required using the number of moles and the molar mass:
\[ \text{mass of Na₂CO₃} = \text{moles} \times \text{molar mass} \]
\[ \text{mass of Na₂CO₃} = 0.05 \, \text{moles} \times 106 \, \text{g/mol} = 5.3 \, \text{g} \]

### Conclusion
The mass of Na₂CO₃ required to prepare 500 ml of a 0.1 M solution is **5.3 grams**.

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