The amount of anhydrous `Na_2 CO_3` present in 250 ml of 0.25 M solution is

Options

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

Answer

To find the amount of anhydrous Na₂CO₃ present in 250 mL of a 0.25 M solution, we can follow these steps:

### Step 1: Understand the formula for molarity
Molarity (M) is defined as:
\[ M = \frac{\text{moles of solute}}{\text{liters of solution}} \]
From this, we can rearrange the formula to find the moles of solute:
\[ \text{moles of solute} = M \times \text{liters of solution} \]

### Step 2: Convert the volume of the solution from mL to L
Given that the volume of the solution is 250 mL, we need to convert this to liters:
\[ \text{Volume in L} = \frac{250 \text{ mL}}{1000} = 0.250 \text{ L} \]

### Step 3: Calculate the moles of Na₂CO₃
Now we can use the molarity to find the moles of Na₂CO₃:
\[ \text{moles of Na₂CO₃} = 0.25 \, \text{M} \times 0.250 \, \text{L} = 0.0625 \, \text{moles} \]

### Step 4: Calculate the molecular weight of Na₂CO₃
To find the mass of Na₂CO₃, we need its molecular weight. The molecular weight is calculated as follows:
- Sodium (Na): 23 g/mol (2 atoms)
- Carbon (C): 12 g/mol (1 atom)
- Oxygen (O): 16 g/mol (3 atoms)

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

### Step 5: Calculate the mass of Na₂CO₃
Now, we can find the mass of Na₂CO₃ using the number of moles and the molecular weight:
\[ \text{mass of Na₂CO₃} = \text{moles} \times \text{molecular weight} \]
\[ = 0.0625 \, \text{moles} \times 106 \, \text{g/mol} \]
\[ = 6.625 \, \text{g} \]

### Final Answer
The amount of anhydrous Na₂CO₃ present in 250 mL of 0.25 M solution is **6.625 grams**.

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