The mass of anhydrous `Na_2CO_3` present in 100ml of 0.25M solution is

Options

A. 2.65g
B. 22.5g
C. 24.65g
D. 12g

Explanation

To find the mass of anhydrous Na₂CO₃ present in 100 ml of a 0.25 M solution, we can follow these steps:

### Step 1: Calculate the number of moles of Na₂CO₃
The number of moles can be calculated using the formula:

\[ \text{Moles} = \text{Molarity} \times \text{Volume (in liters)} \]

Given:
- Molarity (M) = 0.25 M
- Volume = 100 ml = 0.1 L (since 1000 ml = 1 L)

Now, substituting the values:

\[ \text{Moles} = 0.25 \, \text{mol/L} \times 0.1 \, \text{L} = 0.025 \, \text{moles} \]

### Step 2: Calculate the mass of Na₂CO₃
To find the mass, we use the formula:

\[ \text{Mass} = \text{Number of moles} \times \text{Molar mass} \]

The molar mass of Na₂CO₃ (Sodium Carbonate) can be calculated as follows:
- Sodium (Na) = 23 g/mol (2 Na)
- Carbon (C) = 12 g/mol (1 C)
- Oxygen (O) = 16 g/mol (3 O)

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:

\[ \text{Mass} = 0.025 \, \text{moles} \times 106 \, \text{g/mol} = 2.65 \, \text{g} \]

### Final Answer
The mass of anhydrous Na₂CO₃ present in 100 ml of 0.25 M solution is **2.65 grams**.

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