The mass of anhydrous `Na_2CO_3` present in 500ml of 0.25M solution is

Options

A. 66.25g
B. 6.625g
C. 656g
D. 13.25g

Explanation

To find the mass of anhydrous \( \text{Na}_2\text{CO}_3 \) present in a 500 mL solution with a molarity of 0.25 M, we can follow these steps:

### Step 1: Calculate the number of moles of \( \text{Na}_2\text{CO}_3 \)

We use the formula for molarity:

\[ \text{Molarity (M)} = \frac{\text{Number of moles}}{\text{Volume in liters}} \]

Rearranging this formula gives us:

\[ \text{Number of moles} = \text{Molarity} \times \text{Volume in liters} \]

Given:
- Molarity = 0.25 M
- Volume = 500 mL = 0.5 L (since 1 L = 1000 mL)

Now, substituting the values:

\[ \text{Number of moles} = 0.25 \, \text{mol/L} \times 0.5 \, \text{L} = 0.125 \, \text{moles} \]

### Step 2: Calculate the mass of \( \text{Na}_2\text{CO}_3 \)

To find the mass, we use the formula:

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

The molar mass of \( \text{Na}_2\text{CO}_3 \) 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)

Calculating the molar mass:

\[ \text{Molar mass of } \text{Na}_2\text{CO}_3 = (2 \times 23) + (1 \times 12) + (3 \times 16) = 46 + 12 + 48 = 106 \, \text{g/mol} \]

Now substituting the number of moles and the molar mass into the mass formula:

\[ \text{Mass} = 0.125 \, \text{moles} \times 106 \, \text{g/mol} = 13.25 \, \text{grams} \]

### Final Answer

The mass of anhydrous \( \text{Na}_2\text{CO}_3 \) present in 500 mL of 0.25 M solution is **13.25 grams**.

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