The mass of anhydrous `Na_2CO_3` present in 350 ml of 0.25 M solution is

Options

A. 6.625g
B. 9.275g
C. 4.758g
D. 66.25g

Explanation

To find the mass of anhydrous Na₂CO₃ present in a 350 ml of 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 (V) = 350 ml = 0.350 liters (since 1 L = 1000 ml)

Now, substituting the values:
\[ \text{Moles of Na₂CO₃} = 0.25 \, \text{M} \times 0.350 \, \text{L} = 0.0875 \, \text{moles} \]

### Step 2: Calculate the mass of Na₂CO₃
To calculate 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 = 46 g/mol
- Carbon (C) = 12 g/mol × 1 = 12 g/mol
- Oxygen (O) = 16 g/mol × 3 = 48 g/mol

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

Now, substituting the values to find the mass:
\[ \text{Mass of Na₂CO₃} = 0.0875 \, \text{moles} \times 106 \, \text{g/mol} = 9.275 \, \text{grams} \]

### Final Answer:
The mass of anhydrous Na₂CO₃ present in 350 ml of 0.25 M solution is **9.275 grams**.

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