Calculate the mass of anhydrous `Ca_3(PO_4)_2` present in 500 ml of 0.25 M solution.

Options

A. 38.75g
B. 35g
C. 31.45g
D. 32g

Explanation

To calculate the mass of anhydrous \( \text{Ca}_3(\text{PO}_4)_2 \) present in 500 ml of a 0.25 M solution, we can follow these steps:

### Step 1: Calculate the number of moles of \( \text{Ca}_3(\text{PO}_4)_2 \)

The number of moles can be calculated using the formula:

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

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

Now, substituting the values:

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

### Step 2: Calculate the molar mass of \( \text{Ca}_3(\text{PO}_4)_2 \)

To find the molar mass, we need to sum the atomic masses of all the atoms in the formula \( \text{Ca}_3(\text{PO}_4)_2 \):

- Calcium (Ca): 40.08 g/mol
- Phosphorus (P): 30.97 g/mol
- Oxygen (O): 16.00 g/mol

Calculating the molar mass:

\[ \text{Molar mass} = (3 \times 40.08) + (2 \times 30.97) + (8 \times 16.00) \]

Calculating each part:

- Calcium: \( 3 \times 40.08 = 120.24 \, \text{g/mol} \)
- Phosphorus: \( 2 \times 30.97 = 61.94 \, \text{g/mol} \)
- Oxygen: \( 8 \times 16.00 = 128.00 \, \text{g/mol} \)

Now, summing these values:

\[ \text{Molar mass} = 120.24 + 61.94 + 128.00 = 310.18 \, \text{g/mol} \]

### Step 3: Calculate the mass of \( \text{Ca}_3(\text{PO}_4)_2 \)

Now, we can calculate the mass using the number of moles and the molar mass:

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

Substituting the values:

\[ \text{Mass} = 0.125 \, \text{mol} \times 310.18 \, \text{g/mol} = 38.77 \, \text{g} \]

### Final Answer

The mass of anhydrous \( \text{Ca}_3(\text{PO}_4)_2 \) present in 500 ml of 0.25 M solution is approximately **38.77 grams**.

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