Calculate the mass of anhydrous `Ca_3(PO_4)_2` present in 250 ml of 0.25 M solution.

Options

A. 15.42g
B. 21.89g
C. 16.65g
D. 19.37g

Explanation

To solve the problem of calculating the mass of anhydrous \( \text{Ca}_3(\text{PO}_4)_2 \) present in 250 ml of a 0.25 M solution, we can follow these steps:

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

To find the number of moles, we can use the formula:

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

Given:
- Molarity = 0.25 M
- Volume = 250 ml = 0.250 L (since 1 L = 1000 ml)

Now, substituting the values:

\[ \text{Number of moles} = 0.25 \, \text{mol/L} \times 0.250 \, \text{L} = 0.0625 \, \text{moles} \]

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

To calculate the molar mass, we need to sum the atomic masses of all the atoms in the formula:

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

The formula \( \text{Ca}_3(\text{PO}_4)_2 \) contains:
- 3 Calcium atoms
- 2 Phosphorus atoms
- 8 Oxygen atoms (since there are 4 oxygen atoms in each phosphate group and there are 2 phosphate groups)

Calculating the molar mass:

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

Calculating each part:

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

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

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

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

Substituting the values:

\[ \text{Mass} = 0.0625 \, \text{moles} \times 310.18 \, \text{g/mol} \approx 19.37 \, \text{grams} \]

### Final Answer

The mass of anhydrous \( \text{Ca}_3(\text{PO}_4)_2 \) present in 250 ml of 0.25 M solution is approximately **19.37 grams**.

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