Calculate the mass of anhydrous `Ca_3(PO_4)_2` present in 1L of 0.25 M solution.

Options

A. 7.75g
B. 70.45g
C. 25.15g
D. 77.5g

Explanation

To calculate the mass of anhydrous \( \text{Ca}_3(\text{PO}_4)_2 \) present in 1 L of a 0.25 M solution, 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:

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

Given:
- Molarity = 0.25 M
- Volume = 1 L

Substituting the values:

\[ \text{Number of moles} = 0.25 \, \text{mol/L} \times 1 \, \text{L} = 0.25 \, \text{moles} \]

### 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 (3 atoms)
- Phosphorus (P): 30.97 g/mol (2 atoms)
- Oxygen (O): 16.00 g/mol (8 atoms)

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 formula:

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

Substituting the values:

\[ \text{Mass} = 0.25 \, \text{moles} \times 310.18 \, \text{g/mol} = 77.545 \, \text{g} \]

Rounding to three significant figures, the mass of anhydrous \( \text{Ca}_3(\text{PO}_4)_2 \) is approximately:

\[ \text{Mass} \approx 77.5 \, \text{g} \]

### Final Answer:
The mass of anhydrous \( \text{Ca}_3(\text{PO}_4)_2 \) present in 1 L of a 0.25 M solution is **77.5 g**.

---