Determine the amount of `CaCl_(2)` (i = 2.47) dissolved in 2.5 L of water sucjh that its osmotic pressure is 0.75 atm at `27^(@)C`.

Explanation

To determine the amount of calcium chloride (CaCl₂) that should be dissolved in 2.5 L of water to achieve an osmotic pressure of 0.75 atm at 27°C, we can use the formula for osmotic pressure:

### Step-by-Step Solution:

1. **Understand the Formula for Osmotic Pressure:**
The osmotic pressure (π) is given by the formula:
\[ \pi = iCRT \]
where:
- \( \pi \) = osmotic pressure (atm)
- \( i \) = van 't Hoff factor (for CaCl₂, \( i = 2.47 \))
- \( C \) = molarity (moles of solute per liter of solution)
- \( R \) = ideal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature in Kelvin

2. **Convert Temperature to Kelvin:**
The temperature given is 27°C. To convert to Kelvin:
\[ T = 27 + 273 = 300 \text{ K} \]

3. **Rearrange the Formula to Find Molarity (C):**
Rearranging the osmotic pressure formula to solve for molarity \( C \):
\[ C = \frac{\pi}{iRT} \]

4. **Substitute the Known Values:**
Substitute the values into the equation:
- \( \pi = 0.75 \) atm
- \( i = 2.47 \)
- \( R = 0.0821 \) L·atm/(K·mol)
- \( T = 300 \) K

Thus,
\[ C = \frac{0.75}{2.47 \times 0.0821 \times 300} \]

5. **Calculate Molarity (C):**
Calculate the denominator:
\[ 2.47 \times 0.0821 \times 300 = 60.77 \]
Now calculate \( C \):
\[ C = \frac{0.75}{60.77} \approx 0.01234 \text{ mol/L} \]

6. **Calculate Moles of CaCl₂:**
The volume of the solution is 2.5 L, so the number of moles \( n \) is:
\[ n = C \times V = 0.01234 \text{ mol/L} \times 2.5 \text{ L} \approx 0.03085 \text{ mol} \]

7. **Calculate the Molar Mass of CaCl₂:**
The molar mass of CaCl₂ is calculated as follows:
- Calcium (Ca) = 40.08 g/mol
- Chlorine (Cl) = 35.45 g/mol (2 Cl atoms)
\[ \text{Molar mass of CaCl₂} = 40.08 + 2 \times 35.45 = 111.98 \text{ g/mol} \]

8. **Calculate the Mass of CaCl₂:**
To find the mass \( m \) of CaCl₂:
\[ m = n \times \text{molar mass} = 0.03085 \text{ mol} \times 111.98 \text{ g/mol} \approx 3.45 \text{ g} \]

### Final Answer:
The amount of calcium chloride (CaCl₂) that should be dissolved in 2.5 L of water to achieve an osmotic pressure of 0.75 atm at 27°C is approximately **3.45 grams**.