What amount of `CaCl_2` (i=2.47) is dissolved in 2 litres of water so that its osmotic pressure is 0.5 atm at `27^@` C ?

Options

A. 3.42 g
B. 9.24 g
C. 2.834 g
D. 1.820 g

Explanation

To solve the problem of determining the amount of `CaCl_2` that needs to be dissolved in 2 liters of water to achieve an osmotic pressure of 0.5 atm at 27°C, we can follow these steps:

### Step 1: Write the formula for osmotic pressure
The osmotic pressure (π) is given by the formula:
\[ \pi = i \cdot C \cdot R \cdot T \]
where:
- \( \pi \) = osmotic pressure (in atm)
- \( i \) = van 't Hoff factor (for \( CaCl_2 \), \( i = 2.47 \))
- \( C \) = concentration in mol/L
- \( R \) = universal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature in Kelvin

### Step 2: Convert temperature to Kelvin
The temperature given is 27°C. To convert this to Kelvin:
\[ T = 27 + 273 = 300 \, K \]

### Step 3: Rearrange the formula to solve for concentration (C)
Rearranging the osmotic pressure formula gives:
\[ C = \frac{\pi}{i \cdot R \cdot T} \]

### Step 4: Substitute the known values into the equation
Substituting the known values into the equation:
- \( \pi = 0.5 \, \text{atm} \)
- \( i = 2.47 \)
- \( R = 0.0821 \, \text{L·atm/(K·mol)} \)
- \( T = 300 \, K \)

We get:
\[ C = \frac{0.5}{2.47 \cdot 0.0821 \cdot 300} \]

### Step 5: Calculate the concentration (C)
Calculating the denominator:
\[ 2.47 \cdot 0.0821 \cdot 300 \approx 60.81 \]
Now, substituting this back into the equation for concentration:
\[ C = \frac{0.5}{60.81} \approx 0.00823 \, \text{mol/L} \]

### Step 6: Calculate the number of moles needed for 2 liters
To find the number of moles (n) needed for 2 liters:
\[ n = C \cdot V = 0.00823 \, \text{mol/L} \cdot 2 \, \text{L} = 0.01646 \, \text{mol} \]

### Step 7: Calculate the mass of \( CaCl_2 \)
The molar mass of \( CaCl_2 \) is approximately 110 g/mol. Therefore, the mass (m) can be calculated as:
\[ m = n \cdot \text{molar mass} = 0.01646 \, \text{mol} \cdot 110 \, \text{g/mol} \approx 1.81 \, \text{g} \]

### Final Answer
The amount of \( CaCl_2 \) that needs to be dissolved is approximately **1.81 grams**.

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