The osmotic pressure of 2.22% (w/v) CaCl₂ solution at 27°C is (R = 0.08 L-atm/K-mol)
(1)4.8 J
(2)4.8 atm
(3)14.4 J
(4)14.4 atm

To determine the osmotic pressure of a 2.22% (w/v) CaCl₂ solution at 27°C, we can use the formula for osmotic pressure, which is given by the equation:

Understanding Osmotic Pressure

Osmotic pressure (π) can be calculated using the formula:

π = iCRT

• i = van 't Hoff factor (number of particles the solute dissociates into)
• C = molarity of the solution (mol/L)
• R = ideal gas constant (0.08 L·atm/K·mol)
• T = temperature in Kelvin (K)

Step 1: Calculate the van 't Hoff Factor

For calcium chloride (CaCl₂), it dissociates into three ions in solution: one calcium ion (Ca²⁺) and two chloride ions (2 Cl⁻). Therefore, the van 't Hoff factor (i) is:

i = 1 + 2 = 3

Step 2: Convert the percentage concentration to molarity

A 2.22% (w/v) solution means there are 2.22 grams of CaCl₂ in 100 mL of solution. To find the molarity, we first need to convert grams to moles:

The molar mass of CaCl₂ is approximately 110.98 g/mol (40.08 for Ca + 2 × 35.45 for Cl).

Now, we calculate the number of moles in 2.22 grams:

Number of moles = mass (g) / molar mass (g/mol) = 2.22 g / 110.98 g/mol ≈ 0.0200 mol

Since this is in 100 mL, we convert it to liters:

Molarity (C) = moles / volume (L) = 0.0200 mol / 0.1 L = 0.200 mol/L

Step 3: Convert temperature to Kelvin

The temperature in Celsius is 27°C. To convert this to Kelvin:

T(K) = 27 + 273.15 = 300.15 K

Step 4: Calculate the osmotic pressure

Now we can plug the values into the osmotic pressure formula:

π = iCRT = 3 × 0.200 mol/L × 0.08 L·atm/K·mol × 300.15 K

Calculating this gives:

π = 3 × 0.200 × 0.08 × 300.15 ≈ 14.4 atm

Final Result

The osmotic pressure of the 2.22% (w/v) CaCl₂ solution at 27°C is approximately 14.4 atm. Therefore, the correct answer is option (4) 14.4 atm.