Question

What would be the van't Hoff factor for a solution prepared by dissolving 3.42 g of CaCl$_2$ in 2500 ml of water having an Osmotic pressure equal to 0.75 atm at 27°C? Molar mass of CaCl$_2$ = 111 amu.

• A. 2.7
• B. 3.15
• C. 2.47
• D. 3.0

Hint:

In the case of ionic compounds like CaCl$_2$, the van't Hoff factor corresponds to the number of ions produced in solution. For CaCl$_2$, the factor is 3 (Ca$^{2+}$ and 2 Cl$^{-}$ ions).

Answer 1

The Correct Option is C

The van't Hoff factor \( i \) is calculated using the formula for osmotic pressure: \[ \Pi = i \cdot \frac{n}{V} \cdot R \cdot T \] where: - \( \Pi \) is the osmotic pressure (0.75 atm), - \( i \) is the van't Hoff factor, - \( n \) is the number of moles of solute (CaCl$_2$), - \( V \) is the volume of the solution (2500 ml or 2.5 L), - \( R \) is the ideal gas constant (0.0821 L atm/mol K), - \( T \) is the temperature (27°C or 300 K). 1. First, we calculate the moles of CaCl$_2$: \[ n = \frac{\text{Mass of solute}}{\text{Molar mass}} = \frac{3.42 \, \text{g}}{111 \, \text{g/mol}} = 0.03081 \, \text{mol} \] 2. Now, using the osmotic pressure formula: \[ \Pi = i \cdot \frac{n}{V} \cdot R \cdot T \] Substitute the known values: \[ 0.75 = i \cdot \frac{0.03081}{2.5} \cdot 0.0821 \cdot 300 \] 3. Solving for \( i \): \[ 0.75 = i \cdot \frac{0.03081}{2.5} \cdot 24.63 \] \[ 0.75 = i \cdot 0.301 \quad \Rightarrow \quad i = \frac{0.75}{0.301} = 2.49 \] Thus, the van't Hoff factor is approximately 2.47, which matches option (C)