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

The osmotic pressure formula is given by $\pi = iCRT$, where $i$ is the van't Hoff factor, $C$ is the molar concentration, $R$ is the gas constant, and $T$ is the temperature in Kelvin.

Given values:
Mass of $\text{CaCl}_{2} = 3.42 \text{ g}$
Molar mass of $\text{CaCl}_{2} = 111 \text{ g/mol}$
Volume of solution $V = 2500 \text{ ml} = 2.5 \text{ L}$
Osmotic pressure $\pi = 0.75 \text{ atm}$
Temperature $T = 27^{\circ}\text{C} = 300 \text{ K}$
Gas constant $R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1}$

Calculate the number of moles of $\text{CaCl}_{2}$:
$n = \dfrac{3.42}{111} \approx 0.03081 \text{ mol}$

Calculate the molar concentration $C$:
$C = \dfrac{n}{V} = \dfrac{0.03081}{2.5} = 0.012324 \text{ M}$

Using the osmotic pressure equation $\pi = iCRT$:
$0.75 = i \times 0.012324 \times 0.0821 \times 300$

$0.75 = i \times 0.303586$

$i = \dfrac{0.75}{0.303586} \approx 2.47$

Answer: 2.47