The osmotic pressure of a \(0.010 \mathrm{M}\) aqueous solution of \(\mathrm{CaCl}_{2}\) is found to be \(68.3 \mathrm{kPa}\) at $25^{\circ} \mathrm{C}\(. Calculate the van't Hoff factor, \)i$, for the solution.

The temperature given is in Celsius, we need to convert it to Kelvin for the osmotic pressure formula. \(T_{\text{Kelvin}}=T_{^\circ\text{C}}+273.15\) \[ T_{\text{Kelvin}}=25^\circ\mathrm{C}+273.15=298.15\; \mathrm{K} \]

The theoretical osmotic pressure is calculated using the given molarity and assuming that the solution does not dissociate, i.e., \(i=1\). Use the osmotic pressure equation and the ideal gas constant, \(R=8.314 \frac{\mathrm{J}}{\mathrm{mol} \cdot \mathrm{K}}\) (converted to \(kPa\)). \[ \Pi_{\text{theoretical}}=i_{\text{assumed}}MRT \] \[ \Pi_{\text{theoretical}}=(1)(0.010 \; \mathrm{M})(0.0821 \; \mathrm{\frac{L \cdot kPa}{mol \cdot K}})(298.15 \; \mathrm{K}) \] \[ \Pi_{\text{theoretical}}=24.5 \; \mathrm{kPa} \]

Now that we have both the experimental and theoretical osmotic pressure, we can calculate the van't Hoff factor. Rearrange the osmotic pressure equation to solve for \(i\). \[ i= \frac{\Pi}{MRT} \] Substitute the experimental osmotic pressure (\(68.3 \; \mathrm{kPa}\)) and the previously calculated values for \(M\) and \(T\). \[ i= \frac{68.3 \; \mathrm{kPa}}{(0.010 \; \mathrm{M})(0.0821 \; \mathrm{\frac{L \cdot kPa}{mol \cdot K}})(298.15 \; \mathrm{K})} \] \[ i \approx 2.79 \] The van't Hoff factor, \(i\), for the \(0.010 \; M\) aqueous solution of calcium chloride is approximately \(2.79\). This indicates that calcium chloride dissociates into ions in the solution, leading to higher osmotic pressure than theoretically expected for a non-dissociating solute.