Question

if a 0.510 $m$ aqueous solution freezes at $-2.90\\ ^\circ\text{c}$, what is the vant hoff factor, $i$, of the solute?
consult the table of $k_f$ values.
$i = $

Explanation:

Step1: Recall freezing point depression formula

The freezing point depression equation is $\Delta T_f = i \times K_f \times m$, where $\Delta T_f$ is the change in freezing point, $i$ is the van't Hoff factor, $K_f$ is the cryoscopic constant for water, and $m$ is molality.

Step2: Calculate $\Delta T_f$

Pure water freezes at $0^\circ\text{C}$, so $\Delta T_f = 0^\circ\text{C} - (-2.90^\circ\text{C}) = 2.90^\circ\text{C}$

Step3: Identify $K_f$ for water

For aqueous solutions, $K_f = 1.86^\circ\text{C}/m$

Step4: Rearrange formula to solve for $i$

Rearrange to $i = \frac{\Delta T_f}{K_f \times m}$

Step5: Substitute values and calculate

$i = \frac{2.90^\circ\text{C}}{1.86^\circ\text{C}/m \times 0.510\ m}$
$i = \frac{2.90}{0.9486} \approx 3.06$

Answer:

$3.06$