Question

if a 0.500 m aqueous solution freezes at -2.40 °c, what is the vant hoff factor, i, of the solute? consult the table of kf values.

Explanation:

Step1: Recall freezing - point depression formula

The formula for freezing - point depression is $\Delta T_f = iK_fm$, where $\Delta T_f$ is the change in freezing point, $i$ is the van't Hoff factor, $K_f$ is the cryoscopic constant of the solvent, and $m$ is the molality of the solution. For water, $K_f= 1.86^{\circ}C/m$. The normal freezing point of water is $0^{\circ}C$, so $\Delta T_f=0 - (- 2.40^{\circ}C)=2.40^{\circ}C$, and $m = 0.500m$.

Step2: Rearrange the formula to solve for $i$

From $\Delta T_f = iK_fm$, we can solve for $i$ by the formula $i=\frac{\Delta T_f}{K_fm}$.

Step3: Substitute the known values

Substitute $\Delta T_f = 2.40^{\circ}C$, $K_f = 1.86^{\circ}C/m$, and $m = 0.500m$ into the formula for $i$.
$i=\frac{2.40^{\circ}C}{1.86^{\circ}C/m\times0.500m}$
$i=\frac{2.40}{1.86\times0.500}$
$i=\frac{2.40}{0.93}\approx2.58$

Answer:

$i\approx2.58$