Question

The freezing point of an aqueous 0.050 m cacl2solution is −0.27 °c. what is the van’t hoff factor (i) for cacl2at this concentration? how does it compare to the expected value of i?

Answer 1

Then the answer is 2.9

$\begin{aligned} \Delta T_{f} &=i \times m \times k_{f} \\ i &=\frac{\Delta T_{f}}{m \times k_{f}} \end{aligned}$

$=\frac{0.27^{\circ} \AC}{0.050 \times \frac{1.86^{\circ} \mathrm{C}}{\mathrm{m}}}$

$\therefore i^{2}=2.9$

Using van’t hoff factor the answer has no unit, as, expected since i is a ratio. The magnitude is about right since it is close to the value ace would expect upon the complete dissociation $\mathrm{CaCl}_{2}$.

What is Van't Hoff factor?

• The Van't Hoff factor is always positive and can never be negative. When the solute remains completely undissociated in solution, the Van't Hoff factor is one; it is greater than one for salts and acids and less than one for the solute that associates when dissolved to form a solution.

• The van't Hoff factor is defined as the ratio of the observed colligative property produced by a given concentration of electrolyte solution to the observed colligative property produced by the same concentration of non-electrolyte solution.

To learn more about van't Hoff factor visit: brainly.com/question/24598605

#SPJ4