Chapter 16, Problem 13STP

Interpretation Introduction

Interpretation : The slopes of the three lines are to be compared and their difference is to be stated.

Concept Introduction : Molarity mainly explains the concentration of a solute and Freezing point depression is a colligative property that increases with an increase in concertation of the solute.

Answer to Problem 13STP

The freezing point depression vs molarity plot for ${\text{CaCl}}_2$ , shows a maximum increase among all three lines, and the plot for $C_2{h}_{6}O$ shows a minimum increase in freezing point depression with an increase in molarity.

Explanation of Solution

The compounds $\text{NaCl}$ and ${\text{CaCl}}_2$ are ionic compounds that dissolve in water to form ions; $C_2{h}_{6}O$ is a covalent compound and does not dissociate into ions on dissolving in water. So, $\text{NaCl}$ will form 2 ions, ${\text{CaCl}}_2$ will form 3 ions and $C_2{h}_{6}O$ remains as a single ion.

The higher the number of the ions in the solution more will be the increase in depression in the freezing point of the solution. Hence, the plot of ${\text{CaCl}}_2$ shows the maximum increase in depression in the freezing point. As ${\text{CaCl}}_2$ has the maximum number of ions in the solution and the colligative properties depend on the number of solute particles.

The dissociation of the given compound is as follows:

  $\text{NaCl}\to {\text{Na}}^{+}(aq)+{\text{ Cl}}^{-}(aq)$

  ${\text{CaCl}}_2\to C{\text{a}}^{2+}(aq)+{\text{ 2Cl}}^{-}(aq)$

The data is given as:

Molarity	Freezing point depression
	$\text{NaCl}$	${\text{CaCl}}_2$	${\text{C}}_2{H}_{6}O$
$0.5$	$1.7$	$2.6$	$0.95$
$1.0$	$3.5$	$5.6$	$2.0$
$1.5$	$5.3$	$8.3$	$3.0$
$2.0$	$7.2$	$11.2$	$4.1$
$2.5$	$9.4$	$14.0$	$5.3$

The required graph is plotted as:

  

Conclusion

The plot of freezing point depression vs molarity for ${\text{CaCl}}_2$ shows maximum increases in freezing point depression among all three due to having the maximum number of ions.