Chapter 16, Problem 14STP

Interpretation Introduction

Interpretation : The new line approximate is to be determined if similar data is taken and the fourth line is added to the graph.

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

Answer to Problem 14STP

The freezing point depression vs molarity plot for $\text{KOH}$ , will be similar to the plot of $\text{NaCl}$ among all three lines of the graph.

Explanation of Solution

The compound $\text{NaCl}$ and ${\text{CaCl}}_2$ are ionic compounds, on dissolving in water form ions while $C_2{h}_{6}O$ 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 compound $\text{KOH}$ is also an ionic compound and it will dissociate into 2 ions, that is similar to that of $\text{NaCl}$ . Hence, the freezing point depression vs molarity plot for $\text{KOH}$ , will be similar to the plot of $\text{NaCl}$ among all three lines of the graph.

The dissociation of the given compound is as follows:

• $\text{NaCl}\to {\text{Na}}^{+}(aq)+{\text{ Cl}}^{-}(aq)$
• $\text{KOH}\to K^{+}(aq)+{\text{ OH}}^{-}(aq)$

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

No. of ions given by $\text{NaCl}$ in the solution is $2$ .

  $C_2{h}_{6}O$ remains as a single ion

No. of ions given by $\text{KOH}$ in the solution is $2$ .

No. of ions given by ${\text{CaCl}}_2$ in the solution is $2$ .

The freezing point depression vs molarity plot for $\text{KOH}$ , will be similar to the plot of $\text{NaCl}$ among all three lines of the graph.