Chapter 16, Problem 16.91P

Interpretation Introduction

Interpretation:

The number of ions produced per formula unit when the indicated substance is dissolved in aqueous solution to produce a $1.00m$ solution has to be calculated.

Formula	$\Delta T/K$
$PtC{l}_2.4N{H}_3$	$5.58$
$PtC{l}_2.3N{H}_3$	$3.72$
$PtC{l}_2.2N{H}_3$	$1.86$
$KPtC{l}_3.N{H}_3$	$3.72$
$K_{2}PtC{l}_4$	$5.58$

Concept Introduction:

Freezing point depression:

The lowering of the vapor pressure of a solvent by a solute leads to a lowering of the freezing point of the solution relative to that of the pure solvent.  This effect is called as the freezing point depression.  The mathematical relationship is given below.

  $\Delta T_f=T_f{}^{\circ }-T_f=K_f{m}_c=K_f\left(i\times m\right)$

Where, $\Delta T_f$ is freezing point depression, $T_f$ is the freezing point of solution, $T_f{}^{\circ }$ is the freezing point of pure solvent, $K_f$ is the freezing point depression constant, $i$ is van’t Hoff factor, and $m_c$ is the colligative molality.

Answer to Problem 16.91P

The number of solute particles per formula unit of $H_{2}S{O}_4\left(aq\right)$ is $2.01.$

Explanation of Solution

The value of $K_f$ for water is $1.86K.m_c{}^{-1}.$

For $PtC{l}_2.4N{H}_3:$

Given that, the freezing point depression of a $1.00m$ solution of $PtC{l}_2.4N{H}_3$ is $5.58K.$

Now, using the freezing point depression expression, the van’t Hoff factor for $PtC{l}_2.4N{H}_3$ solution can be calculated.

  $\begin{array}{l}\Delta T_f=T_f{}^{\circ }-T_f=K_f{m}_c=K_f\left(i\times m\right)\\ \\ i=\frac{\Delta T_f}{K_f\times m}=\frac{5.58K}{1.86K.m_c{}^{-1}\times 1.00m}=3\end{array}$

Therefore, the number of ions produced per formula unit of $PtC{l}_2.4N{H}_3$ is $3.$

For $PtC{l}_2.3N{H}_3:$

Given that, the freezing point depression of a $1.00m$ solution of $PtC{l}_2.3N{H}_3$ is $3.72K.$

Now, using the freezing point depression expression, the van’t Hoff factor for $PtC{l}_2.3N{H}_3$ solution can be calculated.

  $\begin{array}{l}\Delta T_f=T_f{}^{\circ }-T_f=K_f{m}_c=K_f\left(i\times m\right)\\ \\ i=\frac{\Delta T_f}{K_f\times m}=\frac{3.72K}{1.86K.m_c{}^{-1}\times 1.00m}=2\end{array}$

Therefore, the number of ions produced per formula unit of $PtC{l}_2.3N{H}_3$ is $2.$

For $PtC{l}_2.2N{H}_3:$

Given that, the freezing point depression of a $1.00m$ solution of $PtC{l}_2.2N{H}_3$ is $1.86K.$

Now, using the freezing point depression expression, the van’t Hoff factor for $PtC{l}_2.2N{H}_3$ solution can be calculated.

  $\begin{array}{l}\Delta T_f=T_f{}^{\circ }-T_f=K_f{m}_c=K_f\left(i\times m\right)\\ \\ i=\frac{\Delta T_f}{K_f\times m}=\frac{1.86K}{1.86K.m_c{}^{-1}\times 1.00m}=1\end{array}$

Therefore, the number of ions produced per formula unit of $PtC{l}_2.2N{H}_3$ is $1.$

For $KPtC{l}_3.N{H}_3:$

Given that, the freezing point depression of a $1.00m$ solution of $KPtC{l}_3.N{H}_3$ is $3.72K.$

Now, using the freezing point depression expression, the van’t Hoff factor for $KPtC{l}_3.N{H}_3$ solution can be calculated.

  $\begin{array}{l}\Delta T_f=T_f{}^{\circ }-T_f=K_f{m}_c=K_f\left(i\times m\right)\\ \\ i=\frac{\Delta T_f}{K_f\times m}=\frac{3.72K}{1.86K.m_c{}^{-1}\times 1.00m}=2\end{array}$

Therefore, the number of ions produced per formula unit of $KPtC{l}_3.N{H}_3$ is $2.$

For $K_{2}PtC{l}_4:$

Given that, the freezing point depression of a $1.00m$ solution of $K_{2}PtC{l}_4$ is $5.58K.$

Now, using the freezing point depression expression, the van’t Hoff factor for $K_{2}PtC{l}_4$ solution can be calculated.

  $\begin{array}{l}\Delta T_f=T_f{}^{\circ }-T_f=K_f{m}_c=K_f\left(i\times m\right)\\ \\ i=\frac{\Delta T_f}{K_f\times m}=\frac{5.58K}{1.86K.m_c{}^{-1}\times 1.00m}=3\end{array}$

Therefore, the number of ions produced per formula unit of $K_{2}PtC{l}_4$ is $3.$

Formula	Number of ions produced per formula unit
$PtC{l}_2.4N{H}_3$	$3$
$PtC{l}_2.3N{H}_3$	$2$
$PtC{l}_2.2N{H}_3$	$1$
$KPtC{l}_3.N{H}_3$	$2$
$K_{2}PtC{l}_4$	$3$