Chapter 13, Problem 105SCQ

Interpretation Introduction

Interpretation: van’t Hoff factor for $NaCl$ and $N{a}_{2}S{O}_4$ have to be determined. Whether the values are following the expectations based on trend in the table $13.4$ has to be identified.

Concept introduction:

Colligative properties: Properties of solutions which having influence on the concentration of the solute in it. Colligative properties are,

• Decrease in the vapor pressure
• Increase in the boiling point
• Decline in the freezing point
• Osmotic pressure

Freezing point depression: The freezing point of the solution varies with the solute concentration.

  $\Delta T_{fp}=K_{fp}{m}_{solute}$

  where,

  $K_{fp}$ is the molal freezing point depression constant.

van’t Hoff factor, i: it is the ration between change in in freezing point measured and change in in freezing point calculated. It indicates the total number of ions that are produced.

  $van\text{'}tHofffactor,i=\frac{measured\Delta {\text{T}}_{fp}}{Calculated\Delta {\text{T}}_{fp}}$

Molality (m): Molality is the number of moles of solute present in one kilogram of solvent.

  $\text{Molality (m) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1kg}\text{of}\text{solvent}}$

The number of moles of any substance can be determined using the equation

  $\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}$

Answer to Problem 105SCQ

• van’t Hoff factor for $NaCl$ is $1.82$
• van’t Hoff factor for $N{a}_{2}S{O}_4$ is $1.98$
• The continuity of trend for the value of van’t Hoff factor for $N{a}_{2}S{O}_4$ is decreased significantly and it is near to the value of van’t Hoff factor for $NaCl$. The value of van’t Hoff factor for $NaCl$ is similar to the other values.

Explanation of Solution

Given,

  $\begin{array}{c}measured\Delta T_{fp}forNaCl=-{3.05}^{0}C\\ measured\Delta T_{fp}forN{a}_{2}S{O}_4=-{1.36}^{0}C\\ {\text{K}}_{fp}\text{=}-\text{1}{\text{.86}}^{\text{0}}\text{C/m}\end{array}$

  Mass of $NaCl$ is $\frac{5}{100}\times 100g=5g$

  Molar mass of $NaCl$ is $\text{58}\text{.44}\text{g/mol}$

  Mass of $N{a}_{2}S{O}_4$ is $\frac{5}{100}\times 100g=5g$

  Molar mass of $N{a}_{2}S{O}_4$ is $\text{142}\text{.04}\text{g/mol}$

  Mass of water is $100g-5g=95g$

The number of moles of any substance can be determined using the equation

  $\begin{array}{c}\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}\\ \text{=}\frac{\text{5}\text{g}}{\text{58}\text{.44}\text{g/mol}}\\ \text{=}\text{0}\text{.086}\text{mol}\end{array}$

Molarity of the $NaCl$ solution is,

  $\begin{array}{c}\text{Molality (m) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{Solvent}\text{in}\text{kg}}\\ \text{=}\frac{\text{0}\text{.086}\text{mol}}{\text{0}\text{.095}\text{kg}}\\ \text{=}\text{0}\text{.90}\text{m}\end{array}$

Molarity of the $NaCl$ solution is $\text{0}\text{.90}\text{m}$

It is assumed that dissociation does not take place.

Depression in freezing point is calculated by using the equation

  $\begin{array}{c}\Delta {\text{T}}_{fp}{\text{= K}}_{fp}{\text{. m}}_{solute}\\ =\left(-\text{1}{\text{.86}}^{\text{0}}\text{C/m}\right)\left(\text{0}\text{.90}\text{m}\right)\\ =-{1.67}^0\text{C}\end{array}$

Depression in freezing point when there is no dissociation is $-{1.67}^0\text{C}$

$van\text{'}tHofffactor,i=\frac{measured\Delta {\text{T}}_{fp}}{Calculated\Delta {\text{T}}_{fp}}$

  $\begin{array}{c}i=\frac{-{3.05}^0\text{C}}{-{1.67}^0\text{C}}\\ =1.82\end{array}$

van’t Hoff factor for $NaCl$ is $1.82$

The number of moles of any substance can be determined using the equation

  $\begin{array}{c}\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}\\ \text{=}\frac{\text{5}\text{g}}{\text{142}\text{.04}\text{g/mol}}\\ \text{=}\text{0}\text{.035}\text{mol}\end{array}$

Molality of the $N{a}_{2}S{O}_4$ solution is,

  $\begin{array}{c}\text{Molality (m) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{Solvent}\text{in}\text{kg}}\\ \text{=}\frac{\text{0}\text{.035}\text{mol}}{\text{0}\text{.095}\text{kg}}\\ \text{=}\text{0}\text{.368}\text{m}\end{array}$

Molality of the $N{a}_{2}S{O}_4$ solution is $\text{0}\text{.368}\text{m}$

It is assumed that dissociation does not take place.

Depression in freezing point is calculated by using the equation

  $\begin{array}{c}\Delta {\text{T}}_{fp}{\text{= K}}_{fp}{\text{. m}}_{solute}\\ =\left(-\text{1}{\text{.86}}^{\text{0}}\text{C/m}\right)\left(\text{0}\text{.368}\text{m}\right)\\ =-{0.6845}^0\text{C}\end{array}$

$van\text{'}tHofffactor,i=\frac{measured\Delta {\text{T}}_{fp}}{Calculated\Delta {\text{T}}_{fp}}$

  $\begin{array}{c}i=\frac{-{1.36}^{0}C}{-{0.6845}^0\text{C}}\\ =1.98\end{array}$

van’t Hoff factor for $N{a}_{2}S{O}_4$ is $1.98$

van’t Hoff factor for $NaCl$ is $1.82$ whereas for $N{a}_{2}S{O}_4$, it is $1.98$. The value of van’t Hoff factor decreases with the increase in concentration of solute. The values are steady with the trend that seen in the given data table. Since the rate is decreasing in both cases. The continuity of trend for the value of van’t Hoff factor for $N{a}_{2}S{O}_4$ is decreased significantly and it is near to the value of van’t Hoff factor for $NaCl$. The value of van’t Hoff factor for $NaCl$ is similar to the other values.

Conclusion

van’t Hoff factor for $NaCl$ and $N{a}_{2}S{O}_4$ were determined.

• van’t Hoff factor for $NaCl$ is $1.82$
• van’t Hoff factor for $N{a}_{2}S{O}_4$ is $1.98$
• The continuity of trend for the value of van’t Hoff factor for $N{a}_{2}S{O}_4$ is decreased significantly and it is near to the value of van’t Hoff factor for $NaCl$. The value of van’t Hoff factor for $NaCl$ is similar to the other values.