Chapter 16, Problem 84A

Interpretation Introduction

Interpretation: The freezing point of the glucose solution is to be calculated when $20.0\text{g}$ of glucose $\left({\text{C}}_6{\text{H}}_{12}{\text{O}}_6\right)$ is dissolved in $500.0\text{g}$ of water.

Concept introduction: Depression in freezing point is a term associated with the decrease in the value of the freezing point of solvents by the addition of solute molecules.

The formula used for calculating the depression in the freezing point is,

  $\Delta T_f=K_{f}m$

Here, $\Delta T_f$ is the freezing point depression, $K_f$ is the molal freezing point depression constant, and $m$ is molality.

The expression for moles is as follows:

  $\text{moles}=\frac{\text{ mass of solute}}{\text{Molar mass of solute}}$   ...... (1)

The expression for molality is as follows:

  $m=\frac{\text{moles of solute}}{\text{Mass of solvent(kg)}}$   ...... (2)

Answer to Problem 84A

The freezing point of the glucose solution when it contains $20.0\text{g}$ glucose and $500.0\text{g}$ water is $-0.41°\text{C}$ .

Explanation of Solution

The mass of solvent water is $500.0\text{g}$ .

Convert $\text{g}$ to $\text{kg}$ .

  $\begin{array}{c}500.0\text{g}=500\times \frac{1\text{kg}}{1000\text{g}}\\ =0.5\text{kg}\end{array}$

By calculating the molality of glucose, the depression in the freezing point can be calculated.

It can be calculated by knowing the number of moles of glucose.

The given mass of glucose is $20.0\text{g}$ and its molecular mass is 180.

Substitute, $20.0\text{g}$ for the mass of $\left({\text{C}}_6{\text{H}}_{12}{\text{O}}_6\right)$ and $\text{ 180 g/mol}$ for molar mass of $\left({\text{C}}_6{\text{H}}_{12}{\text{O}}_6\right)$ in equation (1).

  $\begin{array}{c}\text{moles}\left({\text{C}}_6{\text{H}}_{12}{\text{O}}_6\right)=\frac{\text{ 20 g}}{\text{180 g/mol}}\\ =0.11\text{ mol}\end{array}$

Substitute, $0.11moles$ for $\left({\text{C}}_6{\text{H}}_{12}{\text{O}}_6\right)$ and $\text{0}\text{.5}\text{kg}$ for the mass of solvent in equation (2).

  $\begin{array}{c}m=\frac{0.11}{0.5}\\ =0.22\text{mol}\end{array}$

The molal freezing point depression constant of glucose is $1.86{\text{Kkgmol}}^{-1}$ .

So,

Substitute $1.86{\text{Kkgmol}}^{-1}$ for $K_f$ and $\text{ 0}\text{.22}\text{mol/kg}$ for $m$ in the equation of freezing point depression.

  $\begin{array}{c}\Delta T_f=1.86{\text{Kkgmol}}^{-1}\times \text{ 0}\text{.22}\text{mol/kg}\\ =0.41\text{K}\end{array}$

Now, the freezing point of water is 0°C.

The freezing point of glucose can be calculated by subtracting the depression in the freezing point of the solution from the freezing point of water.

The freezing point of glucose is,

  $0-0.41°\text{C=}-0.41°\text{C}$

Conclusion

The freezing point of glucose is $-0.41°\text{C}$ .