Chapter 14.4, Problem 46PP

Interpretation Introduction

Interpretation:

The boiling point and freezing point of a 0.40 m solution of sucrose in ethanol needs to be determined.

Concept introduction:

Boiling point elevation is calculated as follows:

$\Delta {\mathrm{T}}_b=K_b\times m$

Here, m is molality of the solution.

Similarly,

Depression of freezing point is calculated as follows:

$\Delta T_f=K_f\times m$

Here, m is molality of the solution.

Answer to Problem 46PP

The boiling point and freezing point of a 0.40 m solution of sucrose in ethanol is 79.0 °C and -114.9 °C respectively

Boiling point elevation $\Delta {\mathrm{T}}_b=K_b\times m$

Data given: molality of the sucrose solution = 0.40 m

K_b of ethanol = 1.22 $°C/m$

$\begin{array}{l}\Delta T_b=1.22°C/m\times 0.40m\\ =0.49°C\\ \Delta T_b=T_{b}ofthesucrosesolutioninethanol-T_{b}ofthesolvent(ethanol)\\ T_{b}ofthesucrosesolution=78.5°C+0.49°C=79.0°C\end{array}$

Depression of freezing point $\Delta T_f=K_f\times m$

Data given: molality of the sucrose solution = 0.40 m

K_f of ethanol = 1.99 $°C/m$

$\begin{array}{l}\Delta T_f=1.99°C/m\times 0.40m\\ =0.80°C\\ \Delta T_f=T_{f}ofEthanol-T_{f}ofthesucrosesolutioninethanol\\ T_{f}ofthesucrosesolutioninethanol=-114.1°C-0.80°C=-114.9°C\end{array}$

Explanation of Solution

Boiling point elevation $\Delta {\mathrm{T}}_b=K_b\times m$

Data given: molality of the sucrose solution = 0.40 m

K_b of ethanol = 1.22 $°C/m$

$\begin{array}{l}\Delta T_b=1.22°C/m\times 0.40m\\ =0.49°C\\ \Delta T_b=T_{b}ofthesucrosesolutioninethanol-T_{b}ofthesolvent(ethanol)\\ T_{b}ofthesucrosesolution=78.5°C+0.49°C=79.0°C\end{array}$

Depression of freezing point $\Delta T_f=K_f\times m$

Data given: molality of the sucrose solution = 0.40 m

K_f of ethanol = 1.99 $°C/m$

$\begin{array}{l}\Delta T_f=1.99°C/m\times 0.40m\\ =0.80°C\\ \Delta T_f=T_{f}ofEthanol-T_{f}ofthesucrosesolutioninethanol\\ T_{f}ofthesucrosesolutioninethanol=-114.1°C-0.80°C=-114.9°C\end{array}$