Chapter 14.4, Problem 45PP

Interpretation Introduction

Interpretation:

The boiling point and freezing point of a 0.625 m aqueous solution of any non-volatile, non-electrolyte solute 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 45PP

The boiling point and freezing point of a 0.625 m aqueous solution of any non-volatile, non-electrolyte solute is 100.32 °C and -1.16 °C respectively

Boiling point elevation $\Delta T_b=K_b\times m$

Data given: molality of the solution = 0.625 m

K_b of water = 0.512 $°C/m$

$\begin{array}{l}\Delta T_b=0.512°C/m\times 0.625m\\ =0.320°C\\ \Delta T_b=T_{b}ofthesolution-T_{b}ofthesolvent(Water)\\ T_{b}ofthesolution=100°C+0.320°C=100.32°C\end{array}$

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

Data given: molality of the solution = 0.625 m

K_f of water = 1.86 $°C/m$

$\begin{array}{l}\Delta T_f=1.86°C/m\times 0.625m\\ =1.16°C\\ \Delta T_f=T_{f}ofthesolvent(Water)-T_{f}ofthesolution\\ T_{f}ofthesolution=0°C-1.16°C=-1.16°C\end{array}$

Explanation of Solution

Boiling point elevation $\Delta T_b=K_b\times m$

Data given: molality of the solution = 0.625 m

K_b of water = 0.512 $°C/m$

$\begin{array}{l}\Delta T_b=0.512°C/m\times 0.625m\\ =0.320°C\\ \Delta T_b=T_{b}ofthesolution-T_{b}ofthesolvent(Water)\\ T_{b}ofthesolution=100°C+0.320°C=100.32°C\end{array}$

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

Data given: molality of the solution = 0.625 m

K_f of water = 1.86 $°C/m$

$\begin{array}{l}\Delta T_f=1.86°C/m\times 0.625m\\ =1.16°C\\ \Delta T_f=T_{f}ofthesolvent(Water)-T_{f}ofthesolution\\ T_{f}ofthesolution=0°C-1.16°C=-1.16°C\end{array}$