Chapter 14, Problem 118A

Compare Which of the following solutions has thehighest concentration? Rank the solutions from thegreatest to the smallest boiling point depression. Explainyour answer.
a. 0.10 mol NaBr in 100.0 mL solution
b. 2.1 mol KOH in 1.00 L solution
c. 1.2 mol $KMn{O}_4$ in 3.00 L solution

Interpretation Introduction

Interpretation:

The solution having greatest boiling pointelevationhave to be identified.

Concept introduction:

Elevation of boiling point: The boiling point is a temperature at which temperature the vapour pressure of the liquid is equal to the atmospheric pressure or external pressure. When a non-volatile, non-electrolyte solute is added to a solvent, the boiling point of the resulting solution will increase.

For nonvolatile and nonelectrolyte solutes, the elevation of the boiling point is directly proportional to molality of the solution.

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

Answer to Problem 118A

KOH contributes the greatest number of particles to solution. As, higher the number of solute particles in the solution greater will bethe boiling point elevation. So, KOH has the greatest boiling point elevation and KMnO₄has the smallest boiling point elevation as it has the lowest concentration.

Explanation of Solution

For a solution, elevation of boiling point is a colligative property and it depends on the number of solute particles or the concentration of the solution. Upon addition of a non-volatile, non-electrolyte solute to the solvent, the boiling point of the resulting solution will increase. So, first the concentration of the individual solution have to calculate.

1. 0.10 mol NaBr in 100.0 mL solution

$\begin{array}{l}MolarityofNaBrsolution=\frac{Numberofmolesof NaBr}{ Volume of the solution(inlitre)}\\ =\frac{0.10mole}{ 0.10litre}=1.0M\end{array}$

2. 2.1 mol KOH in 1.00 L solution

$\begin{array}{l}MolarityofKOHsolution=\frac{Numberofmolesof KOH}{ Volume of the solution(inlitre)}\\ =\frac{2.1mole}{ 1.0litre}=2.1M\end{array}$

3. 1.2 mol KMnO₄ in 3.00 L solution

$\begin{array}{l}MolarityofKMn{O}_{4}solution=\frac{Numberofmolesof KMn{O}_4}{ Volume of the solution(inlitre)}\\ =\frac{1.2mole}{ 3.0litre}=0.4M\end{array}$

Because the concentration of KOH solution is highest among them. KOH contributes the greatest number of particles to solution. As, higher the number of solute particles in the solution greater will bethe boiling point elevation. So, KOH has the greatest boiling point elevation and KMnO₄has the smallest boiling point elevation as it has the lowest concentration.