Chapter 14, Problem 106A

Which solute has the greatest effect on the boiling pointof 1.00 kg of water: 50.0 g of strontium chloride $(SrC{l}_2)$ or 150.0 g of carbon tetrachloride $(CC{l}_4)$ ? Justify youranswer.

Interpretation Introduction

Interpretation:

The solute that shows the greatest effect on the boiling point have to be justified.

Concept introduction:

For electrolyte solutes, the elevation of the boiling point is directly proportional to molalityof the solution.

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

Where,

$\Delta {\mathrm{T}}_b$ is the elevation of boiling point = the boiling point difference between a solution and the pure solvent.

$i$ is the van’t Hoff factor = for strong electrolyte $i$ is equal to number of dissociate ions.

$K_b$ is the molal boiling point elevation constant.

$m$ is the molality of the solution.

Answer to Problem 106A

150 gm of CCl₄has the greater effect than 50.0 gram of SrCl₂.

Explanation of Solution

Given information:

Mass of ${\mathrm{SrCl}}_2$ = 50 gm

Mass of water = 1 kg

Mass of ${\mathrm{CCl}}_4$ = 150 gm

Here, SrCl₂is a strong electrolyte, it dissociates completelyin the solution to givethree ions (Sr²+ and 2 Cl- ion) and i = 3 while of CCl₄is a nonelectrolyte.

For $SrC{l}_2$ solution, $\Delta {\mathrm{T}}_b=i\times K_b\times m$

$\begin{array}{l}NumberofmolesofSrC{l}_2=\frac{MassofSrC{l}_2}{MolecularweightofSrC{l}_2}\\ =\frac{50gm}{158.53gm/mol}=0.315mole\end{array}$

(Note: Molecular weight of $SrC{l}_2$ = 158.53 gm/mol)

$\begin{array}{l}MolalityofSrC{l}_{2}solution=\frac{Numberofmolesof SrC{l}_2}{ Mass of water(inKg)}\\ =\frac{0.315mole}{ 1kg}=0.315m\end{array}$

$\begin{array}{l}\Delta T_b=i\times K_b\times m\\ =3\times 0.512°C/m\times 0.315m\\ =0.484°C\end{array}$

Calculatethe boiling point of SrCl₂ solution.

$\begin{array}{c}{T}_b=100°\mathrm{C}+0.484°\mathrm{C}\\ =100.484°\mathrm{C}\end{array}$

$\begin{array}{l}NumberofmolesofCC{l}_4=\frac{MassofCC{l}_4}{MolecularweightofCC{l}_4}\\ =\frac{150gm}{154gm/mol}=0.974mole\end{array}$

(Note: Molecular weight of CCl₄ = 154 gm/mol)

$\begin{array}{l}MolalityofCC{l}_{4}solution=\frac{Numberofmolesof CC{l}_4}{ Mass of water(inKg)}\\ =\frac{0.974mole}{ 1kg}=0.974m\end{array}$

$\begin{array}{l}\Delta T_b=K_b\times m\\ =0.512°C/m\times 0.974m\\ =0.499°C\end{array}$

Calculate theboiling point of CCl₄ solution.

$\begin{array}{c}{T}_b=100°C+0.499°C\\ =100.499°C\end{array}$

By calculatingof the boiling point of two different solutions, we conclude that 150 gm of CCl₄has the greater effect than 50.0 gram of SrCl₂.