Chapter 14.4, Problem 51SSC

Interpretation Introduction

Interpretation:

The mass of calcium chloride needs to be calculated which is dissolved in 1000 g of solvent and boils at 101.3°C.

Concept introduction:

Boiling point elevation is calculated as follows:

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

Here,

$\Delta T_b=T_{b}ofthesolution-T_{b}ofthesolvent(Water)$

Molality of solution is calculated as follows:

$\mathrm{Molality}ofsolution=\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{Mass of solvent}}$

Also,

$\mathrm{Number}ofmolesofsolute=\frac{Massofsolute}{molecularweightofsolute}$

Answer to Problem 51SSC

The mass of calcium chloride is 0.0936 kg which dissolved in 1000 g of solvent which boils at 101.3°C.

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

Data given:

K_b of water = 0.512 $°C/m$

$T_{b}ofthesolution$₌101.3°C

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

$\Delta T_b=T_{b}ofthesolution-T_{b}ofthesolvent(Water)$

$\Delta T_b=101.3°C-100°C=1.3°C$

$1.3=0.512\times m$

m=2.53

Molality of solution = 2.53 molal

$\mathrm{Molality}ofsolution=\frac{Numberofmolesofsolute}{Molecularweightofsolute}$

Again, $\mathrm{Number}ofmolesofsolute=\frac{Massofsolute}{molecularweightofsolute}$

Given,

$\begin{array}{l}2.53=MassofCaC{l}_2\\ 2.53=\frac{MassofCaC{l}_2\times 3}{110.98}\end{array}$

Mass of solvent=1000g

So putting the values in equation

$2.53=\frac{NumberofmolesofCaC{l}_2\times 1000(kg)}{Molecularweightofsolvent(g)}$

$2.53=\frac{NumberofmolesofCaC{l}_2\times 1000(kg)}{1000}$

$\mathrm{Number}ofmolesofsolute=\frac{Massofsolute}{molecularweightofsolute}$

$\mathrm{Number}ofmolesofCaC{l}_2=\frac{MassofCaC{l}_2}{MolecularweightofCaC{l}_2}$

$MolecularweightofCaC{l}_2$ = 110.98g/mol

Putting the values in equation:

$2.53=\frac{MassofCaC{l}_2\times 3}{110.98}$

$MassofCaC{l}_2$ = 0.0936 kg

Explanation of Solution

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

Data given:

K_b of water = 0.512 $°C/m$

$T_{b}ofthesolution$₌101.3°C

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

$\Delta T_b=T_{b}ofthesolution-T_{b}ofthesolvent(Water)$

$\Delta T_b=101.3°C-100°C=1.3°C$

$1.3=0.512\times m$

m=2.53

Molality of solution = 2.53 molal

$\mathrm{Molality}ofsolution=\frac{Numberofmolesofsolute}{Molecularweightofsolute}$

Again, $\mathrm{Number}ofmolesofsolute=\frac{Massofsolute}{molecularweightofsolute}$

Given,

$\begin{array}{l}2.53=MassofCaC{l}_2\\ 2.53=\frac{MassofCaC{l}_2\times 3}{110.98}\end{array}$

Mass of solvent=1000g

So putting the values in equation

$2.53=\frac{NumberofmolesofCaC{l}_2\times 1000(kg)}{Molecularweightofsolvent(g)}$

$2.53=\frac{NumberofmolesofCaC{l}_2\times 1000(kg)}{1000}$

$\mathrm{Number}ofmolesofsolute=\frac{Massofsolute}{molecularweightofsolute}$

$\mathrm{Number}ofmolesofCaC{l}_2=\frac{MassofCaC{l}_2}{MolecularweightofCaC{l}_2}$

$MolecularweightofCaC{l}_2$ = 110.98g/mol

Putting the values in equation:

$2.53=\frac{MassofCaC{l}_2\times 3}{110.98}$

$MassofCaC{l}_2$ = 0.0936 kg