Chapter 14.2, Problem 28PP

Interpretation Introduction

Interpretation:

Amount of $\mathrm{Ba}{\left(OH\right)}_2$ is needed to make a 1.0 m aqueous solution have to be calculated.

Concept introduction:

The concentrations of solutions are expressed in different ways, all of them involved in the quantity of solute and the quantity of solution or solvent. Among them, molality is most appropriate unit to express the concentration of solution.Molality: Molality is defined as the number of moles of solute per kilogram of solvent. Molality is denoted by ‘m’. The volume of a solution changes with temperature, this change in volume alters the molarity of the solution. Masses, however, do not change with temperature. Molality does not change with change of temperature.

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

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

Answer to Problem 28PP

171gm of $\mathrm{Ba}{\left(OH\right)}_2$ is needed to make a 1.0 m aqueous solution.

Given information:

The molecular weight of $\mathrm{Ba}{\left(OH\right)}_2$ = 171 gm/mol The molality of the solution = 1.0 m

The mass of solvent = 1 kg

Explanation:

Write the expression ofMolarity of the so

$Molalityofthesolution\left(m\right)=\frac{Numberofmolesofchoice}{kgofsolvent}$

Calculating the number of moles of the solute from above equation:

$\begin{array}{l}Numberofmolesofthesolute=Molalityofthesolution\times kgofsolvent\hfill \\ =1.0m\times 1kg\hfill \\ =1.0mol\hfill \end{array}$

Write the expression for number of a solute.

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

$\mathrm{Number}ofmolesof\mathrm{Ba}{\left(OH\right)}_2=\frac{Massof\mathrm{Ba}{\left(OH\right)}_2}{Molecularweightof\mathrm{Ba}{\left(OH\right)}_2}$

$\begin{array}{c}Massof\mathrm{Ba}{\left(OH\right)}_2=1.0mol\times 171gm/mol \\ =171gm\end{array}$

Thus, 171gm of $\mathrm{Ba}{\left(OH\right)}_2$ is needed to make a 1.0 m aqueous solution.

Explanation of Solution

Given information:

The molecular weight of $\mathrm{Ba}{\left(OH\right)}_2$ = 171 gm/mol The molality of the solution = 1.0 m

The mass of solvent = 1 kg

Write the expression ofMolarity of the so

$Molalityofthesolution\left(m\right)=\frac{Numberofmolesofchoice}{kgofsolvent}$

Calculating the number of moles of the solute from above equation:

$\begin{array}{l}Numberofmolesofthesolute=Molalityofthesolution\times kgofsolvent\hfill \\ =1.0m\times 1kg\hfill \\ =1.0mol\hfill \end{array}$

Write the expression for number of a solute.

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

$\mathrm{Number}ofmolesof\mathrm{Ba}{\left(OH\right)}_2=\frac{Massof\mathrm{Ba}{\left(OH\right)}_2}{Molecularweightof\mathrm{Ba}{\left(OH\right)}_2}$

$\begin{array}{c}Massof\mathrm{Ba}{\left(OH\right)}_2=1.0mol\times 171gm/mol \\ =171gm\end{array}$

Thus, 171gm of $\mathrm{Ba}{\left(OH\right)}_2$ is needed to make a 1.0 m aqueous solution.