Chapter 14.2, Problem 27PP

Interpretation Introduction

Interpretation:

Molality of ${\mathrm{Na}}_{2}S{O}_4$ 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 27PP

Molality of ${\mathrm{Na}}_{2}S{O}_4$₄ solution is 0.0704.

Given information:

Mass of ${\mathrm{Na}}_{2}S{O}_4$ = 10 gm

Mass of solvent = 1000 gm = 1 kg

Molecular weight of Na₂SO₄ = 142.04 gm/mol

Explanation:

Write the expression of molarity of solution,

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

This can be written as

$\mathrm{Number}ofmolesofsolute=\frac{Massofsolute}{molecularweightofsolute}$$\mathrm{Number}ofmolesofN{a}_{2}S{O}_4=\frac{MassofN{a}_{2}S{O}_4}{MolecularweightofN{a}_{2}S{O}_4}$

Substituting the values in the above equation:

$\begin{array}{c}\mathrm{Number}ofmolesofN{a}_{2}S{O}_4=\frac{10}{142.04}\\ =0.0704\end{array}$

Write the expression for the molarity of solution:

$\begin{array}{c}\mathrm{Molality}ofsolution=\frac{Numberofmolesofsolute}{Molecularweightofsolute}\\ =\frac{0.074}{1}\\ =0.0704m\end{array}$

Thus, Molality of ${\mathrm{Na}}_{2}S{O}_4$₄ solution is 0.0704.

Explanation of Solution

Given information:

Mass of ${\mathrm{Na}}_{2}S{O}_4$ = 10 gm

Mass of solvent = 1000 gm = 1 kg

Molecular weight of Na₂SO₄ = 142.04 gm/mol

Write the expression of molarity of solution,

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

This can be written as

$\mathrm{Number}ofmolesofsolute=\frac{Massofsolute}{molecularweightofsolute}$$\mathrm{Number}ofmolesofN{a}_{2}S{O}_4=\frac{MassofN{a}_{2}S{O}_4}{MolecularweightofN{a}_{2}S{O}_4}$

Substituting the values in the above equation:

$\begin{array}{c}\mathrm{Number}ofmolesofN{a}_{2}S{O}_4=\frac{10}{142.04}\\ =0.0704\end{array}$

Write the expression for the molarity of solution:

$\begin{array}{c}\mathrm{Molality}ofsolution=\frac{Numberofmolesofsolute}{Molecularweightofsolute}\\ =\frac{0.074}{1}\\ =0.0704m\end{array}$

Thus, Molality of ${\mathrm{Na}}_{2}S{O}_4$₄ solution is 0.0704.