Chapter 12, Problem 12.89QP

Interpretation Introduction

Interpretation:

An aqueous solution is $\text{14 \%}$ ${\text{NH}}_4\text{Cl}$ by mass.  Given the density of the solution, molality, mole fraction and molarity of ${\text{NH}}_4\text{Cl}$ in solution has to be determined.

Concept Introduction:

Molality is one of the many parameters that is used to express concentration of a solution.  It is expressed as,

$\text{Molality = }\frac{number\text{ of }molesof\text{ solute}}{mass\text{ of solvent in kg}}$

Molarity is one of the many parameters that is used to express concentration of a solution.  It is expressed as,

$\text{Molarity = }\frac{number\text{ of }molesof\text{ solute}}{volume\text{ of solution in L}}$

A solution is at least made up of two components.  Mole fraction of a component in the solution correlates to the ratio of number of moles of that component to the total number of moles.  It is expressed as,

$\text{Mole fraction = }\frac{number\text{ of moles of a component}}{total\text{ number of moles in the solution}}$

Mass percent is one of the many parameters that is used to express concentration of a solution.  It is expressed as,

$\text{Mass percent = }\frac{mass\text{ of solute}}{mass\text{ of solution}}\times \text{ 100\%}$

Answer to Problem 12.89QP

Molality of $14\%{\text{ NH}}_4\text{Cl}$ is determined as $3.04m.$

Mole fraction of $N{H}_{4}Cl$ in $14\%{\text{ NH}}_4\text{Cl}$ is determined as $0.520.$

Molarity of $14\%{\text{ NH}}_4\text{Cl}$ is determined as $2.72M.$

Explanation of Solution

Assume that volume of solution is $1.000\text{ L}$ which is equivalent to $1.040\text{ kg}$ as density of the solution is $1.040\text{ g/mL}\text{.}$ Then mass of ammonium chloride used to prepare $\text{14 \%}$ ${\text{NH}}_4\text{Cl}$ solution is calculated as,

${\text{Mass percent of NH}}_4\text{Cl = }\frac{mass{\text{ of NH}}_4\text{Cl}}{mass\text{ of solution}}\times \text{ 100\%}$

Rearranging the above expression,

${\text{mass of NH}}_4\text{Cl = }Mass\text{ percent}\times \text{ mass of solution}$

Substitute the known values,

$\begin{array}{l}{\text{mass of NH}}_4\text{Cl }\text{= 14\% }\times \text{ 1}\text{.040 kg}\\ \text{= 0}\text{.1456 kg}\end{array}$

As we know,

$\begin{array}{l}Mass\text{ of solution = mass of solute + mass of solvent}\\ {\text{ = mass of NH}}_4\text{Cl + mass of water}\end{array}$

Mass of the solute, ${\text{NH}}_4\text{Cl}$ is calculated above. Therefore,

$\begin{array}{l}\text{1}\text{.040 kg of solution }\text{= 0}\text{.1456 kg + mass of water}\\ \text{mass of water }\text{= 1}\text{.040 kg - 0}\text{.1456 kg = 0}\text{.8944 kg}\end{array}$

Number of moles of ammonium chloride and water are calculated as,

$\begin{array}{l}no.of{\text{ moles of NH}}_4\text{Cl}\text{= }\frac{mass{\text{ of NH}}_4\text{Cl}}{molar{\text{ mass of NH}}_4\text{Cl}}\\ =\frac{145.6\text{ g}}{53.49\text{ g/mol}}\text{ = 2}\text{.72 mol}\\ \\ \text{No}\text{.of moles of water}\text{= }\frac{mass{\text{ of H}}_2\text{O}}{molar{\text{ mass of H}}_2\text{O}}\\ =\frac{894.4\text{ g}}{18.02\text{ g/mol}}\text{ = 49}\text{.65 mol}\end{array}$

$\begin{array}{l}Totalnumberofmolesinsolution,\\ \text{= }no.of{\text{ moles of NH}}_4\text{Cl + no}{\text{.of moles of H}}_2\text{O}\\ \text{= 2}\text{.72 mol + 49}\text{.65 mol}\\ \text{= 52}\text{.37 mol}\end{array}$

Molality of the solution is calculated as,

$\begin{array}{l}{\text{Molality of NH}}_4\text{Cl}\text{= }\frac{number\text{ of }molesof{\text{ NH}}_4\text{Cl}}{mass{\text{ of H}}_2\text{O in kg}}\\ =\frac{2.72\text{ mol}}{0.8944\text{ kg}}\\ =\text{ 3}\text{.043 m }\approx 3.04\text{ m}\end{array}$

Molarity of the solution is calculated as,

$\begin{array}{l}{\text{Molarity of NH}}_4\text{Cl}\text{= }\frac{number\text{ of }molesN{H}_{4}Cl}{volume\text{ of solution in L}}\\ =\frac{2.72\text{ mol}}{1.000\text{ L}}\text{ = 2}\text{.72 M}\end{array}$

Mole fraction of ammonium chloride is calculated as,

$\begin{array}{l}{\text{Mole fraction of NH}}_4\text{Cl}\text{= }\frac{number\text{ of }molesN{H}_{4}Cl}{total\text{ number of moles in solution}}\\ =\frac{2.72\text{ mol}}{52.37\text{ mol}}\text{ = 0}\text{.0519 = 0}\text{.0520}\end{array}$

Conclusion

Given the density of aqueous solution of $\text{14 \%}$ ${\text{NH}}_4\text{Cl}$, the molality, mole fraction and molarity of ${\text{NH}}_4\text{Cl}$ in solution are determined by calculating the number of moles of ${\text{NH}}_4\text{Cl}$ in solution.