Chapter 12, Problem 12.90QP

Interpretation Introduction

Interpretation:

An aqueous solution is $\text{27 \%}$ $\text{LiCl}$ 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 12.90QP

Molality of $27\%\text{ LiCl}$ is determined as $8.72m.$

Mole fraction of $LiCl$ in $27\%\text{ LiCl}$ is determined as $0.136.$

Molarity of $27\%\text{ LiCl}$ is determined as $7.2M.$

Explanation of Solution

Assume that volume of solution is $1.000\text{ L}$ which is equivalent to $1.127\text{ kg}$ as density of solution is $\text{1}\text{.127 g/mL}$. Then mass of $LiCl$ used to prepare $27\%\text{ LiCl}$ solution is,

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

Rearranging the above expression,

$\text{mass of LiCl = }Mass\text{ percent }\times \text{ mass of solution}$

Substitute the known values,

$\begin{array}{l}\text{mass of LiCl }\text{= }27\%\times \text{ 1}\text{.127 kg }\\ \text{= 0}\text{.3043 kg}\\ =\text{ 304}\text{.3 g LiCl}\end{array}$

As we know,

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

Mass of the solute, $\text{LiCl}$ is calculated above. Therefore,

$\begin{array}{l}\text{1}\text{.127 kg of solution }\text{= 0}\text{.3043 kg of LiCl + mass of water}\\ \text{mass of water }\text{= 1}\text{.127 kg - 0}\text{.3043 kg }\\ \text{= 0}\text{.8227 kg}\end{array}$

Number of moles of LiCl and water are calculated as,

$\begin{array}{l}no.of\text{ moles of LiCl}\text{= }\frac{mass\text{ of LiCl}}{molar\text{ mass of LiCl}}\\ =\frac{304.3\text{ g}}{42.39\text{ g/mol}}\text{ = 7}\text{.178 mol}\\ \\ \text{No}\text{.of moles of water}\text{= }\frac{mass{\text{ of H}}_2\text{O}}{molar{\text{ mass of H}}_2\text{O}}\\ =\frac{822.7\text{ g}}{18.02\text{ g/mol}}\text{ = 45}\text{.67 mol}\end{array}$

$\begin{array}{l}Totalnumberofmolesinsolution,\\ \text{= }no.of\text{ moles of LiCl + no}{\text{.of moles of H}}_2\text{O}\\ \text{= 7}\text{.178 mol + 45}\text{.67 mol}\\ \text{= 52}\text{.845 mol}\end{array}$

Molality of the solution is calculated as,

$\begin{array}{l}\text{Molality of LiCl}\text{= }\frac{number\text{ of }molesof\text{ LiCl}}{mass{\text{ of H}}_2\text{O in kg}}\\ =\frac{7.178\text{ mol}}{0.8227\text{ kg}}\\ =\text{ 8}\text{.7249 m }\approx 8.72\text{ m}\end{array}$

Molarity of the solution is calculated as,

$\begin{array}{l}\text{Molarity of LiCl}\text{= }\frac{number\text{ of }moles\text{ Li}Cl}{volume\text{ of solution in L}}\\ =\frac{7.178\text{ mol}}{1.000\text{ L}}\\ \text{= 7}\text{.178 M = 7}\text{.2 M}\end{array}$

Mole fraction of Lithium chloride is calculated as,

$\begin{array}{l}\text{Mole fraction of LiCl}\text{= }\frac{number\text{ of }moles\text{ Li}Cl}{total\text{ number of moles in solution}}\\ =\frac{7.178\text{ mol}}{52.845\text{ mol}}\\ \text{= 0}\text{.1358 = 0}\text{.136}\end{array}$

Conclusion

Given the density of aqueous solution of $\text{27 \%}$ $LiCl$, the molality, mole fraction and molarity of $\text{LiCl}$ in solution are determined by calculating the number of moles of $\text{LiCl}$ in solution.