9.7: Molality
- Page ID
- 288494
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\dsum}{\displaystyle\sum\limits} \)
\( \newcommand{\dint}{\displaystyle\int\limits} \)
\( \newcommand{\dlim}{\displaystyle\lim\limits} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\(\newcommand{\longvect}{\overrightarrow}\)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- To calculate molarity and use it in calculations
- To compare and contrast molarity and molality
For many purposes, molarity is very convenient concentration unit. However, when we want to know the concentration of solute present in situations where there are temperature changes, molarity will not work. The volume of the solution will change somewhat with temperature, enough to affect the accuracy of data observations and calculations. Another unit is needed, one not affected by the temperature of the material we are studying.
Molality
Another way to express the concentration of a solution is by its molality. The molality \(\left( \textit{m} \right)\) of a solution is the moles of solute divided by the kilograms of solvent. A solution that contains \(1.0 \: \text{mol}\) of \(\ce{NaCl}\) dissolved into \(1.0 \: \text{kg}\) of water is a "one-molal" solution of sodium chloride. The symbol for molality is a lower-case \(\textit{m}\) written in italics.
\[\text{Molality} \left( \textit{m} \right) = \frac{\text{moles of solute}}{\text{kilograms of solvent}} = \frac{\text{mol}}{\text{kg}}\nonumber \]
Molality differs from molarity only in the denominator. While molarity is based on the liters of solution, molality is based on the kilograms of solvent. Concentrations expressed in molality are used when studying properties of solutions related to vapor pressure and temperature changes. Molality is used because its value does not change with changes in temperature. The volume of a solution, on the other hand, is slightly dependent upon temperature. This can be helpful when studying the human body since molarity would be different at room temperature vs body temperature but the molality of a solution would be the same at both temperatures.
*Careful! There is only one letter that is different in the words molarity and molality, but they are different units.
| Aspect/Property | Molarity | Molality |
|---|---|---|
| Unit for numerator | moles | moles |
| Substance for numerator | solute | solute |
| Property and unit for denominator | volume in L | mass in kg |
| Substance for denominator | solution (both solute and solvent) | solvent |
| Benefit | It is usually easy to measure the volume of solution in a lab, even for someone who did not prepare the sample themselves. | Volume changes with temperature but mass and moles do not, so molality stays constant at high and low temperatures. |
| Summary | moles of solute per L of solution | moles of solute per kg of solvent |
Example \(\PageIndex{1}\)
Determine the molality of a solution prepared by dissolving \(28.60 \: \text{g}\) of glucose \(\left( \ce{C_6H_{12}O_6} \right)\) into \(250 \: \text{g}\) of water.
Solution
Step 1: List the known quantities and plan the problem.
Known
- Mass solute \(= 28.60 \: \text{g} \: \ce{C_6H_{12}O_6}\)
- Mass solvent \(= 250 \: \text{g} = 0.250 \: \text{kg}\)
- Molar mass \(\ce{C_6H_{12}O_6} = 180.18 \: \text{g/mol}\)
Unknown
Convert grams of glucose to moles and divide by the mass of the water in kilograms.
Step 2: Solve.
Convert mass of solute to moles of solute:
\[ 28.60 \: \text{g} \: \ce{C_6H_{12}O_6} \times \frac{1 \: \text{mol} \: \ce{C_6H_{12}O_6}}{180.18 \: \text{g} \: \ce{C_6H_{12}O_6}} = 0.1587 \: \text{mol} \: \ce{C_6H_{12}O_6} \nonumber \]
Solve for the molality:
\[ Molality = \dfrac{moles\:solute}{mass\:solvent} = \frac{0.1587 \: \text{mol} \: \ce{C_6H_{12}O_6}}{0.250 \: \text{kg} \: \ce{H_2O}} = 0.63 \: \textit{m} \: \ce{C_6H_{12}O_6} \nonumber \]
Step 3: Think about your result.
The answer represents the moles of glucose per kilogram of water and has two significant figures.
What is the mass of salt, NaCl, in a 0.98 m solution prepared with 1.5 kg of water?
- Answer
-
\[ Molality = \dfrac{moles\:solute}{mass\:solvent}\nonumber\]
This can be rearranged to:
\[moles = Molality \times mass\:solvent = (0.98\:\textit{m})(1.5\:kg) = 1.47\:mol\:NaCl \nonumber \]
Finally, convert moles to mass using the molar mass of NaCl (calculated from values on the periodic table):
\[1.47\:mol\:NaCl\times \dfrac{58.44\:g\:NaCl}{1\:mol\:NaCl} = 86\:g\: NaCl \nonumber \]
Alternatively, the problem could be solved using unit analysis. Start with the simple unit (kg). Remember that m stands for mol/kg so it is a compound unit and can be split into two lines to make a conversion factor.
\[1.5\:kg \times \dfrac{0.98\:mol\:NaCl}{1\:kg}\times \dfrac{58.44\:g\:NaCl}{1\:mol\:NaCl} = 86\:g\: NaCl \nonumber \]
Molality and molarity are closely related in value for dilute aqueous solutions because the density of those solutions is relatively close to \(1.0 \: \text{g/mL}\). This means that \(1.0 \: \text{L}\) of solution has nearly a mass of \(1.0 \: \text{kg}\). As the solution becomes more concentrated, its density will not be as close to \(1.0 \: \text{g/mL}\) and the molality value will be different than the molarity. For solutions with solvents other than water, the molality will be very different than the molarity. Make sure that you are paying attention to which quantity is being used in a given problem.
Summary
- The molality \(\left( \textit{m} \right)\) of a solution is the moles of solute divided by the kilograms of solvent.
- The value of molality does not change with temperature.