[merged with Libre] 12.4 Solution Concentration

(a) The mole fraction of ethylene glycol may be computed by first deriving molar amounts of both solution components and then substituting these amounts into the unit definition.

$\begin{array}{l}\\ \text{mol}{\mathrm{C}}_2{\mathrm{H}}_4{\left(\mathrm{OH}\right)}_2=2220\text{g}\times {\displaystyle \frac{1\text{mol}{\mathrm{C}}_2{\mathrm{H}}_4{\left(\mathrm{OH}\right)}_2}{62.07\text{g}{\mathrm{C}}_2{\mathrm{H}}_4{\left(\mathrm{OH}\right)}_2}}=35.8\text{mol}{\mathrm{C}}_2{\mathrm{H}}_4{\left(\mathrm{OH}\right)}_2\\ \text{mol}{\mathrm{H}}_2\mathrm{O}=2000\text{g}\times {\displaystyle \frac{1\text{mol}{\mathrm{H}}_2\mathrm{O}}{18.02\text{g}{\text{ H}}_2\mathrm{O}}}=11.1\text{mol}{\mathrm{H}}_2\mathrm{O}\\ X_{\text{ethylene }\text{glycol}}={\displaystyle \frac{35.8\text{mol}{\mathrm{C}}_2{\text{H}}_4{\left(\mathrm{OH}\right)}_2}{(35.8+11.1)\text{mol total}}}=0.763\end{array}$

Notice that mole fraction is a dimensionless property, being the ratio of properties with identical units (moles).
(b) To find molality, we need to know the moles of the solute and the mass of the solvent (in kg).
First, use the given mass of ethylene glycol and its molar mass to find the moles of solute:

$2220\text{g}{\mathrm{C}}_2{\mathrm{H}}_4{\left(\mathrm{OH}\right)}_2\left({\displaystyle \frac{\text{mol}{\mathrm{C}}_2{\mathrm{H}}_2{\left(\mathrm{OH}\right)}_2}{62.07\text{g}}}\right)=35.8\text{mol}{\mathrm{C}}_2{\mathrm{H}}_4{\left(\mathrm{OH}\right)}_2$

Then, convert the mass of the water from grams to kilograms:

$\text{2000 g}{\mathrm{H}}_2\mathrm{O}\left({\displaystyle \frac{1\text{kg}}{1000\text{g}}}\right)=\text{2 kg}{\mathrm{H}}_2\mathrm{O}$

Finally, calculate molarity per its definition:

$\begin{array}{l}\\ \hfill \text{molality}& =& {\displaystyle \frac{\text{mol solute}}{\text{kg solvent}}}\\ \hfill \text{molality}& =& {\displaystyle \frac{35.8\text{mol}{\mathrm{C}}_2{\mathrm{H}}_4{\left(\mathrm{OH}\right)}_2}{2\text{kg}{\mathrm{H}}_2\mathrm{O}}}\\ \hfill \text{molality}& =& 17.9m\end{array}$

Converting from one concentration unit to another is accomplished by first comparing the two unit definitions. In this case, both units have the same numerator (moles of solute) but different denominators. The provided molal concentration may be written as:

${\displaystyle \frac{3.0\text{mol NaCl}}{1.0\text{kg}{\mathrm{H}}_2\mathrm{O}}}$

The numerator for this solution’s mole fraction is, therefore, $3.0mol\mathrm{NaCl}$. The denominator may be computed by deriving the molar amount of water corresponding to $1.0kg$

$1.0\text{kg}{\mathrm{H}}_2\mathrm{O}\left({\displaystyle \frac{1000\text{g}}{1\text{kg}}}\right)\left({\displaystyle \frac{\text{mol}{\mathrm{H}}_2\mathrm{O}}{18.02\text{g}}}\right)=55\text{mol}{\mathrm{H}}_2\mathrm{O}$

and then substituting these molar amounts into the definition for mole fraction.
Deleted:

$\begin{array}{ccc}\hfill X_{{\text{H}}_2\text{O}}& =& {\displaystyle \frac{\text{mol}{\text{ H}}_2\text{O}}{\text{mol NaCl}+\text{mol}{\text{ H}}_2\text{O}}}\hfill \\ \hfill X_{{\text{H}}_2\text{O}}& =& {\displaystyle \frac{55\text{mol}{\text{ H}}_2\text{O}}{3.0\text{mol NaCl}+55\text{mol}{\text{ H}}_2\text{O}}}\hfill \\ \hfill X_{{\text{H}}_2\text{O}}& =& 0.95\hfill \\ \hfill X_{\text{NaCl}}& =& {\displaystyle \frac{\text{mol NaCl}}{\text{mol NaCl}+\text{mol}{\text{ H}}_2\text{O}}}\hfill \\ \hfill X_{\text{NaCl}}& =& {\displaystyle \frac{3.0\text{mol}{\text{ H}}_2\text{O}}{3.0\text{mol NaCl}+55\text{mol}{\text{ H}}_2\text{O}}}\hfill \\ \hfill X_{\text{NaCl}}& =& 0.052\hfill \end{array}$

The provided molal concentration may be explicitly written as: $M=0.556\text{mol glucose/ 1 L solution}$

Consider the definition of molality:

$$m=\text{mol solute/ kg solvent}$$

The amount of glucose in $1L$ of this solution is $0.556mol$, so the mass of water in this volume of solution is needed.

First, compute the mass of $1.00L$ of the solution:

$$(1.0\text{L soln})(1.04\text{g/mL})(1000\text{mL/L})(1\text{kg}/1000\text{g})=1.04\text{kg soln}$$

This is the mass of both the water and its solute, glucose, and so the mass of glucose must be subtracted. Compute the mass of glucose from its molar amount:

$$(0.556\text{mol glucose})(180.2\text{g/1 mol})=100.2\text{g or }0.1002\text{kg}$$

Subtracting the mass of glucose yields the mass of water in the solution:

$$1.04\text{kg solution}–0.1002\text{kg glucose}=0.94\text{kg water}$$

Finally, the molality of glucose in this solution is computed as:

$$m=0.556\text{mol glucose}/0.94\text{kg water}=0.59\text{m}$$