# Concentration Units

## Weight Ratio Percentage

• The common form of solubility
• Units: g solute / 100 g solvent

$\mathrm{weight~ratio~\%} = \dfrac{\mathrm{g~solute}}{100~\mathrm{g~solvent}}$

## Mass Concentration

• Given as $$\rho$$
• Units: g L–1
• Temperature dependent

$\rho = \dfrac{\mathrm{g~solute}}{\mathrm{L~solution}}$ Mass concentration is the density of a component in a mixture (mass over volume), hence the use of the Greek letter $$\rho$$ to represent mass concentration as it is used to represent density,

## Molarity

• Given as M
• Units: mol L–1
• Temperature dependent

$M = \dfrac{\mathrm{mol~solute}}{\mathrm{L~solution}}$ Molarity is derived from mass concentration by converting grams of solute into moles of solute via molar mass.

## Molality

• Given as m
• Units: mol kg–1
• Temperature independent

$m = \dfrac{\mathrm{mol~solute}}{\mathrm{kg~solvent}}$

## Mole Fraction

• Given as $$\chi$$
• Unitless

$\chi_{\mathrm{solute}} = \dfrac{n_{\mathrm{solute}}}{n_{\mathrm{solution}}}$ $\chi_{\mathrm{solvent}} = \dfrac{n_{\mathrm{solvent}}}{n_{\mathrm{solution}}}$

## Mole Percent

• Given as mol %
• Units: %

$\mathrm{mol~\%~solute} = \chi_{\mathrm{solute}} \times 100\%$ $\mathrm{mol~\%~solvent} = \chi_{\mathrm{solvent}} \times 100\%$

## Mass Fraction

• Given as $$\omega$$
• Unitless

$\omega_{\mathrm{solute}} = \dfrac{m_{\mathrm{solute}}}{m_{\mathrm{solution}}}$ $\omega_{\mathrm{solvent}} = \dfrac{m_{\mathrm{solvent}}}{m_{\mathrm{solution}}}$

## Mass Percent

• Given as mass %, wt %, % wt, percent by mass, weight-weight percentage, wt/wt %, w/w %
• Units: %

$\mathrm{mass~\%~solute} = \omega_{\mathrm{solute}} \times 100\%$ $\mathrm{mass~\%~solvent} = \omega_{\mathrm{solvent}} \times 100\%$

## Parts by Mass

$\dfrac{m_\mathrm{{solute}}}{m_\mathrm{{solution}}} \times \mathrm{~multiplication~factor}$

Unit and multiplication factor

• Parts per hundred (percent by mass; %) – 100
• Parts per million (ppm) – 106
• Parts per billion (ppb) – 109

## Parts by Volume

$\dfrac{V_\mathrm{{solute}}}{V_\mathrm{{solution}}} \times \mathrm{~multiplication~factor}$

Unit and multiplication factor

• Parts per hundred (percent by volume; %) – 100
• Parts per million (ppm) – 106
• Parts per billion (ppb) – 109

## Practice Problems

### Problem 1

A solution is prepared by dissolving 17.2 g ethylene glycol (C2H6O2; m.w. = 62.07 g mol–1) in 0.500 kg of water at 25 $$^{\circ}$$C. The final solution is 515.0 mL. Calculate the following concentrations:

1. M
2. m
3. % by mass of solute
4. mol fraction solute
5. mol % solute
6. mol fraction solvent
7. mol % solvent

Solution

Givens:

1. $$V_{\mathrm{solv}} = 500.0~\mathrm{g}$$
2. $$V_{\mathrm{soln}} = 515.0~\mathrm{mL}$$
3. $$m_{\mathrm{solu}} = 17.2~\mathrm{g}$$

Find moles of solute: \begin{align*} n_{\mathrm{C_2H_6O_2}} &= 17.2~\mathrm{g} \times \dfrac{\mathrm{mol}}{62.07~\mathrm{g}} = 0.2771~\mathrm{mol} \end{align*}

1.) molarity \begin{align*} M &= \dfrac{n_{\mathrm{solute}}}{\mathrm{L~solution}} \\[2ex] &= \dfrac{0.2771~\mathrm{mol}}{0.5150~\mathrm{L}} = 0.538 \end{align*}

2.) molality \begin{align*} m &= \dfrac{n_{\mathrm{solute}}}{\mathrm{kg~solvent}} \\[2ex] &= \dfrac{0.2771~\mathrm{mol}}{0.500~\mathrm{kg}} = 0.554 \end{align*}

3.) % by mass of solute \begin{align*} \%~\mathrm{by~mass} &= \dfrac{\mathrm{mass~solute}}{\mathrm{mass~soln.}} \times 100\% \\[2ex] &= \dfrac{17.2~\mathrm{g}}{17.2~\mathrm{g} + 500~\mathrm{g}} \times 100\% \\[2ex] &= 3.33\% \end{align*}

4.) mol fraction solute

Find moles of solvent: \begin{align*} n_{\mathrm{H_2O}} &= 500~\mathrm{g~H_2O} \left ( \dfrac{\mathrm{mol}}{18.02~\mathrm{g}} \right ) = 27.75~\mathrm{mol} \end{align*}

\begin{align*} \chi_{\mathrm{solute}} &= \dfrac{n_{\mathrm{solute}}}{n_{\mathrm{solution}}} \\[2ex] &= \dfrac{0.2771~\mathrm{mol}}{0.2771~\mathrm{mol} + 27.75~\mathrm{mol}} \\[2ex] &= 9.89\times 10^{-3} \end{align*}

5.) mol % solute \begin{align*} \mathrm{mol~\%~solute} &= \chi_{\mathrm{solute}} \times 100\% \\[2ex] &= (9.89\times 10^{-3}) (100\%) \\[2ex] &= 0.989\% \end{align*}

6.) mol fraction solvent \begin{align*} \chi_{\mathrm{solvent}} &= \dfrac{n_{\mathrm{solvent}}}{n_{\mathrm{solution}}} \\[2ex] &= \dfrac{27.75~\mathrm{mol}}{0.2771~\mathrm{mol} + 27.75~\mathrm{mol}} \\[2ex] &= 9.90\times 10^{-1} \end{align*}

7.) mol % solvent \begin{align*} \mathrm{mol~\%~solvent} &= \chi_{\mathrm{solvent}} \times 100\% \\[2ex] &= (9.90\times 10^{-1}) (100\%) \\[2ex] &= 99.0\% \end{align*}

### Problem 2

A 100 g aqueous solution is 20% (by mass) NaCl (m.w. = 58.44 g mol–1). What is the mole fraction of NaCl?

Solution

\begin{align*} m_{\mathrm{NaCl}} &= 100~\mathrm{g~solution} \times 0.2 = 20~\mathrm{g} \\ m_{\mathrm{water}} &= 100~\mathrm{g~solution} - 20~\mathrm{g~NaCl} = 80.0~\mathrm{g} \\[2ex] n_{\mathrm{NaCl}} &= 20~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{58.44~\mathrm{g}} \right ) = 0.34223~\mathrm{mol}\\ n_{\mathrm{water}} &= 80~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{18.02~\mathrm{g}} \right ) = 4.4395~\mathrm{mol}\\[2ex] \chi_{\mathrm{NaCl}} &= \dfrac{n_{\mathrm{NaCl}}}{n_{\mathrm{tot}}} = \dfrac{0.34223}{4.78198} = 0.072 \end{align*}

### Problem 3

A 200 g aqueous solution is 20% (by mass) NaCl (m.w. = 58.44 g mol–1). What is the mole fraction of NaCl?

Solution

\begin{align*} m_{\mathrm{NaCl}} &= 200~\mathrm{g~solution} \times 0.2 = 40~\mathrm{g} \\ m_{\mathrm{water}} &= 200~\mathrm{g~solution} - 40~\mathrm{g~NaCl} = 160.0~\mathrm{g} \\[2ex] n_{\mathrm{NaCl}} &= 40~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{58.4~\mathrm{g}} \right ) = 0.685~\mathrm{mol}\\ n_{\mathrm{water}} &= 160~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{18.02~\mathrm{g}} \right ) = 8.88~\mathrm{mol}\\[2ex] \chi_{\mathrm{NaCl}} &= \dfrac{n_{\mathrm{NaCl}}}{n_{\mathrm{tot}}} = \dfrac{0.685}{9.565} = 0.072 \end{align*}

### Problem 4

A 5 g aqueous solution is 20% (by mass) NaCl (m.w. = 58.44 g mol–1). What is the mole fraction of NaCl?

Solution

\begin{align*} m_{\mathrm{NaCl}} &= 5.00~\mathrm{g~solution} \times 0.2 = 1.0~\mathrm{g} \\ m_{\mathrm{water}} &= 5.00~\mathrm{g~solution} - 1.0~\mathrm{g~NaCl} = 4.0~\mathrm{g} \\[2ex] n_{\mathrm{NaCl}} &= 1.0~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{58.4~\mathrm{g}} \right ) = 0.0171~\mathrm{mol}\\ n_{\mathrm{water}} &= 4.0~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{18.02~\mathrm{g}} \right ) = 0.222~\mathrm{mol}\\[2ex] \chi_{\mathrm{NaCl}} &= \dfrac{n_{\mathrm{NaCl}}}{n_{\mathrm{tot}}} = \dfrac{0.0171}{0.2391} = 0.072 \end{align*}

### Problem 5

What is the mole fraction of NaCl (m.w. = 58.44 g mol–1) in a 20% (by mass) aqueous solution?

Solution

Assume a mass of the solution. 100 g is convenient.

\begin{align*} m_{\mathrm{NaCl}} &= 100~\mathrm{g~solution} \times 0.2 = 20~\mathrm{g} \\ m_{\mathrm{water}} &= 100~\mathrm{g~solution} - 20~\mathrm{g~NaCl} = 80.0~\mathrm{g} \\[2ex] n_{\mathrm{NaCl}} &= 20~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{58.4~\mathrm{g}} \right ) = 0.34247~\mathrm{mol}\\ n_{\mathrm{water}} &= 80~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{18.02~\mathrm{g}} \right ) = 4.4395~\mathrm{mol}\\[2ex] \chi_{\mathrm{NaCl}} &= \dfrac{n_{\mathrm{NaCl}}}{n_{\mathrm{tot}}} = \dfrac{0.34247}{4.78198} = 0.072 \end{align*}

### Problem 6

This problem illustrates determining the concentration of 1 teaspoon sugar in a typical cup of tea.

A typical cup of tea (150.0 mL) is prepared. One teaspoon of glucose (C6H12O6; m.w. = 180.156 g mol–1) is added (approximately 4.0 g of sugar). The resulting volume of solution is 152.6 mL, your standard cup of tea. Treating the sugar as the solute and the tea as water, the solvent (i.e. treat tea as water; dtea = dsolvent = dwater = 1.00 g mL–1), calculate the following concentrations:

1. M
2. m
3. % by mass of solute
4. mol fraction solute
5. mol % solute
6. mol fraction solvent
7. mol % solvent

Solution

Givens:

1. $$V_{\mathrm{solv}} = 150.0~\mathrm{mL}$$
2. $$\mathrm{m.w.~solute} = 180.156~\mathrm{g~mol^{-1}}$$
3. $$m_{\mathrm{solu}} = 4.0~\mathrm{g}$$
4. $$V_{\mathrm{soln}} = 152.6~\mathrm{mL}$$
5. $$d_{\mathrm{solv}} = 1.00~\mathrm{g~mL^{-1}}$$

Find moles of solute: \begin{align*} n_{\mathrm{C_2H_6O_2}} &= 4.0~\mathrm{g} \times \dfrac{\mathrm{mol}}{180.156~\mathrm{g}} = 0.0222~\mathrm{mol} \end{align*}

1.) molarity \begin{align*} M &= \dfrac{n_{\mathrm{solute}}}{\mathrm{L~solution}} \\[2ex] &= \dfrac{0.0222~\mathrm{mol}}{0.1526~\mathrm{L}} \\[2ex] &= 0.146 \end{align*}

2.) molality

Find mass of tea (treat it like water according to the question):

\begin{align*} d &= \dfrac{m}{V} \\[2ex] m_{\mathrm{tea}} &= dV \\[2ex] &= \left ( \dfrac{1.00~\mathrm{g}}{\mathrm{mL}}\right ) \left ( 150.0 ~\mathrm{mL} \right )\\[2ex] &= 150.0~\mathrm{g} = 0.150~\mathrm{kg} \end{align*}

Find molality: \begin{align*} m &= \dfrac{n_{\mathrm{solute}}}{\mathrm{kg~solvent}} \\[2ex] &= \dfrac{0.0222~\mathrm{mol}}{0.150~\mathrm{kg}} \\[2ex] &= 0.148 \end{align*}

3.) % by mass of solute \begin{align*} \%~\mathrm{by~mass} &= \dfrac{\mathrm{mass~solute}}{\mathrm{mass~soln.}} \times 100\% \\[2ex] &= \dfrac{4.00~\mathrm{g}}{4.00~\mathrm{g} + 150.0~\mathrm{g}} \times 100\% \\[2ex] &= 2.60\% \end{align*}

4.) mol fraction solute

Find moles of solvent: \begin{align*} n_{\mathrm{H_2O}} &= 150.0~\mathrm{g~H_2O} \left ( \dfrac{\mathrm{mol}}{18.02~\mathrm{g}} \right ) = 8.32~\mathrm{mol} \end{align*}

Find mol fraction of solute: \begin{align*} \chi_{\mathrm{solute}} &= \dfrac{n_{\mathrm{solute}}}{n_{\mathrm{solution}}} \\[2ex] &= \dfrac{0.0222~\mathrm{mol}}{0.0222~\mathrm{mol} + 8.32~\mathrm{mol}} \\[2ex] &= 2.66\times 10^{-3} \end{align*}

5.) mol % solute \begin{align*} \mathrm{mol~\%~solute} &= \chi_{\mathrm{solute}} \times 100\% \\[2ex] &= (2.66\times 10^{-3}) (100\%) \\[2ex] &= 0.266\% \end{align*}

6.) mol fraction solvent \begin{align*} \chi_{\mathrm{solvent}} &= \dfrac{n_{\mathrm{solvent}}}{n_{\mathrm{solution}}} \\[2ex] &= \dfrac{8.32~\mathrm{mol}}{0.0222~\mathrm{mol} + 8.32~\mathrm{mol}} \\[2ex] &= 9.97\times 10^{-1} \end{align*}

7.) mol % solvent \begin{align*} \mathrm{mol~\%~solvent} &= \chi_{\mathrm{solvent}} \times 100\% \\[2ex] &= (9.97\times 10^{-1}) (100\%) \\[2ex] &= 99.73\% \end{align*}

### Problem 7

This problem illustrates determining the concentration of 2 teaspoons sugar in a typical cup of tea.

A typical cup of tea (150.0 mL) is prepared. Two teaspoons of glucose (C6H12O6; m.w. = 180.156 g mol–1) is added (approximately 8.0 g of sugar). The resulting volume of solution is 155.2 mL, your standard cup of tea. Treating the sugar as the solute and the tea as water, the solvent (i.e. treat tea as water; dtea = dsolvent = dwater = 1.00 g mL–1), calculate the following concentrations:

1. M
2. m
3. % by mass of solute
4. mol fraction solute
5. mol % solute
6. mol fraction solvent
7. mol % solvent

Solution

Givens:

1. $$V_{\mathrm{solv}} = 150.0~\mathrm{mL}$$
2. $$\mathrm{m.w.~solute} = 180.156~\mathrm{g~mol^{-1}}$$
3. $$m_{\mathrm{solu}} = 8.0~\mathrm{g}$$
4. $$V_{\mathrm{soln}} = 155.2~\mathrm{mL}$$
5. $$d_{\mathrm{solv}} = 1.00~\mathrm{g~mL^{-1}}$$

Find moles of solute: \begin{align*} n_{\mathrm{C_2H_6O_2}} &= 8.0~\mathrm{g} \times \dfrac{\mathrm{mol}}{180.156~\mathrm{g}} = 0.0444~\mathrm{mol} \end{align*}

1.) molarity \begin{align*} M &= \dfrac{n_{\mathrm{solute}}}{\mathrm{L~solution}} \\[2ex] &= \dfrac{0.0222~\mathrm{mol}}{0.1552~\mathrm{L}} \\[2ex] &= 0.286 \end{align*}

2.) molality

Find mass of tea (treat it like water according to the question):

\begin{align*} d &= \dfrac{m}{V} \\[2ex] m_{\mathrm{tea}} &= dV \\[2ex] &= \left ( \dfrac{1.00~\mathrm{g}}{\mathrm{mL}}\right ) \left ( 150.0 ~\mathrm{mL} \right )\\[2ex] &= 150.0~\mathrm{g} = 0.150~\mathrm{kg} \end{align*}

Find molality: \begin{align*} m &= \dfrac{n_{\mathrm{solute}}}{\mathrm{kg~solvent}} \\[2ex] &= \dfrac{0.0444~\mathrm{mol}}{0.150~\mathrm{kg}} \\[2ex] &= 0.296 \end{align*}

3.) % by mass of solute \begin{align*} \%_{\mathrm{solute}}~\mathrm{by~mass} &= \dfrac{\mathrm{mass~solute}}{\mathrm{mass~soln.}} \times 100\% \\[2ex] &= \dfrac{8.00~\mathrm{g}}{8.00~\mathrm{g} + 150.0~\mathrm{g}} \times 100\% \\[2ex] &= 5.06\% \end{align*}

4.) mol fraction solute

Find moles of solvent: \begin{align*} n_{\mathrm{H_2O}} &= 150.0~\mathrm{g~H_2O} \left ( \dfrac{\mathrm{mol}}{18.02~\mathrm{g}} \right ) = 8.32~\mathrm{mol} \end{align*}

Find mol fraction of solute: \begin{align*} \chi_{\mathrm{solute}} &= \dfrac{n_{\mathrm{solute}}}{n_{\mathrm{solution}}} \\[2ex] &= \dfrac{0.0444~\mathrm{mol}}{0.0444~\mathrm{mol} + 8.32~\mathrm{mol}} \\[2ex] &= 5.31\times 10^{-3} \end{align*}

5.) mol % solute \begin{align*} \mathrm{mol~\%~solute} &= \chi_{\mathrm{solute}} \times 100\% \\[2ex] &= (5.31\times 10^{-3}) (100\%) \\[2ex] &= 0.531\% \end{align*}

6.) mol fraction solvent \begin{align*} \chi_{\mathrm{solvent}} &= \dfrac{n_{\mathrm{solvent}}}{n_{\mathrm{solution}}} \\[2ex] &= \dfrac{8.32~\mathrm{mol}}{0.0444~\mathrm{mol} + 8.32~\mathrm{mol}} \\[2ex] &= 9.95\times 10^{-1} \end{align*}

7.) mol % solvent \begin{align*} \mathrm{mol~\%~solvent} &= \chi_{\mathrm{solvent}} \times 100\% \\[2ex] &= (9.95\times 10^{-1}) (100\%) \\[2ex] &= 99.5\% \end{align*}

VIDEO

Question 1

1. 0.538 M
2. 0.554 m
3. 3.33%
4. 0.00989
5. 0.99
6. 0.990
7. 99.0%

Question 2

1. 5.69 M
2. 6.90 m
3. 30.0%
4. 0.11
5. 11.1%
6. 0.889
7. 88.9%

Question 3

1. 0.562 M
2. 0.584 m
3. 16.7%
4. 0.0104
5. 1.04%
6. 0.990
7. 99.0%

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