Concentration Units


g/100g

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

\[r = \dfrac{\mathrm{g~solute}}{100~\mathrm{g~solvent}}\]

Mass Concentration

  • Given as \(\rho\)
  • Units: g L–1
  • Temperature dependent
  • Used when molecular weights are unknown but masses are known

\[\rho = \dfrac{\mathrm{g~solute}}{\mathrm{L~solution}}\] 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
  • Used for quantitative solution reactions and titrations

\[M = \dfrac{\mathrm{mol~solute}}{\mathrm{L~solution}}\] 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
  • Used in colligative property calculations

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

Mole Fraction

  • Given as \(\chi\)
  • Unitless
  • Used for characterizing the partial pressures of gases and vapor pressures of solutions

\[\chi_{\mathrm{solute}} = \dfrac{n_{\mathrm{solute}}}{n_{\mathrm{solution}}}\] \[\chi_{\mathrm{solvent}} = \dfrac{n_{\mathrm{solvent}}}{n_{\mathrm{solution}}}\]

Mole Percent

  • Given as mol %
  • Units: %
  • Used for characterizing the partial pressures of gases and vapor pressures of solutions

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

Mass Fraction

  • Given as \(\omega\)
  • Unitless
  • Generally used for solid- and liquid-phase solutions

\[\omega_{\mathrm{solute}} = \dfrac{m_{\mathrm{solute}}}{m_{\mathrm{solution}}}\] \[\omega_{\mathrm{solvent}} = \dfrac{m_{\mathrm{solvent}}}{m_{\mathrm{solution}}}\]

Mass Percent

  • Given as mass %, wt %, % wt, percent by mass, weight-weight percentage, wt/wt %, w/w %
  • Units: %
  • Generally used for solid- and liquid-phase solutions
  • Note: Assume a volume of solution (100g) if starting with this.

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

Mass Volume Percent

  • Given as m/v %, (m/v%)
  • Also called weight volume % – w/v %, (w/v%)
  • Ratio of solute mass (in g) to volume (in mL) of solution multiplied by 100%
  • Generally used for liquid- and gas-phase solutions

\[m/v~\% = \dfrac{m_{\mathrm{solute~(g)}}}{V_{\mathrm{solution~(mL)}}} \times 100\%\]

Parts by Mass

  • Generally used for dilute to very dilute solutions

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

Unit and multiplication factor

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

Parts by Volume

  • Generally used for dilute to very dilute solutions

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

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 °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. \(m_{\mathrm{solv}} = 500.0~\mathrm{g}\)
  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 \[\begin{align*} \mathrm{mol~\%~solvent} &= \chi_{\mathrm{solvent}} \times 100\% \\[2ex] &= (9.95\times 10^{-1}) (100\%) \\[2ex] &= 99.5\% \end{align*}\]


Problem 8

An aqueous solution of NaCl (f.w. = 58.44 g mol–1) is 26.45 % by mass and has a density of 1.1944 g mL–1. The density of NaCl is 2.16 g cm–1. Determine the following concentrations.

  1. g/100g
  2. Molar concentration (M)
  3. Molality (m)
  4. mol fraction
  5. mol percent
  6. mass fraction
  7. mass concentration (g L–1)
  8. parts by mass (ppm)
  9. parts by volume (ppm)
  10. mass volume percent (m/v%)
Solution
  1. 36.96g/100g
  2. 5.406 M
  3. 6.154 m
  4. 0.0998
  5. 9.98 %
  6. 0.2645
  7. 315.98 g L–1
  8. 264,500 ppm (by mass)
  9. 146,254 ppm (by volume)
  10. 31.59 m/v%


Timed Assessments


Concentration Units

VIDEO

Answers

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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