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Wize High School Grade 11 Chemistry Textbook > Quantities in Chemistry

Concentrations

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Concentration

Amount concentration

  • Amount concentration is a quantitative measure of the amount of solute present in a solution
c=nsolute(mol)Vsolution(L)\boxed{c = \dfrac{n_\text{solute} (\text{mol})}{V_\text{solution} (L)}}
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Percent Concentrations

  • When it comes to consumer products, often times concentration is expressed in terms of percentages
% Volume by Volume (% V/V)% Weight by Volume (% w/V) % Weight by Weight (% w/w) c=VsoluteVsolution×100%c=msolute(g)Vsolution(mL)×100%c=msolutemsolution×100%\def\arraystretch{2.5} \begin{array}{c:c:c} \footnotesize \text {\% Volume by Volume (\% V/V)}& \footnotesize\text{\% Weight by Volume (\% w/V) } & \footnotesize \text{\% Weight by Weight (\% w/w) } \\ \hline c=\dfrac{V_\text{solute}}{V_\text{solution}}\times100\% &c=\dfrac{m_\text{solute}(g)}{V_\text{solution}(mL)}\times100\% & c=\dfrac{m_\text{solute}}{m_\text{solution}}\times100\% \end{array}


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Dilute Solutions

  • When working with very dilute solutions, we can express their concentrations using parts-per notation
  • We can make the assumption that very dilute aqueous solutions have a density equal to the density of water or 1g/mL, then we can use the mass percent equation to get the part-per concentration
Parts per Million (ppm)Parts per Billion (ppb)Parts per Trillion (ppt)cppm =msolutemsolution×106cppb =msolutemsolution×109cppt =msolutemsolution×1012\def\arraystretch{2.5} \begin{array}{c:c:c} \footnotesize \text {Parts per Million (ppm)}& \footnotesize\text{Parts per Billion (ppb)} & \footnotesize \text{Parts per Trillion (ppt)} \\ \hline c_{ppm}\ =\dfrac{m_\text{solute}}{m_\text{solution}}\times10^6&c_{ppb}\ =\dfrac{m_\text{solute}}{m_\text{solution}}\times10^9& c_{ppt}\ =\dfrac{m_\text{solute}}{m_\text{solution}}\times10^{12} \end{array}

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Example: Calculation of Concentration

Calculate the amount concentration of a solution of 24g NaOH in 150mL of water.

Let's first calculate the molar mass of NaOH:
MNaOHM_{NaOH}
=MNa+MO+MH=M_{Na}+M_O+M_H
=22.990g/mol+15.999g/mol+1.008g/mol=22.990g/mol+15.999g/mol+1.008g/mol
=39.997g/mol=39.997g/mol

Now, we convert the mass to moles:
nNaOHn_{NaOH}

=24g39.997g/mol=\frac{24g}{39.997g/mol}

=24g1÷39.997gmol=\frac{24g}{1}\div\frac{39.997g}{mol}

=24g1×mol39.997g=\frac{24g}{1}\times\frac{mol}{39.997g}

=0.6000 mol=0.6000\ mol (in 150mL)

Finally, we convert the moles in given volume to moles in 1 L
0.6000 mol150 mL\frac{0.6000\ mol}{150\ mL}

=0.6000 mol150 mL×1000 mL1 L=\frac{0.6000\ mol}{150\ mL}\times\frac{1000\ mL}{1\ L}

=4 mol/L=4\ mol/L

Therefore concentration = 4 mol/Lmol/L

Practice: Amount Concentration

To make a 2.00mol/L solution, how many moles of solute will be needed if 4.0 liters of solution are required? Give your answer to one decimal place; do not include units.

Practice: Percent Volume by Volume

How many mL of hydrogen peroxide are needed to make a 8.5% solution by volume of hydrogen peroxide if you want to make 450mL of solution?

Practice: Dilute Concentrations

Symptoms of mercury poisoning become apparent after a person has accumulated 20mg of mercury. If a person ingested 30mg of mercury, what concentration of mercury in parts per million, are in his body? Assume the person has a mass of 65kg.