Concentration and dilutions of solutions
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Concentration refers to the amount of solute present in a given quantity (known volume) of a solution.
Scientists use different methods to express concentration. Depending on the method, the units differ.
Different methods of expressing concentration

Percent concentration (%)

Parts per million (ppm)

Parts per billion (ppb)

Molarity (M)
Percent concentration
There are three ways of expressing concentration as a percentage.

mass/mass percent concentration, (m/m)% \[ \frac{\textrm{mass of solute}\left(g\right)}{\textrm{mass of solution$\left(g\right)$ }}\times100 \] Example: What is the percentage by mass concentration of a sodium chloride solution prepared by dissolving \(3.25\textrm{g}\) of sodium chloride in \(42.50\textrm{g}\) of water?
The solute and the solvent masses are given, and we are asked to calculate the percent concentration of sodium chloride by mass.

volume/volume percent concentration, (v/v)% \[ \frac{\textrm{volume of solute}\left(ml\right)}{\textrm{volume of solution}\left(ml\right)}\times100 \] Example: If \(8.00ml\) of ethanol is dissolved in enough water to give \(80.00ml\) of solution, what is the percent by volume concentration of ethanol in the resulting solution?
We are asked to calculate the ethanol concentration in percentage by volume. The solute and the solution volumes are given. \[\begin{align*} = & \frac{\textrm{volume of solute}\left(ml\right)}{\textrm{volume of solution}\left(ml\right)}\times100\\ = & \frac{8.00ml}{80.00ml}\times100\\ = & 10.0\% \end{align*}\]

mass/volume percent concentration, (m/v)% \[ \frac{\textrm{mass of solute}\left(g\right)}{\textrm{volume of solution}\left(ml\right)}\times100 \] Example: If \(5.0g\) of \(\ce{MgCl}_{2}\) is dissolved in enough water to give \(200\textrm{ml}\) of solution, calculate the concentration of \(\ce{MgCl}_{2}\) in (m/v)%.
We are asked to calculate the concentration of \(\ce{MgCl}_{2}\) in (m/v)%. Mass of the \(\ce{MgCl}_{2}\) and volume of the resulting solution are given. \[\begin{align*} = & \frac{\textrm{mass of solute}\left(g\right)}{\textrm{volume of solution}\left(ml\right)}\times100\\ = & \frac{5.0g}{200ml}\times100\\ = & 2.5\% \end{align*}\]
Parts Per Million and Parts Per Billion
Parts per million  number of parts (mass or volume) present in one million parts (mass or volume).
Parts per billion  number of parts (mass or volume) present in one billion parts (mass or volume).
Parts can be in either mass or volume units. Both solute and solvent must have the same units. For example, one gram of solute is present in a million grams of solution, one millilitre of benzene is present in a million millilitres of air. \[ 1ppm=\frac{1ml}{1000000ml}\times1000000 \]
Example: Drinking water contains \(0.002g\) of \(\ce{Mg}^{+}\)ions in \(100g\) of water. What is this concentration in parts per million?
We are asked to calculate the given concentration in ppm. \[\begin{align*} = & \frac{0.002g}{100g}\times1000000\\ = & 20ppm \end{align*}\]
Molarity (M)
Molarity is the number of solute moles present in one litre of a solution. \[ \textrm{Molarity(M)}=\frac{\textrm{moles of solute}}{\textrm{litres of solution}} \]
Example: If \(2.50\textrm{mol}\) of \(\ce{CuSO}_{4}\) are dissolved in enough water to give a \(500.0\mathscr{\textrm{ml}}\) solution, what is the molarity of the resulting solution?
We are given the number of moles present in \(500\textrm{ml}\) of a solution and asked to calculate the molarity. Molarity means the number of moles present in \(1000\textrm{ml}\) \(\left(1\textrm{L}\right)\) of solution. \[\begin{align*} = & \frac{2.50\textrm{mol}}{500.0\textrm{ml}}\times1000\textrm{ml}\\ = & 5.00\textrm{M} \end{align*}\]
Calculations using molarity as a conversion factor
 Calculating the number of moles or the mass of a solute present in a known volume of a solution.
Example 1: How many moles of glucose are present in \(250\textrm{ml}\) of a \(1.50\textrm{M}\) glucose solution?
We are given the molarity of the solution and asked to calculate the number of moles present in \(250\textrm{ml}\) of the same solution. Molarity indicates the number of moles present in \(1000\textrm{ml}\). So, we can use molarity as a conversion factor to calculate the number of moles present in a given volume as follows: \[\begin{align*} = & \frac{1.50\textrm{mol}}{1000\textrm{ml}}\times250\textrm{ml}\\ = & 0.375\textrm{mol} \end{align*}\]
Example 2: How many grams of glucose are present in \(250\textrm{ml}\) of a \(1.5\textrm{M}\) glucose solution? The molar mass of glucose is \(180.15\textrm{g/mol}\).
Here, we are given the molarity of the solution and asked to calculate the mass of the solute present in a given volume. As in the previous example, we can use molarity as the conversion factor to get the number of moles of the solute present in the given volume. Then, use the \(n=\frac{m}{M}\) equation to convert moles to grams.
The number of moles of glucose present in \(250\textrm{ml}\) of solution: \[\begin{align*} = & \frac{1.50\textrm{mol}}{1000\textrm{ml}}\times250\textrm{ml}\\ = & 0.375\textrm{mol} \end{align*}\]
Grams of glucose present in \(250\textrm{ml}\) solution \[\begin{align*} n & =\frac{m}{M}\\ n\times M & =m\\ 0.375\textrm{mol}\times180.15\textrm{g/mol} & =m\\ m & =67.6\textrm{g} \end{align*}\]
 Calculating the volume of a solution from a given quantity of a solute.
Example: How many millilitres of a \(1.00\textrm{M}\) \(\ce{HCl}\) solution would be required to obtain \(7.3\textrm{g}\) of \(\ce{HCl}\)? The molar mass of \(\ce{HCl}\) is \(36.458\textrm{g/mol}\).
We are given the molarity of the solution and asked to calculate the volume that the given grams of the solute would present. First, convert the grams of the solute into moles using \(n=\frac{m}{M}\). Then use molarity as the conversion factor to find the volume that the calculated moles would present.
Number of moles of \(\ce{HCl}\) corresponds to \(7.3\textrm{g}\): \[\begin{align*} n & =\frac{m}{M}\\ n & =\frac{7.3\textrm{g}}{36.458\textrm{g/mol}}\\ n & =0.20\textrm{mol} \end{align*}\]
The volume that the \(0.20\textrm{mol}\) of \(\ce{HCl}\) would present: \[\begin{align*} \frac{1000\textrm{ml}}{1.00\textrm{mol}}\times0.20\textrm{mol} & =200\textrm{ml} \end{align*}\]
Dilution of Solutions
Dilution is a process used to lower the concentration of the original solution by adding more solvent. The concentrated solution is known as the stock solution. However, the amount of solute present in the solution remains constant before and after the dilution; only the volume changes. The following equation shows the relationship between concentration and volume before and after dilution. This equation is built upon the fact that the same amount of solute is present in the solution before (stock solution) and after (diluted solution) the dilution. \[ C_{1}V_{1}=C_{2}V_{2} \]
Where;
\(C_{1}\)= Concentration of the stock solution
\(V_{1}\)= Volume of the stock solution
\(C_{2}\)= Concentration of the diluted solution
\(V_{2}\)= Volume of the diluted solution
Example: A chemist prepared \(10\textrm{ml}\) of a \(1.00\textrm{M}\) \(\ce{HCl}\) solution by adding water to the \(3.0\textrm{M}\) stock solution of \(\ce{HCl}\). How much stock solution should be taken for the dilution?
We are given the concentration of the diluted solution, the volume of the diluted solution, concentration of the stock solution, and we are asked to calculate the volume of the stock solution used for the dilution. As the number of solute moles remains constant before and after the dilution, we can use the equation \(C_{1}V_{1}=C_{2}V_{2}\) to calculate the volume of the stock solution. \[\begin{align*} C_{1}V_{1} & =C_{2}V_{2}\\ 3.0\textrm{M}\times V_{1} & =1.0\textrm{M}\times10\textrm{ml}\\ V_{1} & =\frac{1.0\textrm{M}\times10\textrm{ml}}{3.0\textrm{M}}\\ V_{1} & =3.3\textrm{ml} \end{align*}\]