Solutions
2.0 Methods of expressing concentration of a Solution
2.0 Methods of expressing concentration of a Solution
Concentration of solution is the amount of solute present in given solution or solvent. Concentration can be measured qualitatively and quantitatively. In this chapter, we will express concentration quantitatively.
Several ways of expressing concentration are as follow:
1. Mass percentage ($\% w/w$)
Mass percentage of any component in solution is the mass of component present in $100 gram$ of solution. $$Mass\ \%\ of\ component = (mass\ of\ component/total\ mass\ of\ solution) \times 100$$
Total mass of solution = mass of solute + mass of solvent
Example: $20\% $ of $NaCl$ in water means $20g$ of $NaCl$ is present in $80g$ of water.
2. Volume percentage ($\%v/v$)
Volume percentage of any component is the volume of component present in $100 ml$ of solution. $$volume\ \% of\ component = (volume\ of\ component\ in\ ml/total\ volume\ of\ solution\ in\ ml) \times 100$$
$35\% $ $(\frac{v}{v})$ solution of ethylene glycol in water means $35 ml$ of ethylene glycol is disolved in water such that total volume of solution is $100ml$ (this solution of ethylene glycol is used as antifreeze in vehicles).
3. Mass by Volume percentage ($\%w/v$) $$\% w/v = (mass\ of\ solute\ in\ gram/volume\ of\ solution\ in\ ml) \times 100$$
4. Parts per million ($ppm$) $$parts\;per\;million = \left( {\frac{{number\;of\;parts\;of\;component}}{{total\;number\;of\;parts\;of\;all\;component}}} \right) \times {10^6}$$
The number of parts can be measured by mass or volume.
If parts are to be measured by mass then, $$parts\;per\;million = \left( {\frac{{mass\;of\;component}}{{total\;mass\;of\;solution}}} \right) \times {10^6}$$
Hardness of water and concentration of pollutant in air or water is calculated in $ppm$.
5. Mole fraction ($x$)
Mole fraction of the component is the number of moles of component divided by total number of moles of solution.
Mathematically, $$mole\;fraction = \left( {\frac{{no\;of\;moles\;of\;component}}{{total\;number\;of\;moles\;of\;solution}}} \right) \times {10^6}$$
Consider, binary solution of $A$ and $B$ having $n$ and $m$ moles respectively, then
$$\begin{equation} \begin{aligned} \Rightarrow {x_A} = \frac{n}{{m + n}} \\ \Rightarrow {x_B} = \frac{m}{{m + n}} \\ \Rightarrow {x_a} + {x_b} = 1 \\\end{aligned} \end{equation} $$
6. Molarity ($M$)
It is the number of moles of solute dissolved in one litre of solution. $$M = \left( {\frac{{moles\;of\;solute}}{{volume\;of\;solution\;in\;litres(L)}}} \right)$$
$0.35M$ of solution of $NaCl$ means $0.35mol$ of $NaCl$ is present in one litre of solution.
7. Molality ($m$)
It is the number of moles of solute dissolved in $1$ $kg$ of solvent. $$m = \left( {\frac{{mole\;of\;solute}}{{weight\;of\;solvent\;in\;kg}}} \right)$$
Relation between mole fraction and molality when solution is diluted. $${x_{solute}} = \left( {\frac{{m \times {M_{solvent}}}}{{1000}}} \right)$$
Relation between Molality and Molarity when solution is aquous. $$\frac{m}{M} = \left( {\frac{{1000 + m \times {M_{solute}}}}{{1000 \times \rho }}} \right)$$
where $\rho $ is the density of solution.
If solution is dilute aqueous then $\left( {m \times {M_{solute}}} \right) < < 1\quad \& \quad \rho = 1$ then, $$m=M$$
8. Normality ($N$)
It is number of gram equivalent of solute present in one litre of solution. $$N = \left( {\frac{{no.\;of\;gram\;equivalent}}{{volume\;of\;solution\;in\;litre}}} \right)$$
where, $$\begin{equation} \begin{aligned} no.\;of\;gram\;equivalent = \left( {\frac{{given\;mass\;of\;solute}}{{equivalent\;weight\;of\;solute}}} \right) \\ \\ equivalent\;weight = \frac{{molecular\;weight}}{{n\;factor}} \\\end{aligned} \end{equation} $$
So, $$\begin{equation} \begin{aligned} no.\;of\;gram\;equivalent = moles \times n\;factor \\ N = \left( {\frac{{moles\;of\;solute \times n\;factor \times 1000}}{{volume\;of\;solution\;in\;ml}}} \right) \\\end{aligned} \end{equation} $$
Normailty on dilution, $${N_1}{V_1} = {N_2}{V_2}$$
Concentration terms involving volume are dependent on temperature.
So, mass percentage, $ppm$, mole fraction and molality are temperature independent while molarity is temperature dependent, this is because volume depends on temperature.
All these concentration terms are basically used for calculating molar mass of solute.
So, for Molarity $(M)$, $$\begin{equation} \begin{aligned} M = \left( {\frac{n}{V} \times 1000} \right) \\ M = \left( {\frac{{w \times 1000}}{{M \times V}}} \right) \\ M = \left( {\frac{{w \times \rho \times 1000}}{{M \times M'}}} \right)\quad \left( {As,V = \frac{{M'}}{\rho }} \right) \\ M = \left( {\frac{{y \times \rho \times 10}}{M}} \right)\quad \quad \left( {As,y = \frac{w}{{M'}} \times 100} \right) \\\end{aligned} \end{equation} $$
Similarly for Normality $(N)$, $$\begin{equation} \begin{aligned} N = \left( {\frac{{y \times \rho \times 10}}{E}} \right) \\ N = \left( {\frac{{y \times \rho \times 10}}{M}} \right) \times n\;factor\quad \left( {As,E = \frac{M}{{n\;factor}}} \right) \\ N = M \times n\;factor \\\end{aligned} \end{equation} $$
where,
$y$ is the mass $\%$ of solute in solution
$\rho$ is the density of solution
$M'$ is the weight of solvent
$M$ is the molecular mass of solute
$w$ is the weight of solute
