Some Basic Concepts of Chemistry • Topic 3 of 3

Concentration of Solutions

A solution is a homogeneous mixture of a solute (the dissolved substance) and a solvent (the dissolving medium). Its concentration expresses how much solute is present in a given amount of solution or solvent. Chemists use several measures, each suited to a particular purpose.

  • Mass percent $= \dfrac{\text{mass of solute}}{\text{mass of solution}}\times100$. Simple and temperature-independent.
  • Mole fraction $x_A = \dfrac{n_A}{n_A + n_B}$; the sum of all mole fractions in a solution is $1$. Used in thermodynamics and gas laws.
  • Molarity $M = \dfrac{\text{moles of solute}}{\text{volume of solution in litres}}$, in $mol\,L^{-1}$. The most common lab measure, but it changes slightly with temperature because volume expands on heating.
  • Molality $m = \dfrac{\text{moles of solute}}{\text{mass of solvent in kg}}$, in $mol\,kg^{-1}$. Independent of temperature because it uses mass, so it is preferred for studying colligative properties.
  • Parts per million (ppm) $= \dfrac{\text{mass of solute}}{\text{mass of solution}}\times10^{6}$. Used for very dilute solutions such as pollutants in water.

Dilution adds solvent without changing the moles of solute, so molarity falls while moles stay constant. This gives the dilution equation $M_1V_1 = M_2V_2$, where subscripts $1$ and $2$ denote the concentrated and diluted states. It is the basis of preparing standard solutions from a stock.

The concentration terms are interconvertible if the density of the solution is known. For instance, molality can be obtained from molarity using the solution density and the molar mass of the solute. Note that molarity always uses the volume of the whole solution, whereas molality uses the mass of solvent only — a frequent source of error. Because volume depends on temperature but mass does not, molality is the more reliable measure when temperature varies.

Comparison of common concentration terms
TermSymbol & UnitDefinitionTemperature dependent?
Mass percent% w/w(mass solute / mass solution) x 100No
Mole fractionx (dimensionless)moles of component / total molesNo
MolarityM (mol/L)moles solute / litres of solutionYes
Molalitym (mol/kg)moles solute / kg of solventNo
ppmppm(mass solute / mass solution) x 10^6No
1
Worked Example
Calculate the molarity of a solution containing $4\,g$ of $NaOH$ dissolved in water to make $500\,mL$ of solution.
Solution
  1. Molar mass of $NaOH = 40\,g\,mol^{-1}$; moles $= 4/40 = 0.1\,mol$.
  2. Volume $= 500\,mL = 0.5\,L$.
  3. $M = \dfrac{0.1}{0.5} = 0.2\,mol\,L^{-1}$.

Answer: $0.2\,M$.

2
Worked Example
What is the molality of a solution made by dissolving $9.8\,g$ of $H_2SO_4$ in $250\,g$ of water?
Solution
  1. Molar mass of $H_2SO_4 = 98\,g\,mol^{-1}$; moles $= 9.8/98 = 0.1\,mol$.
  2. Mass of solvent $= 250\,g = 0.25\,kg$.
  3. $m = \dfrac{0.1}{0.25} = 0.4\,mol\,kg^{-1}$.

Answer: $0.4\,m$.

3
Worked Example
Calculate the mole fraction of ethanol ($C_2H_5OH$) in a solution containing $46\,g$ ethanol and $54\,g$ water.
Solution
  1. Moles ethanol $= 46/46 = 1\,mol$; moles water $= 54/18 = 3\,mol$.
  2. Total moles $= 1 + 3 = 4$.
  3. $x_{ethanol} = \dfrac{1}{4} = 0.25$.

Answer: Mole fraction of ethanol $= 0.25$.

4
Worked Example
How much water must be added to $100\,mL$ of $2\,M$ $HCl$ to dilute it to $0.5\,M$?
Solution
  1. Use $M_1V_1 = M_2V_2$: $2\times100 = 0.5\times V_2$.
  2. $V_2 = \dfrac{200}{0.5} = 400\,mL$.
  3. Water to add $= V_2 - V_1 = 400 - 100 = 300\,mL$.

Answer: Add $300\,mL$ of water.

5
Worked Example
A water sample contains $5\,mg$ of dissolved oxygen per litre. Express this in ppm (take solution density $= 1\,g\,mL^{-1}$).
Solution
  1. $1\,L$ of solution $= 1000\,g$ (since density $= 1\,g\,mL^{-1}$).
  2. $5\,mg = 0.005\,g$ of solute.
  3. $ppm = \dfrac{0.005}{1000}\times10^{6} = 5\,ppm$.

Answer: $5\,ppm$.

6
Worked Example
Calculate the mass percent of solute in a solution prepared by dissolving $20\,g$ of glucose in $180\,g$ of water.
Solution
  1. Mass of solution $= 20 + 180 = 200\,g$.
  2. Mass percent $= \dfrac{20}{200}\times100$.
  3. $= 10\%$.

Answer: $10\%$ by mass.

Key Points

  • Molarity M = moles of solute / litres of solution; it varies with temperature.
  • Molality m = moles of solute / kg of solvent; it is temperature-independent, ideal for colligative properties.
  • Mole fraction is the ratio of a component's moles to the total moles; all mole fractions sum to 1.
  • Mass percent and ppm are mass-based ratios; ppm suits very dilute solutions like pollutants.
  • Dilution keeps moles of solute constant, giving M1V1 = M2V2; density links molarity and molality.
Tap an option to check your answer0 / 4
Q1.The molarity of a solution with $0.5\,mol$ solute in $250\,mL$ solution is:
Explanation: $M = 0.5\,mol / 0.25\,L = 2\,mol\,L^{-1}$.
Q2.Which concentration term is independent of temperature?
Explanation: Molality uses mass of solvent (kg), which does not change with temperature, unlike volume-based molarity.
Q3.On diluting $50\,mL$ of $4\,M$ $HCl$ to $200\,mL$, the new molarity is:
Explanation: $M_1V_1 = M_2V_2 \Rightarrow 4\times50 = M_2\times200 \Rightarrow M_2 = 1\,M$.
Q4.The sum of the mole fractions of all components in a solution is always:
Explanation: Mole fractions are normalised ratios; by definition they add up to exactly 1.