Concentrations of the solution can be expressed as
1. Molarity: No: of moles of solute present in one liter solution. Units: mol L¯¹
\(M=\frac{Moles\text{ }of\text{ }solute}{Volume\text{ }of\text{ }solution\text{ }in\text{ }liter}=\frac{n}{v\left( ml \right)}\times 1000=\frac{wt}{G.M.W}\times \frac{1000}{v\left( ml \right)}\)Ex: 0.5 M ethanol means 0.5 moles of ethanol dissolved in one liter.
2. Mass percentage (w/w) = \(\frac{Mass\text{ }of\text{ }the\text{ }component\text{ }in\text{ }the\text{ }solution}{Total\text{ }mass\text{ }of\text{ }the\text{ }solution}\times 100\)
Ex: a solution of 20% salt in water by mass means: 20g of salt + 80g water = 100g solution.
3. Volume percentage (v/v) = \(\frac{\text{Volume }of\text{ }the\text{ }component\text{ }in\text{ }the\text{ }solution}{Total\text{ }volume\text{ }of\text{ }the\text{ }solution}\times 100\)
Ex: a solution of 20% ethanol in water by volume or 20% (v/v) ethanol in water means: 20ml ethanol + 80ml water = 100ml solution.
4. Mass by volume percentage (w/v) = mass of solute dissolved in 100ml solution. Units: g/ml
5. Parts per million (ppm) = \(\frac{No:\text{ }of\text{ }parts\text{ }of\text{ }the\text{ }components~}{Total\text{ }no:\text{ }of\text{ }parts\text{ }of\text{ }all\text{ }the\text{ }components}\times {{10}^{6}}\)
So, 20ppm of NaOH in sea water means, 20mg in 10⁶ mg of water i.e. 1kg of sea water
6. Normality: No. of gram equivalents present in one Liter of solution. Units: g/l
\(N=\frac{n}{v\left( lit \right)}=\frac{n}{v\left( ml \right)}\times 1000=\frac{wt}{G.E.W}\times \frac{1000}{v\left( ml \right)}\)n = no. of gram equivalents = wt/ GEW
G.E.W = gram equivalent weight
G.E.W = gram molecular weight/ n factor
n factor:
=> For acids, n-factor is defined as the number of H⁺ ions replaced by 1 mole of acid in a reaction.
=> For bases, n-factor is defined as the number of OH¯ ions replaced by 1 mole of base in a reaction.
=> For salts, total number of positive or negative charge.
=> In case of redox reactions, for oxidizing or reducing agent = change in oxidation state per molecule of reactant (or) total number of electrons transferred per mole of reactant.
7. Molality: – No. of moles of solute present in 1 Kg of the solvent. Units: mol/Kg
\(m=\frac{Moles\text{ }of\text{ }solute}{Mass\text{ }of\text{ }solvent\text{ }in\text{ }kg}=\frac{n}{w\left( gm \right)}\times 1000=\frac{wt}{GMW}\times \frac{1000}{w\left( gm \right)}\)8. Mole Fraction: It is the ratio of the number of moles of one compound to the total number of moles of the solution.
\({{x}_{A}}=\frac{{{n}_{A}}}{{{n}_{A}}+{{n}_{B}}};{{x}_{B}}=\frac{{{n}_{B}}}{{{n}_{A}}+{{n}_{B}}}\)xA→ Mole Fraction of solvent
xB → Mole fraction of solute
nA → No: of moles of solvent.
nB → No: of moles of solute.
Mole Fraction of the solution = xA + xB = 1