Class 12 Chemistry

Solutions

Solution: A homogenous mixture of two or more than two substances is called solution. The component that is present in the largest quantity is called the solvent. The component other than the solvent is called solute. A solution which is composed of only two components is called binary solution.

Expressing Concentration of Solution

Mass Percentage (w/w): The mass percentage of a component of a solution is defined as the percentage of mass of the component in total mass of the solution.

Mass % of a component `=text(Mass of component)/text(Total mass of solution)xx100`


Volume Percentage (V/V): The volume percentage of a component is defined as the percentage of volume of the component in the total volume of the solution.

Volume % of a component `=text(Volume of component)/text(Total volume of solution)xx100`

Mass by Volume Percentage (w/V): The mass of the solute dissolved in 100 mL of solution is called mass by volume percentage.

Parts Per Million (ppm): When a solute is present in trace quantities, it is convenient to express the concentration in parts per million. Concentration in parts per million can also be expressed as mass to mass, volume to volume and mass to volume.


Mole Fraction (X): Mole fraction of a component is defined as the ratio of number of moles of the component to total number of moles of all the components.

Mole fraction of a component `=text(Number of moles of a component)/text(Total number of moles of all the components)`

For example, in a binary mixture if the number of moles of A and B are nA and nB respectively then mole fraction of A will be given as follows:

`X_A=(n_A)/(n_A+n_B)`

Example: Calculate the mole fraction of ethylene glycol (C2H6O2) in a solution containing 20% of C2H6O2 by mass.

Answer: Let us assume that we have 100 g of solution. Then mass of C2H6O2 in this solution will be 20 g (20% of 100 g). Similarly, mass of water in this solution will be 80 g.

Molar mass of C2H6O2

`=2xx12+6xx1+2xx16=24+6+32=62` g mol-1

Molar mass of H2O `=2xx1+16=18` g mol-1

Moles of ethylene glycol `=(20g)/(62g\mol^(-1))=0.322` mol

Moles of water `=(80g)/(18g\mol^(-1))=4.44` mol

Now, mole fraction can be calculated as follows:

`X_(glycol)=(n_g)/(n_g+n_w)`

`=(0.322)/(0.322+4.444)=0.068`

Molarity (M): Number of moles of solute dissolved in one litre of solution is called molarity of the solute.

Molarity `=text(Moles of solute)/text(Volume of solution in litre)`

Example: Calculate the molarity of a solution containing 5 g of NaOH in 450 mL solution.

Answer: Molar mass of NaOH `=23+16+1=40` g mol-1

Moles of NaOH in given solution `=(5g)/(40g\mol^(-1))=0.125` mol

Molarity `=(0.125mol)/(450mL)xx1000L=0.278` M = 0.278 mole L-1


Molality (m): The number of moles of the solute per kilogram of the solvent is called molality. It can be expressed as follows:

Molality (m) `=text(Moles of solute)/text(Mass of solvent in kg)`

Example: Calculate molality of 2.5 g of ethanoic acid (CH3COOH) in 75 g of benzene.

Answer: Molar mass of C2H4O2 `=12 xx 2 + 1 xx 4 + 16 xx 2 = 60` g mol–1

Moles of C2H4O2 `= (2.5g)/(60g\mol^(-1))=0.0417` mol

Mass of benzene in kg `= (75 g)/(1000 g\ kg^(–1)) = 75 xx 10^(–3)` kg

Molality of C2H4O2 `=(0.0417 mol)/(75xx10^(-3)kg)=0.556` mol kg-1



Molality

Solubility

Liquid in Liquid

Solid in Liquid

Boiling Point

Osmosis

InText Questions

NCERT Solution