Methods of Expressing the concentration of solution

Methods of Expressing the concentration of solution

The concentration of a solution refers to the amount of solute present in the given quantity of solution or solvent. The concentration of a solution may be expressed quantitatively in any of the following ways.

1.  Strength

The Strength of a solution is defined as the amount of the solute in grams, present in one litre of the solution. It is expressed in g L^-1

Strength  = Mass of solute in grams / Volume of solution in litres

If X gram of solute is present in V cm3 of a given solution then

Strength  = ( X x 1000) / V

2.Molarity (M)

Molarity of a solution is defined as the number of gram-moles of solute dissolved in 1 litre of a solution

Molarity  = No. of moles of solute / Volume of Solution in litres

If `X' grams of the solute is present in V cm³ of a given solution, then,

Molarity   = (X    / Mol. mass) / x( 1000 / V )

Molarity is represented by the symbol M. Molarity can also be calculated from the strength as follows

Molarity  = Strength in grams per litre / Molecular mass of the solute

Example

A 0.1M solution of Sugar, C₁₂H₂₂O₁₁  (mol.mass = 342), means that g of sugar is present in one litre (1000 cm³) of the solution.

3.Normality

Normality of a solution is defined as the number of gram equivalents of the solute dissolved per litre of the given solution.

Normality = Number of gram-equivalents of solute / Volume of Solution in litre

If X grams of the solute is present in V cm3 of a given solution, then,

Normality =  ( X / Eq.mass ) x ( 1000 mL / V )

Normality is represented by the symbol N. Normality can also be calculated from strength as follows

Normality  = Strength in grams per litre / Eq.mass of the solute

Example

A 0.1N (or decinormal) solution of H2SO4 (Eq.mass = 49), means that 4.9 g of H₂SO₄ is present in one litre (1000 cm³) of the solution.

4.  Molality (m)

Molality of a solution is defined as the number of gram-moles of solute dissolved in 1000 grams (or 1 kg) of a Solvent. Mathematically,

Molality  = Number of moles of solute / Mass of solvent in kilograms

"If X grams of the solute is dissolved in b grams of the solvent", then

Molality =  ( X / Mol. mass ) x  ( 1000g/bg )

Molality is represented by the symbol 'm'.

Example

A 0.1m Solution of glucose C₆H₁₂O₆ (Mol.mass = 180), means that 18g of glucose is present in 1000g (or one kilogram) of water.

5. Mole Fraction

Mole fraction is the ratio of number of moles of one component (Solute or Solvent) to the total number of moles of all the components (Solute and Solvent) present in the Solution. It is denoted by X. Let us suppose that a solution contains 2 components A&B and suppose that nA moles of A and nB moles of B are present in the solution. Then,

Mole fraction of A, XA= nA / (nA + nB)   .....(1)

Mole fraction of B, XB = nB   /  (nA + nB) ......(2)

Adding 1 & 2 we get,    

XA + XB = ( nA / nA+nB  ) + ( nB / nA+nB )   =  nA+nB / nA+Nb

Thus, sum of the two mole fractions is one. Therefore, if mole fraction of one component in a binary solution is known, that of the other can be calculated.

Problems for practice

1.     Calculate the volume of 14.3m NH₃, solution needed to prepare 1L of 0.1M solution.

Ans:-6.77 mL

2.     How would you make up 425 mL of 0.150M HNO₃ from 68.0% HNO₃? The density of 68.0% HNO₃ is1.41g/mL.

Ans: 4.25 mL

3.     Calculate the molarity of a solution obtained by mixing 100 mL of M H₂SO₄ and 200 mL of 1.5M H₂SO₄

Ans:1.1M

4.     4. Calculate the molality of a solution by dissolving 0.850g of ammonia (NH3) in 100g of water.

Ans:0.5m