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.

**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 cm^{3} 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, C_{12}H_{22}O_{11} (mol.mass = 342), means that g of sugar is
present in one litre (1000 cm^{3}) 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 H_{2}SO_{4} is present in
one litre (1000 cm^{3}) 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 C_{6}H_{12}O_{6}
(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 NH_{3},
solution needed to prepare 1L of 0.1M solution.

Ans:-6.77 mL

2.
How would you make up 425 mL of 0.150M HNO_{3}
from 68.0% HNO_{3}? The density of 68.0% HNO_{3} is1.41g/mL.

Ans: 4.25 mL

3.
Calculate the molarity of a solution obtained by
mixing 100 mL of M H_{2}SO_{4} and 200 mL of 1.5M H_{2}SO_{4}

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

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