Methods of Expressing the Concentration of a Solution

The concentration of a solution can be expresses in a number of ways. The important methods are:

Mass/Weight Percentage or Per cent by Mass/Weight :

It is defined as the amount of solute in grams present in 100 grams of the solution.

$\small Mass\ Percentage=\frac{Mass\ of\ Solute}{Mass\ of \ Solution}\times 100$

$\small =\frac{Mass\ of\ Solute}{Mass\ of \ Solute+Mass\ of\ Solvent}\times 100$

$\small =\frac{Mass\ of\ Solute}{Volume\ of \ Solution+Density\ of\ Solution}\times 100$

The ratio mass of solute to the mass of solvent is termed as mass fraction.

Thus, Mass percentage of solute = Mass fraction × 100
10% solution of sugar by mass means that 10 grams of sugar is present in 100 grams of the solution, i.e., 10 grams of sugar has been dissolved in 90 grams of water. e

Example 1:

Question:

What is the weight percentage of urea solution in which 10 gm of urea is dissolved in 90 gm water.

Solution

Weight percentage of urea = (weight of urea/ weight of solution) $\small \times$ 100

= 10/(90+10) $\small \times$ 100 = 10% urea solution (w/W)

Volume Percentage

It is defined as the volume of solute in mL present in 100 mL solution.
$\small Volume\ Percentage=\frac{Volume\ of\ Solute}{Volume \ of \ Solution}\times 100$
10% solution of HCl by volume means that 10 mL of liquid HCl is present in 100 mLof the solution.

Mass by Volume Percentage

It is defined as the mass of solute present in 100 mL of solution.

$\small \small Mass\ by\ Volume\ Percentage=\frac{Mass\ of\ Solute}{Volume \ of \ Solution}\times 100$

A 10% mass by volume solution means that 10 gm solute is present in 100 mL of solution.

Molarity

The molarity of a solution gives the number of gram molecules of the solute present in one litre of the solution.
$\small Molarity(M)=\frac{Number\ of\ moles\ of\ solute}{Volume\ of\ Solution\ in\ L}$
Thus, if one gram molecule of a solute is present in 1 litre of the solution, the concentration of the solution is said to be one molar.
Units of molarity: mol L^-1

Molarity of dilution:

Before dilution After dilution

M₁V₁ = M₂V₂

Molarity of mixing:

Let there be three samples of solution (containing same solvent and solute) with their molarity M₁, M₂, M₃ and volumes V₁, V₂, V₃ respectively. These solutions are mixed; molarity of mixed solution may be given as:

M₁V₁ + M₂V₂ + M₃V₃ = M_R(V₁ + V₂ + V₃)

Where M_R = resultant molarity

V₁ + V₂ + V₃ = resultant molarity

Molarity is dependent on volume; therefore, it depends on temperature.

1 M → Molar solution, i.e., molarity is 1
0.5 M or M/2 → Semimolar
0.1 M or M/10 → Decimolar
0.01 M or M/100 → Centimolar
0.001 M or M/1000 → Millimolar

Example 2:

Question:

3.65 gm of HCL gas is present in 100 mL of its aqueous solution.What is the molarity?

Solution

Molarity = $\small \frac{w}{M}\times \frac{1000}{V\ (mL)}$ = (3.65/36.5) $\small \times$ 1000/100 = 1M

Molality

Molality of a solution is defined as the number of moles of solute dissolved in 1 Kg of the solvent.
$\small Molality\ (m)=\frac{Number\ of\ moles\ of\ solute}{Mass\ of\ Solvent\ in\ kg}$
Thus, if one gram molecule of a solute is present in 1 kg of the solvent, the concentration of the solution is said to be one molal.
Units of molarity: mol kg^-1???
Molality is the most convenient method to express the concentration because it involves the mass of liquids rather than their volumes. It is also independent of the variation in temperature.
Molality and solubity are related by the following relation.

Molality = Solubility×10/Molecular mass of the solute

[Solubility = Mass of solute in grams/Mass of solvent in grams × 100]

Relationship Between Molality and Molarity:

$\small \frac{Molarity}{Molality}= \frac{\frac{No.\ of\ moles\ of\solute}{Volume\ of\ solution}{\frac{No.\ of\ moles\ of\solute}}{Mass\ of\ solvent}}}$ $\small \frac{No.\ of\ moles\ of\ solute}{Volume\ of\ solution\ in\ L }\times \frac{Mass\ of\ solvent\ in\ Kg}{Number\ of\ moles\ of\ solute}$

$\small =\frac{Mass\ of\ solvent\ in\ Kg}{Volume\ of \ solution\ in\ L}$ $\small =\frac{W}{V}$

Let the density of the solution be d. Unit= g mL⁻¹

Mass of solution = V × d

Mass of solute = number of moles × molecular mass of solute = n m_A

Mass of solvent, W = mass of solution – mass of solute = V × d – n × m_A

Thus,
$\small \frac{Molality}{Molarity}$

$\small \frac{M}{m}=V\times d-n\times\frac{m_A}{V}$

Where m_A is molecular mass of solvent.

Example 3:

Question:

The density of a 3M sodium thiosulphate solution is 1.25 gm cm^–3. Calculate

i) the molalities of Na⁺ and S₂O₃^2– ions

ii) percentage of weight of solution

Solution

3 M Na₂S₂O₃ (Sodium thiosulphate) solution means

3 moles Na₂S₂O₃ is present in 1 L or, 1000 ml solution

Wt. of solute Na₂S₂O₃ = 3×158

wt. of solution = v × d

= 1000 ml × 1.25 gm/ml

= 1000 × 1.25 gm

Wt. of solvent = (1000 × 1.25 – 3 ×158) gm H₂O

Molality = no. of moles of solute per 1000 gm solvent

$\small =3\times \frac{1000}{1000\times 1.25-3\times 158}$

= 3.865 mol kg^–1 solvent

Now, Na₂S₂O₃ 2Na⁺ +S₂O₃^-

a) Hence molality of Na⁺ = 2 ×3.865 mol kg^-1

= 7.73 mol kg^–1

Hence molality = 1× 3.865 mol kg^–1

= 3.865 mole/kg

b) % of wt. of solution

1 L i.e. 1000 ml solution containing 3 moles Na₂S₂O₃

1000 × 1.25 gm solution containing 3 ×158 gm Na₂S₂O₃

= 37.92%

Normality:

The normality of a solution gives the number of gram equivalents of the solute present in one litre of the solution.
$\small Normality\ (N)=\frac{Number\ of\ gram\ equivalents\ of\ solute}{Volume\ of\ Solution\ in\ L}$ ?

Number of gram equivalent of solute = Mass of solute in gram/ equivalent weight of solute

Equivalent weight of solute (E) = Molar mass of solute/ Valence factor

Valence factor for base = acidity of base
Valence factor for acid = basicity of acid
Valence factor for element = valency

Thus, if one gram equivalent of a solute is present in one litre of the solution, the concentration of the solution is said to be one normal.
1N = Normal = One gram equivalent of the solute per litre of solution = Normality is 1
N/2 = Seminormal = 0.5 g equivalent of the solute per litre of solution = Normality is 0.5
N/10 = Decinormal = 0.1 g equivalent of the solute per litre of solution = Normality is 0.1
N/100 = Centinormal = 0.01 g equivalent of the solute per litre of solution = Normality is 0.01
N/1000 = Millinormal = 0.001 g equivalent of the solute per litre of solution = Normality is 0.001

Relationship between normality and molarity:

We know that

Molarity × Molecular mass = Strength of solution (g/L)

Similarly,

Normality × Equivalent mass = Normality of the solution (g/L)

Hence,

Molarity × Molecular mass = Normality × Equivalent mass

So, Normality = n × Molarity

__________________________________________

Let ‘d’ is the density of solution in g/mL

and

x is the percentage of the solute by mass.

Then,

$\small N=\frac{x\times d\times 10}{Equivalent\ Mass\ of\ Solute}$

Mole Fraction

The mole fraction of any component in a solution is the ratio of the number of moles of that component to the total number of moles of all components .
Total mole fraction of all the components of any solution is 1.
For a binary solution of A and B

Mole Fraction of A (X_A) =

$\small \frac{n_A}{n_A+n_B}$

Mole Fraction of B (X_B) =

$\small \frac{n_B}{n_A+n_B}$

And, X_A+X_B = 1

Relation between mole fraction and Molality:

X_A = n/N+n and X_B = N/N+n

X_A/X_B = n/N = Moles of solute/Moles of solvent = w_A/m_B/w_B×m_A

X_A×1000/X_B×m_B = w_A×1000/w_B×m_A = m

$\small \frac{X_A\times 1000}{1-X_A} = m$

Refer to the Following Video for Molarity, Molality and Mole Fraction

Parts per million (ppm):

When a solute is present in trace quantities, it is convenient to express concentration in parts per million (ppm)
$\small ppm =\frac{Number\ of\ parts\ of\ the\ component}{Total\ number\ of\ parts\ of\all\ the\ components\ in\ the\ solution}\times 10^6$
In case of mass it may be expressed as : (Mass of solute/Mass of solution )× 106
In case of volume it may be expressed as: (Volume of solute/Volume of solution) × 106
So, concentration in parts per million can be expressed as mass to mass, volume to volume and mass to volume form.
Atmospheric pollution in cities is also expressed in ppm by volume. It refers to the volume of the pollutant in 10⁶ units of volume. 10 ppm of SO₂ in air means 10 mL of SO₂ is present in 10⁶ mL of air.

Formality

It is the number of formula mass in grams present per litre of solution. In case formula mass is equal to molecular mass, formality is equal to molarity. Like molarity and normality, the formality is also dependent on temperature. It is used for ionic compounds in which there is no existence of molecule. Mole of ionic compounds is called formole and molarity as formality.

$\small Formality=\frac{Weight\ of\ solute(gm)}{Formula\ weight\ of\ solute}\times \frac{1}{Volume\ of\ solutions (L)}$

$\small F=\frac{w}{f}\times \frac{1}{V\ (L)} ...................(i)$

$\small F=\frac{w}{f}\times \frac{100}{V\ (mL)} ...................(ii)$

Where,

w = weight of solute,

f = formula weight of solute

V= volume of solution

n_f= no. of gram formula weight

$\small F=n_f\times \frac{1}{V\ (L)} ...................(iii)$

Example 3:

Question:

CH₃COOH exists as dimer in benzene. 1.2 gm of the acid was dissolved and the volume was made up to one litre using benzene. What is the formality?

Solution

Molar mass of CH₃COOH = 60

Formula weight of the associated molecule of the acid = 2 $\small \times$ 60 = 120

Weight of CH₃COOH = 1.2 gm

Volume of solution = 1 L

Formality = 1.2/120 $\small \times$ 1/1 = 0.01 F

Question 1: How much water would be present in 100 mL of 20% aqueous solution of sugar by volume?

a. 20 mL

b. 80 mL

c. 90 mL

d. 100 mL

Question 2: What would be concentration of the solution formed by adding two moles of solute in 1 kg water?

a. 2 M

b. 2 m

c. 20 M

d. 20 m

Question 3: What will be mole fraction of solute of a binary solution if that of its solvent is 0.2?

a. 0.8

b. 0.1

c. 0.7

d. 1.0

Question 4: Unit of molarity is

a. mol L^-1

b.mol kg^-1

c. mol^-1 L

d. mol^-1kg

Q.1	Q.2	Q.3	Q.4
b	b	a	a

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