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Course by: Sunil Kumar Singh. E-mail the author

# Molar concentration

Module by: Sunil Kumar Singh. E-mail the author

The idea of mole concept is simple. It neatly relates various reactants and products in the ratio of whole numbers. There is, however, one hitch in applying this concept. The reactants or products may not necessarily be participating in its pure form. Consider the reaction :

M g + 2 H C l M g C l 2 + H 2 M g + 2 H C l M g C l 2 + H 2

Here, magnesium is used in its pure solid form. The hydrogen gas is evolved in pure gaseous form. On the other hand, Hydrochloric acid is used as a solution of a specific concentration in water.

Thus, magnesium and hydrogen exist in pure form. We can find moles of solid magnesium by using its molecular weight. Similarly, we can determine moles of pure hydrogen gas using Avogadro’s hypothesis. In the case of hydrochloric acid, however, the acid in the form of solute participates in reaction – not the solvent i.e. water. Clearly, we need to find a way to connect the concentration of HCl to mass or moles. In this module, we shall exactly do the same by defining different concentration terms of the solution and its relation with mass or moles of the solute as involved in the reaction.

Definition 1: Solution
A solution is a homogeneous mixture of two or more components. The substance present in smaller proportions is called solute and the substance in larger proportion is called the solvent.

Solution comprises of solute and solvent. In expressing concentration of solution, we make use of the fact that mass is conserved,

Mass of solution W S = Mass of solute W B + Mass of solvent W A Mass of solution W S = Mass of solute W B + Mass of solvent W A

The concentration, however, need not be necessarily expressed in terms of mass (w). We may choose volume (v) also depending upon measuring convenience. Sometime, we may express concentration as combination of mass (w) and volume (v).

Concentration of a solution is basically measurement of solute (B) with respect to solution(S) or solvent(A). With the only exception of molality, concentration is expressed as a comparison of the quantity of solute (B) to that of solution (A+B). In the case of molality, it is comparison of solute (B) to that of solvent (A). Further, this comparative ratio is either a number or percentage (%).

There are large numbers of concentration measuring terms. They are basically classified under following three categories :

• Measurement based on mass/volume : mass percentage (w/w), volume percentage (v/v), strength of solution (w/v)
• Measurement based on molecular weight (moles) : molarity (M), molality (m)
• Measurement based on equivalent weight (gram - equivalents) : normality (N)

## Measurements based on mass/volume

### Mass percentage (w/w)

The mass percentage is expressed as :

Mass percentage x = Mass of solute B Mass of solution A + B X 100 Mass percentage x = Mass of solute B Mass of solution A + B X 100

Mass percentage x = W B W S X 100 = W B W A + W B X 100 Mass percentage x = W B W S X 100 = W B W A + W B X 100

If we measure mass in grams, we can rewrite the expression :

Mass percentage x = g B g S X 100 = g B g A + g B X 100 Mass percentage x = g B g S X 100 = g B g A + g B X 100

Sometimes, we may opt to express concentration simply as fraction. In this case, we are not required to multiply ratio by 100,

Mass fraction = W B W A + W B = g B g A + g B Mass fraction = W B W A + W B = g B g A + g B

#### Example 1

Problem : One litre of oxalic acid of density 1.08 gm/cc contains 3.24 gm of oxalic acid. Find its mass percentage.

Solution : The mass of solute (oxalic acid) is given. We need to find the mass of the solution to determine mass percentage.

g S = V S X ρ S = 1000 X 1.08 = 1080 g m g S = V S X ρ S = 1000 X 1.08 = 1080 g m

Mass percentage x = g B g A + g B X 100 = 3.28 1008 X 100 = 0.3 Mass percentage x = g B g A + g B X 100 = 3.28 1008 X 100 = 0.3

#### Example 2

Problem : One kg solution has 0.6 % of urea (w/w) in it. If the molar mass of urea is 60 g m / m o l - 1 g m / m o l - 1 , then determine the moles of urea present in the solution.

Solution : The mass of solute (oxalic acid) is obtained as :

g B = x X g S 100 = 0.6 X 1000 100 = 60 g m g B = x X g S 100 = 0.6 X 1000 100 = 60 g m

n B = g A M O = 60 60 = 1 n B = g A M O = 60 60 = 1

#### Mass percentage of oleum

Sulphuric acid is formed by passing S O 3 S O 3 gas through water in accordance with following chemical equation :

S O 3 + H 2 O H 2 S O 4 S O 3 + H 2 O H 2 S O 4

When all water molecules combine to form sulphuric acid, there remains free S O 3 S O 3 molecules. Oleum is the name given to this mixture of concentrated sulphric acid solution and free S O 3 S O 3 . From the point of view reaction between oleum and other solution, S O 3 S O 3 molecules are as good as sulphuric acid molecule as it reacts with available water molecules to form sulphuric acid molecule.

If an oleum solution has x percent of free S O 3 S O 3 (x gm in 100 gm of oleum), then the equivalent amount of sulphuric acid is calculated using mole concept :

1 mole of S O 3 1 mole of H 2 S O 4 1 mole of S O 3 1 mole of H 2 S O 4

80 gm of S O 3 98 gm of H 2 S O 4 80 gm of S O 3 98 gm of H 2 S O 4

H 2 S O 4 H 2 S O 4 that can be formed from x gm of SO3 is 98 x 80 98 x 80 . Total mass of H 2 S O 4 H 2 S O 4 is sum of H 2 S O 4 H 2 S O 4 in the liquid form and H 2 S O 4 H 2 S O 4 that can be formed when H 2 S O 4 H 2 S O 4 reacts with water :

total mass of H 2 S O 4 = 100 x + 98 x 80 = 100 + 18 x 80 total mass of H 2 S O 4 = 100 x + 98 x 80 = 100 + 18 x 80

Considering equivalent H 2 S O 4 H 2 S O 4 mass in calculation, the mass percentage of H 2 S O 4 H 2 S O 4 in oleum is given as :

y = Total mass of H 2 S O 4 mass of oleum = 100 + 18 x 80 100 X 100 = 100 + 18 x 80 y = Total mass of H 2 S O 4 mass of oleum = 100 + 18 x 80 100 X 100 = 100 + 18 x 80

Clearly, equivalent H 2 S O 4 H 2 S O 4 mass percentage is more than 100 %.

### Volume percentage (v/v)

The volume percentage is expressed as :

Volume percentage x = Volume of solute B Volume of solution A + B X 100 Volume percentage x = Volume of solute B Volume of solution A + B X 100

Volume percentage x = V B V S X 100 = V B V A + V B X 100 Volume percentage x = V B V S X 100 = V B V A + V B X 100

### Strength of solution (w/v)

The strength of solution is expressed as :

Strength of solution S = Mass of solute (B) in grams Volume of solution in litres Strength of solution S = Mass of solute (B) in grams Volume of solution in litres

S = g B V L S = g B V L

The unit of strength is “grams/liters”. Note that strength of solution is not a percentage – rather a number The symbol V L V L denotes volume of solution in litres, whereas V CC V CC denotes volume of solution in cc. If volume is expressed in cc, then the formula of strength is :

S = g B X 1000 V c c S = g B X 1000 V c c

#### Example 3

Problem : Determine the numbers of moles of sulphuric acid present in 500 cc of 392 gm/litre acid solution.

Solution : In order to find the moles of sulphuric acid, we need to find its mass in the given volume.

g B = S X V L = 392 X 0.5 = 196 g m g B = S X V L = 392 X 0.5 = 196 g m

The moles of H 2 S O 4 M 0 = 2 + 32 + 4 X 16 = 98 H 2 S O 4 M 0 = 2 + 32 + 4 X 16 = 98 is :

n B = 196 98 = 2 n B = 196 98 = 2

## Molar concentration

The conversion of concentration of the solution into molar mass of the solute is a two steps process. In the first step, we calculate the mass of the solute and then in second step we divide the mass of the solute by molecular weight to determine the moles of solute present in the solution. We can think of yet a direct measurement of molar concentration. This will enable us to calculate moles of solute in one step.

### Mole fraction

Mole fraction of solute ( χ B χ B ) is defined as :

Molefraction χ B = Mole of solute B Mole of solvent A + Mole of solute B Molefraction χ B = Mole of solute B Mole of solvent A + Mole of solute B

χ B = n B n A + n B χ B = n B n A + n B

Similarly, mole fraction of solvent (B) or other component of solution is :

χ A = n A n A + n B χ A = n A n A + n B

Clearly,

χ A + χ B = 1 χ A + χ B = 1

#### Example 4

Problem : The mass fraction of ethyl alcohol in a sample of 1 kg of aqueous ethyl alcohol solution is 0.23. Determine mole fraction of ethyl alcohol and water in the solution.

Solution : The mass of ethyl alcohol and water are calculated as :

g B = mass fraction X mass of solution = 0.23 X 1000 = 230 g m g B = mass fraction X mass of solution = 0.23 X 1000 = 230 g m

mass of solvent = g A = mass of solution mass of ethyl alcohol = 1000 230 = 770 g m mass of solvent = g A = mass of solution mass of ethyl alcohol = 1000 230 = 770 g m

The moles of of ethyl alcohol and water are :

n B = 230 M C 2 H 5 O H = 230 2 X 12 + 5 X 1 + 16 + 1 = 230 46 = 5 n B = 230 M C 2 H 5 O H = 230 2 X 12 + 5 X 1 + 16 + 1 = 230 46 = 5

n A = 770 M H 2 O = 230 18 = 770 18 = 42. 8 n A = 770 M H 2 O = 230 18 = 770 18 = 42. 8

The mole fraction of ethyl alcohol and water are :

χ B = n B n A + n B = 5 42.8 + 5 = 5 47.8 = 0.104 χ B = n B n A + n B = 5 42.8 + 5 = 5 47.8 = 0.104

χ A = 1 0.104 = 0.896 χ A = 1 0.104 = 0.896

#### Example 5

Problem : In a container, 14 gm nitrogen is mixed with 2 gm of hydrogen gas to form ammonia gas. The amount of ammonia formed is 5.1 gm. Determine mole fractions of ammonia in the container.

Solution : Here, moles of nitrogen and hydrogen present in the container are :

Moles of nitrogen = 14 2 X 14 = 0.5 Moles of nitrogen = 14 2 X 14 = 0.5

Moles of hydrogen = 2 1 X 2 = 1 Moles of hydrogen = 2 1 X 2 = 1

In order to determine mole fraction, we need to know the moles of gases remaining after the reaction. We can know the moles of gas provided we know the extent reaction takes place. The mass of product is known here. Using chemical equation, we can determine the moles of nitrogen and hydrogen used.

Moles of ammonia produced = 5.1 14 + 3 X 1 = 5.1 17 = 0.3 Moles of ammonia produced = 5.1 14 + 3 X 1 = 5.1 17 = 0.3

The chemical equation is :

N 2 + 3 H 2 2 N H 3 N 2 + 3 H 2 2 N H 3

Applying mole concept :

1 mole of N 2 3 moles of H 2 2 moles of N H 3 1 mole of N 2 3 moles of H 2 2 moles of N H 3

Thus, moles of nitrogen and hydrogen gas consumed are :

Moles of N 2 consumed = 1 2 X 0.3 = 0.15 Moles of N 2 consumed = 1 2 X 0.3 = 0.15

Moles of H 2 consumed = 3 2 X 0.3 = 0.45 Moles of H 2 consumed = 3 2 X 0.3 = 0.45

Clearly, reaction does not exhaust either of gases. The moles remaining in the container after reaction are :

Moles of N 2 remaining = 0.5 0.15 = 0.35 Moles of N 2 remaining = 0.5 0.15 = 0.35

Moles of H 2 remaining = 1.0 0.45 = 0.55 Moles of H 2 remaining = 1.0 0.45 = 0.55

Moles of N H 3 produced = 0.3 Moles of N H 3 produced = 0.3

Mole fraction of N H 3 = 0.3 0.3 + 0.35 + 0.55 = 0.3 1.2 = 0.25 Mole fraction of N H 3 = 0.3 0.3 + 0.35 + 0.55 = 0.3 1.2 = 0.25

### Molarity

Molarity of a solution with respect to solute is defined as :

Molarity M = Moles of solute B Volume of solution in litres Molarity M = Moles of solute B Volume of solution in litres

M = n B V L M = n B V L

Its unit is moles/ litres. In case we consider volume in ml i.e. cc, then the expression of molairty is given as :

M = n B V C C X 1000 = Milli-moles of B V C C M = n B V C C X 1000 = Milli-moles of B V C C

If volume (litres) of a solution of known molarity is known, then number of moles of solute is obtained as :

Moles of B n B = Molarity X Volume in litres = M V L Moles of B n B = Molarity X Volume in litres = M V L

Similarly,

Milli-moles of B = Molarity X Volume in litres = M V C C Milli-moles of B = Molarity X Volume in litres = M V C C

#### Example 6

Problem : Determine (i) molarity of 750 cc NaOH solution containing 40 gm NaOH (ii) moles and milli-moles of NaOH in 500 ml of 0.4 M NaOH solution and (iii) mass of NaOH in 500 ml of 0.2 M NaOH solution.

Solution : (i) In the first part, volume is given in cc. Hence, we use the formula :

M = n B V C C X 1000 M = n B V C C X 1000

In order to use this formula, we need to find the moles present. Here, molecular weight of NaOH is 23+16+1 = 40. Hence,

n B = 40 / 40 = 1 n B = 40 / 40 = 1

Putting values in the expression, we have :

M = 1 750 X 1000 = 1.33 M M = 1 750 X 1000 = 1.33 M

(ii) Moles of solute is :

n B = 0.4 X 500 1000 = 0.2 n B = 0.4 X 500 1000 = 0.2

Millimoles of solute = M V C C = 0.4 X 500 = 200 Millimoles of solute = M V C C = 0.4 X 500 = 200

(iii) Moles of solute is :

n B = 0.2 X 500 1000 = 0.1 n B = 0.2 X 500 1000 = 0.1

The mass of NaOH is :

g B = n B X M N a O H = 0.1 X 40 = 4 g m g B = n B X M N a O H = 0.1 X 40 = 4 g m

#### Example 7

Problem : Two litres of a solution is prepared, which contains 44.4 gm of calcium hydro-oxide. Find milli-moles of calcium hydro-oxide in 100 ml sample of the solution.

Solution : Here we first need to determine the molarity of the solution as :

M = n B V L M = n B V L

The moles of calcium hydro-oxide is :

n B = 44.4 M C a O H 2 = 44.4 40 + 2 X 16 + 1 = 44.4 74 = 0.6 n B = 44.4 M C a O H 2 = 44.4 40 + 2 X 16 + 1 = 44.4 74 = 0.6

Putting this in the formula, we have :

M = n B V L = 0.6 2 = 0.3 M M = n B V L = 0.6 2 = 0.3 M

The 100 ml sample has the same molarity as that of the bulk solution. Hence,

Milli-moles = M V C C = 0.3 X 100 = 30 Milli-moles = M V C C = 0.3 X 100 = 30

#### Example 8

Problem : What is molarity of pure water assuming its density 1 gm/cc ?

Solution : Pure water is one component system. According to definition, molarity is given as :

M = n B V L M = n B V L

Let V L V L = 1 litre. The mass of 1 litre water is 1000 gm as density is 1 gm/cc. The number of moles in 1000 gm is :

n B = 1000 18 = 55.56 n B = 1000 18 = 55.56

Putting this value in the expression of molarity, we have :

M = n B V L = 55.56 1 = 55.56 M = n B V L = 55.56 1 = 55.56

#### Molarity and strength of solution

Molarity is defined as :

M = n B V L M = n B V L

Substituting for expression of moles,

M = n B V L = g B M O V L M = n B V L = g B M O V L

But we know that ratio of mass of solute in gram and volume of solution in litres is “strength of solution” :

S = g B V L S = g B V L

Combining two equations, we have :

M = g B M O V L = S M O M = g B M O V L = S M O

S = M M O S = M M O

#### Molarity and mass percentage

Let mass percentage be x. It means that 100 gm of solution contains “x” gm of solute (B). Also let “d” be the density of solution in gm/cc. Then, 1000 cc of solution weighs 1000d gm. This further implies that :

Solute “B” in 1000d gm of solution (i.e. 1 litre solution) = x 100 X 1000 d = 10 x d Solute “B” in 1000d gm of solution (i.e. 1 litre solution) = x 100 X 1000 d = 10 x d

Now molarity is :

M = n B V L = g B M O V L = 10 x d M O X 1 = 10 x d M O M = n B V L = g B M O V L = 10 x d M O X 1 = 10 x d M O

This relation means that we can determine molarity of a solution, if we know (i) molecular weight (ii) mass percentage and (iii) density of the solution.

##### Example 9

Problem : Sulphuric acid solution of density 1.8 gm/cc contains 24.5% acid by weight. What is molarity of the solution?

Solution : Here,

Mass percentage , x = 24.5 Mass percentage , x = 24.5

Density in gm/cc , d = 1.8 Density in gm/cc , d = 1.8

Molecular weight of H 2 S O 4 = 2 X 1 + 32 + 4 X 16 = 98 Molecular weight of H 2 S O 4 = 2 X 1 + 32 + 4 X 16 = 98

Molarity of the solution is :

M = 10 x d M O = 10 X 24.5 X 1.8 98 = 4.5 M M = 10 x d M O = 10 X 24.5 X 1.8 98 = 4.5 M

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