Concentration Terms - Definition, Examples, practice problems & FAQs

You love to wear different outfits for different occasions like parties, running, swimming, suiting & shirting, etc. We know all clothes are made up of yarn, can you guess which unit is used to express different yarn according to their thickness (in layman's language thinner or courser). 

According to industry convention units of any commercial products are used, for yarn generally uses an indirect system called ‘English count’, and other direct systems like ‘Denier & Tex’ are used. We can simply convert one unit to another by a simple numerical method. In a similar way to express the amount of solute present in a solution is expressed in many ways according to its conventions or commercial need. All concentration methods are interconvertible to each other. (to convert a mass-based concentration term to volume-based concentration terms we need the density of the solution). 

E.g- To measure the hardness of water ppm is generally used, for calculation of the molar mass of unknown sample molarity is used in form of osmotic pressure, impurity present in jewellery expressed in etc.

Table of contents:

• Concentration terms
• Molarity
• Molality
• Normality
• Parts per million (ppm)
• 
• 
• 
• Mole fraction
• Strength of solution
• Formality
• Practice problems
• Frequently asked questions-FAQs

Concentration terms

There are different methods to represent the amount of solute present in a solution. E.g- Molarity, molality, normality, etc.

Temperature dependant concentration terms: Volume-dependent concentration terms are temperature-dependent.

• Molarity (M)
• Normality (N)
• 
• 

Temperature independent concentration terms: Mass-dependent concentration terms are temperature-independent.

• Molality (m)
• Mole fraction ()
• 
• Parts per million (ppm)

Molarity (M)

It is defined as the moles of solute present in 1000 mL or 1 L of the solution.

• It varies inversely with temperature. 
• Mathematically: Molarity ()
• Molarity can be expressed as 

Decimolar =

Semiimolar =

Pentimolar =

Milliimolar =

Centimolar =

Case 1: Calculation of molarity, when solution having molarity (), diluted from to

Initial molarity = , Final molarity =

Initial volume = , diluted to volume =

Case 2: Calculation of molarity when solution having molarity () & volume () mixed with a solution having the same solute with molarity () & volume ()

Molality (m)

Molality is defined as the number of moles of solute present in 1 kg or 1000 g of solvent. 

Relation between molarity & molality

is molality of solution

Weight of solute =

Weight of solvent =

Molar mass of solute =

Molar mass of solvent =

Volume of solution = V mL 

Molality () = —-------- (1)

Molarity () = —---------(2)

= —------- (3)

—------- (4)

Weight of solution = weight of solute + weight of solvent

= +

= +

Density of solution () = = =

Density of solution () = = = 

Normality(N)

It is defined as the number of equivalents of solute present in 1000 mL or 1 L of the solution.

Equivalent weight

The equivalent weight of an element is the number of parts by mass of that element that reacts or displaces from a combination 1.008 parts by mass of hydrogen, 8 parts oxygen by mass, or 35.5 parts chlorine by mass

Example: Find the equivalent weight of for its oxide formation.

Answer:

From the above equation we can say,

16 grams of oxygen combined with 24 g of magnesium

1 gram of oxygen combine with g of magnesium

8 gram of oxygen combine with g of magnesium

Equivalent weight of magnesium = 12

For elements; Equivalent weight =

Similarly

For acid; Equivalent weight =

For base; Equivalent weight =

For reducing agents; Equivalent weight =

For oxidizing agents; Equivalent weight =

In general, Equivalent weight =

Relation between normality of solution and molarity

We know that, 

So,

Parts per million(ppm) 

This concentration expression is used when the solute is present in a very small amount. It's the number of parts of the solute in every million parts of the solution. ppm can both be in terms of mass or in terms of moles. If nothing else is mentioned, we assume ppm refers to mass. As a result, a 10 ppm solution contains 10 g of solute per g of solution.

Water hardness expressed in parts per million. Permissible mineral contents in drinking water is represented in ppm.

Similarly, we can express

Percent weight by volume ()

= amount of solute in gram present in 100 mL of solution.

=

20 % solution means 20 g solute present in 100 mL of solution.

Percent volume by volume ()

= amount of solute in mL present in 100 mL of solution.

=

20 % solution means 20 mL solute present in 100 mL of solution.

Percent weight by weight ()

= amount of solute in gram present in 100 g of solution.

=

20 % solution means 20 g solute present in 100 g of solution.

Mole fraction )

The ratio of the number of moles of the solute or solvent present in the solution and the total number of moles present in the solution is known as the mole fraction of substances concerned.

Let, the number of moles of solute in a solution = & the number of moles of solvent in a solution =

Mole fraction () of solute =

Mole fraction () of solvent =

For Binary solution;

Strength of solution

The strength of the solution is the amount of solute in g present in 1000 mL of solution

Formality

It is the number of grams formula in one litre of solution. It is denoted by F.

Formality (F) =

Relation between strength of solution and Molarity

We know, Strength of solution is amount of solute in g present in 1000 mL of solution

Multiply equation (1) both numerator and denominator by Molar mass of solute

(Mole =

So,

Strength of solution ( = Molarity molar mass 

Relation between strength of solution and Normality

We know, Strength of solution is amount of solute in g present in 1000 mL of solution

Multiply equation (1) both numerator and denominator by equivalent weight of solute

(Number of equivalent = )

So,

Strength of solution ( = normality equivalent weight

Video Link:

Solutions and Concentration Terms | Solutions Class 12 Chemistry (Chapter-2) | Target JEE 2023 Exam

[Time: 19:30 to 58:00]

Practice problems:

Q1. Calculate the molality of the solution if the density of 1.25 M solution is 1.23.

1. 2.96 m
2. 1.08 m
3. 1.34 m
4. 1.73 m

Answer: (B)

Solution: Method 1: 

1.25 M , means moles of

Volume of solution = 1000 mL

Molar mass of

Weight of solute () =

Density of solution = 1.23

Weight of solution = 1.23

Weight of solvent = 1230 g - 75 g = 1155 g

Molality of solution =

Method 2: direct formula use of interconversion of molarity to molality

We know, Density of solution () =

Molar mass of =

=

= 0.075 +

= 1.23 - 0.075

=

Q2. Find molality of an aqueous solution of glucose having mole fraction 0.1 and density of the solution is 1.1

1. 6.17 M
2. 3.85 m
3. 5.34 m
4. 4.73 m

Answer: (A)

Molar mass of solute () = 180

Molar mass of solute () = 18

Solution: We know, =  

=  

Molality

Q3. Calculate the concentration of nitrate ion in the resultaent solution of 75 mL of 0.15 M with 200 mL of 0.1 M and 300 mL of 0.25 M

1. 1.25 M
2. 1.32 M
3. 0.12 M
4. 0.25 M

Answer: (B)

Solution: 

Case 1:

Volume of

Molarity of

We know, Number of moles = Molarity of solution volume of solution (L)

Moles of

According to equation 1 ion produced from 1 mole of

So, Moles of

Case 2:

Volume of

Molarity of

, does not contain nitrate ion.

Case 3:

Volume of

Molarity of

We know, Number of moles = Molarity of solution volume of solution (L)

Moles of

According to equation 3 ion produced from 1 mole of

So, Moles of

Total volume of solution = (0.075 L + 0.2 L + 0.3 L) = 0.575 L

Total number of moles of = 0.75 mol + 0.01125 mol = 0.76125 mol

] = =

Q4. Find mole fraction of solute for 1 NaOH solution.

1. 0.887
2. 0.018
3. 0.010
4. 0.035

Answer: (B)

Solution: molar mass of NaOH =

1 NaOH solution means, 1 mole of NaOH present in 1000 g of water

Moles of NaOH = 1 mol

Moles of water = mol

Mole fraction of solute ( =

Q5. If 30 g of gold powder is added to 600 g of water. Calculate % w/w of the solution

1. 4..8
2. 3.92
3. 2.94
4. Can’t be determined

Answer: (D)

Solution: we can’t simply use the mathematical relation to find %w/w from given information. =

Gold powder is insoluble in water at room temperature so it can’t form solution. So, we can’t determine its concentration.

Frequently Asked Questions-FAQs

Q1. Can the molality of any solution convert into molarity without knowing the density of the solution?
Answer: No, for conversion of molality into molarity or molarity into molality, density must be required. Molarity is volume-dependent and molality is only mass-dependent. 

The density of solution () =

Q2. Can we consider molarity equal to molality for an infinitely dilute solution?
Answer: Only for numerical simplicity for a very dilute aqueous solution, we can consider molarity = molality. (only for aqueous solution)

For numerical simplicity, the density of water at STP condition can be taken as . So, for very dilute solutions weight of the solute becomes very small (negligible). So, the volume of solution = 1L becomes equal to 1 kg. The weight of the solution can be treated as the weight of the solvent.

Q3. What is the difference between density and specific density?
Answer: Specific density of material calculated with respect to the water. So, it is unitless and absolute. Density calculated in or other units.

Q4. Is mole fraction affected by temperature?
Answer: No, mole fraction is temperature independent. By measurement of solute & solvent masses, we can calculate the mole fraction of any components.

Q5. How do you change the concentration of a solution?
Answer: Sometimes, by modifying the quantity of solvent, a worker would need to modify the concentration of a solution. Dilution is the process of lowering the concentration of solutes in a solution by adding a solvent. Concentration is the process of removing a solvent from a solution, resulting in an increase in the concentration of the solute in the solution.

Q6. How do you prepare a solution of known concentration?
Answer: Solutions of known concentration can be prepared either by dissolving the known mass of the solvent solution and diluting it to the desired final volume or by diluting it to the desired final volume by diluting the acceptable volume of the more concentrated solution (the stock solution).

Related Topics:

Normality	Mole
Strength of oleum	Volume strength of
Strength of solution	Density