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Matters and Properties of Matters
Atomic Structure with Examples
Periodic Table
The Mole Concept with Examples
Gases with Examples
Chemical Reactions with Examples
Nuclear Chemistry (Radioactivity)
--Solubility and Factors Affecting Solubility
--Concentration with Examples
--Dilution and Density of Solutions
--Concentration of Ions with Examples
--Properties of Solutions and Solution Calculations
--Solutionns Cheat Sheet
Acids and Bases
Rates of Reactions (Chemical Kinetics)
Chemical Equilibrium
Chemical Bonds
Exams and Problem Solutions
Old Version


Concentration with Examples


Concentration is the amount of solute in given solution. We can express concentration in different ways like concentration by percent or by moles.

1) Concentration by Percent:

It is the amount of solute dissolves in 100 g solvent. If concentration of solution is 20 %, we understand that there are 20 g solute in 100 g solution.

Example: 10 g salt and 70 g water are mixed and solution is prepared. Find concentration of solution by percent mass.


Mass of Solute: 10 g

Mass of Solution: 10 + 70 = 80 g

80 g solution includes 10 g solute

100 g solution includes X g solute


X=12,5 g %

Or using formula;

Percent by mass=10.100/80=12,5 %

Example: If concentration by mass of 600 g NaCl solution is 40 %, find amount of solute by mass in this solution.


100 g solution includes 40 g solute

600 g solution includes X g solute


X=240 g NaCl salt dissolves in solution.

Example: If we add 68 g sugar and 272 g water to 160 g solution having concentration 20 %, find final concentration of this solution.


Mass of solution is 160 g before addition sugar and water.

100 g solution includes 20 g sugar

160 g solution includes X g sugar


X=32 g sugar

Mass of solute after addition=32 + 68=100 g sugar

Mass of solution after addition=272 +68 + 160=500 g

500 g solution includes 100 g sugar

100 g solution includes X g sugar


X= 20 % is concentration of final solution.

2) Concentration by Mole:

We can express concentration of solutions by moles. Number of moles per liter is called molarity shown with M.

Example: Using 16 g NaOH, 200 ml solution is prepared. Which ones of the following statements are true for this solution?(Molar mass of NaOH is 40 g)

I. Concentration of solution is 2 molar

II. Volume of the water in solution is 200 ml

III. If we add water to solution, moles of solute decreases.

Solution: Moles of NaOH

I. nNaOH=16/40=0,4 mole

V=200 mL= 0,2 Liters

Molarity=0,4/0,2=2 molar

I is true

II. Since volume of solution is 200 mL, volume of water is smaller than 200 mL. II is false.

III. If we add water to solution, volume of solution increases but moles of solute does not change.

Example: 4,4 g XCl2 salt dissolves in water and form 100 ml 0,4 molar XCl2 solution. Find molar mass of X. (Cl=35)



n=M.V where V=100mL=0,1 L and M=0,4 molar

n=0,1.0,4=0,04 mole

If 0,04 mole XCl2 is 4,4 g

1 mole XCl2 is ? g


?=110 g XCl2

Molar mass of XCl2=X+2.(35)=110

X=40 g/mole

3) Molality:

Molality is the another expression of concentration of solutions. It is denoted with "m" and formula of molality is;

4) Normality:

We can express concentration in another way with normality using equivalents of solutes.

Equivalents can be defined as; number of moles of H+ ion in acids and  OH- ion in base reactions. For example; 1 mole H2SO4 gives 2 H+ ion, equivalent of H2SO4 is 2. We find equivalent weight;

Solutions Exams and  Problem Solutions

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