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## Concentration with Examples

**Concentration**

**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.

**Solution:**

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

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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.

Solution:

100 g solution includes 40 g solute

600 g solution includes X g solute

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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.

**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

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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

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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. **n_{NaOH}=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 XCl_{2} salt dissolves in water and form 100 ml 0,4 molar XCl_{2} solution. Find molar mass of X. (Cl=35)

**Solution:**

Molarity=n/V

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 XCl_{2} is 4,4 g

1 mole XCl_{2} is ? g

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?=110 g XCl_{2}

Molar mass of XCl_{2}=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 H_{2}SO_{4} gives 2 H^{+} ion, equivalent of H_{2}SO_{4} is 2. We find equivalent weight;

The Original Author: Mrs. Şerife (Erden) SARICA

Tags: Factors Affecting Chemical Equilibrium | Solutions Exam1 and Problem Solutions | Solutions Exam2 and Problem Solutions | Solutions Exam3 and Problem Solutions |

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