Counting Atoms - Molar Solutions (IGCSE CHEMISTRY)

CH. 9 COUNTING ATOM
***MOLAR SOLUTION**

KKCS International Program

Chemistry IGCSE 1

Mr. Markus

Molar solution

• The concentration of a solution indicates the amount of solute present in 1 dm3 (1L) of the solution

• One way of stating the concentration of a solution is to state the mass (in gram) of solute present in 1 dm3 of solution, i.e.

     concentration of solution (g dm-3)

=    mass of solute in gram

volume of solution in dm3

.....solution

• In chemistry, the concentration of a solution is usually expressed as molarity (symbol M)

• The molarity of a solution is defined as the number of moles of solute in 1 dm3 of solution

Molarity of solution (mol dm-3)

= amount of solute in moles

  volume of solution in dm3

• a molar solution is one that contains 1 mole of solute in 1 dm3 of solution

example

• 5.6 0 g of potassium hydroxide was dissolved in sufficient water to form 250 cm3 of solution. Calculate the concentration of the solution in

1. g dm-3

2. mol dm-3

Solution:

1. concentration= mass of solute/vol of solution

= 5.60 g/0.25 dm3  = 22.4 g dm-3

(b) Mr of KOH = 39.0 + 16.0 + 1.0

      = 56.0 g mol-1

      number of moles of KOH = mass/Mr

                    = 5.60 g/56.0 g mol-1

        = 0.10 mol

       molarity = moles of solute/ vol of slolution

          = 0.10 mol/0.25 dm3

          = 0.40 mol dm-3

• 0.105 mol of anhydrous sodium carbonate was dissolved in enough water to form 125 cm3 of solution. Calculate the concentration in mol dm-3 and g dm-3

Solution:

Molarity = moles of solute/vol of solution

      = 0.105 mol/0.125 dm3

      = 0.840 mol dm-3

Mr Na2CO3 = 2(23) + 12 + 3(16)  = 106 g mol-1

ma = 11.2 g

Concentration = mass of solute in g/ vol of sol in dm3

    = 11.2 g/0.125 dm3

    = 89.6 g dm-3

ss of 0.106 mol of Na2CO3

= moles x Mr

= 0.106 mol x 106 g mol-1

= 11.2 g

Concentration = mass of solute in g/ vol of sol in dm3

    = 11.2 g/0.125 dm3

    = 89.6 g dm-3

• The molarity of an aqueous sodium hydroxide solution is 0.50 mol dm-3. waht is the concentration of this solution in g dm-3?

Solution:

Mr of NaOH = 23 + 16 + 1 = 40 g mol-1

0.50 mol dm-3 implies 1 dm3 of solution contains 0.50 moles of NaOH

Mass of  NaOH in 1 dm3 solution

= moles x Mr

= 0.50 mol x 40 g mol-1

= 20 g

Concentration = mass of solute/vol of sol in dm3

      = 20 g/1 dm3

      = 20 g dm-3

• The concentration of sulphuric acid is 2.45 g dm-3. waht is the molarity of the acid?

Solution:

Mr H2SO4 = 2(1) + 32 + 4(16) = 98 g mol-1

2.45 g dm-3 implies 1 dm3 of solution contains 2.45 g of H2SO4

Number of moles of H2SO4 in 1 dm3 solution

= 2.45 g/ 98 g mol-1

= 0.025 mol

The molarity of the acid is 0.025 mol dm-3

• To calculate the number of moles of solute in  a given volume of solution use the following expression:

number of moles of solute

= molarity x volume of  solution in dm3   

Example:

Calculate the number of moles of solute in

1. 350 cm3 of 0.25 mol dm-3 nitric acid

2. 2.5 dm3 of 1.5 mol dm-3 aqueous magnesium chloride

Solution:

1. Volume of solution = 350 cm3 = 0.35 dm3

number of moles of HNO3

= molarity   x   vol in dm3

= o.25 mol dm-3 x 0.35 dm3

= 0.088 mol

b) Number of moles of MgCl2

= 2.5 mol dm-3   x   1.5 dm3

= 3.8 mol

practices

1. Calculate the mass of solute to be used to prepare the following solutions:

a) 250 cm3 of 0.20 mol dm-3 aqueous sodium hydroxide

b) 2.0 dm3 of 0.10 mol dm-3 borax solution from Na2B4O7.10H2O

2.   Calculate the number of moles of hydrogen ions in 125 cm3 of 0.25 mol dm-3 sulphuric acid and hence determine the molarity of the hydrogen ions

-3 x 0.35 dm3

= 0.088 mol

b) Number of moles of MgCl2

= 2.5 mol dm-3   x   1.5 dm3

= 3.8 mol