Molar solutions

A molar solution is one whose concentration is known. The SI unit of concentration is Molarity denoted M. Molarity may be defined as the number of moles of solute present in one cubic decimeter of solution. One cubic decimeter is equal to one litre and also equal to 1000cm3. The higher the molarity the higher the concentration and the higher/more solute has been dissolved in the solvent to make one cubic decimeter/ litre/1000cm3 solution.

Examples 2M sodium hydroxide means 2 moles of sodium hydroxide solute is dissolved in enough water to make one cubic decimeter/ litre/1000cm3 uniform solution mixture of sodium hydroxide and water.0.02M sodium hydroxide means 0.02 moles of sodium hydroxide solute is dissolved in enough water to make one cubic decimeter/ litre/1000cm3 uniform solution mixture of sodium hydroxide and water. “2M” is more concentrated than“0.02M”.

Preparation of molar solution Procedure Weigh accurately 4.0 g of sodium hydroxide pellets into a 250cm3 volumetric flask. Using a wash bottle add about 200cm3 of distilled water. Stopper the flask. Shake vigorously for three minutes. Remove the stopper for a second then continue to shake for about another two minutes until all the solid has dissolved. Add more water slowly upto exactly the 250 cm3 mark.

Sample questions

1.Calculate the number of moles of sodium hydroxide pellets present in: (i) 4.0 g.

Molar mass of NaOH = (23 + 16 + 1) = 40g Moles = Mass => 4.0 = 0.1 / 1.0 x 10 -1 moles Molar mass 40

(ii) 250 cm3 solution in the volumetric flask.

    Moles in 250 cm3 =  0.1   /  1.0 x 10 -1 moles

(iii) one decimeter of solution

    Method 1
    Moles in decimeters = Molarity = Moles  x  1000cm3/1dm3
                             Volume of solution

=> 1.0 x 10 -1 moles x 1000cm3 =250cm3 = 0.4 M / 0.4 molesdm-3

Method 2
      250cm3 solution contain 1.0 x 10 -1 moles

1000cm3 solution = Molarity contain 1000 x 1.0 x 10 -1 moles250 cm3 = 0.4 M / 0.4 molesdm-3

Theoretical sample practice

  1. Calculate the molarity of a solution containing:    (i) 4.0 g sodium hydroxide dissolved in 500cm3 solution

Molar mass of NaOH = (23 + 16 + 1) = 40g

Moles = Mass => 4.0 = 0.1 / 1.0 x 10 -1 moles Molar mass 40

Method 1 Moles in decimeters = Molarity = Moles x 1000cm3/1dm3 Volume of solution => 1.0 x 10 -1 moles x 1000cm3 500cm3 = 0.2 M / 0.2 molesdm-3

Method 2
      500 cm3 solution contain 1.0 x 10 -1 moles

1000cm3 solution = Molarity contain 1000 x 1.0 x 10 -1 moles500 cm3 = 0.2 M / 0.2 molesdm-3

(ii) 5.3 g anhydrous sodium carbonate dissolved in 50cm3 solution Molar mass of Na2CO3 = (23 x 2 + 12 + 16 x 3) = 106 g

Moles = Mass => 5.3 = 0.05 / 5. 0 x 10-2 moles Molar mass 106

Method 1 Moles in decimeters = Molarity = Moles x 1000cm3/1dm3 Volume of solution => 1.0 moles x 1000cm3 =50cm3 =1.0 M

Method 2
      50 cm3 solution contain 5.0 x 10 -2 moles

1000cm3 solution = Molarity contain 1000 x 5.0 x 10 -2 moles50 cm3 = 1.0M / 1.0 molesdm-3

(iii) 5.3 g hydrated sodium carbonate decahydrate dissolved in 50cm3 solution

Molar mass of Na2CO3.10H2O = (23 x 2 + 12 + 16 x 3 + 20 x 1 + 10 x 16) =286g

Moles = Mass => 5.3 = 0.0185 / 1.85 x 10 -2 moles Molar mass 286 Method 1 Moles in decimeters = Molarity = Moles x 1000cm3/1dm3 Volume of solution => 1.85 x 10 -2 moles x 1000cm3 =50cm3 = 0.37 M/0.37 molesdm-3 Method 2 50 cm3 solution contain 1.85 x 10 -2 moles 1000cm3 solution = Molarity contain 1000 x 1.85 x 10 -2 moles50 cm3 = 3.7 x 10-1 M / 3.7 x 10-1 molesdm-3

(iv) 7.1 g of anhydrous sodium sulphate(VI)was dissolved in 20.0 cm3 solution. Calculate the molarity of the solution. Method 1 20.0cm3 solution ->7.1 g 1000cm3 solution -> 1000 x 71 = 3550 g dm-3 20

Molar mass Na2SO4 = 142 g

Moles dm-3 = Molarity = Mass 3550 = 2.5 M/ molesdm-3 Molar mass 142

Method 2 Molar mass Na2SO4 = 142 g

Moles = Mass => 7.1 = 0.05 / 5.0 x 10 -2 moles Molar mass 142 Method 2(a) Moles in decimeters = Molarity = Moles x 1000cm3/1dm3 Volume of solution => 5.0 x 10 -2 moles x 1000cm320cm3 = 2.5 M/2.5 molesdm-3 Method 2(b) 20 cm3 solution contain 5.0 x 10 -2 moles 1000cm3 solution = Molarity contain 1000 x 5.0 x 10 -2 moles20 cm3 = 2.5 M/2.5 molesdm-3

(iv) The density of sulphuric(VI) is 1.84gcm-3 Calculate the molarity of the acid.

Method 1 1.0cm3 solution ->1.84 g 1000cm3 solution -> 1000 x 1.84 = 1840 g dm-3 1

Molar mass H2SO4 = 98 g

Moles dm-3 = Molarity = Mass = 1840Molar mass 98

            = 18.7755 M/ molesdm-3

Method 2 Molar mass H2SO4 = 98 g

Moles = Mass => 1.84 = 0.0188 / 1.88 x 10 -2 moles Molar mass 98

Method 2(a) Moles in decimeters = Molarity = Moles x 1000cm3/1dm3 Volume of solution => 1.88 x 10 -2 moles x 1000cm31.0cm3 = 18.8M/18.8 molesdm-3 Method 2(b) 20 cm3 solution contain 1.88 x 10 -2 moles 1000cm3 solution = Molarity contain 1000 x 1.88 x 10 -2 moles1.0 cm3 = 18.8M/18.8 molesdm-3

  1. Calculate the mass of :    (i) 25 cm3 of 0.2M sodium hydroxide solution(Na =23.0.O =16.0, H=1.0)

   Molar mass NaOH = 40g    Moles in 25 cm3 = Molarity x volume => 0.2 x 25 = 0.005/5.0 x 10-3moles    1000 1000

   Mass of NaOH =Moles x molar mass = 5.0 x 10-3 x 40 = 0.2 g

(ii) 20 cm3 of 0.625 M sulphuric(VI)acid (S =32.0.O =16.0, H=1.0)

Molar mass H2SO4 = 98g

Moles in 20 cm3 = Molarity x volume=> 0.625 x 20 = 0.0125/1.25.0 x 10-3moles 1000 1000

Mass of H2SO4 =Moles x molar mass => 5.0 x 10-3 x 40 = 0.2 g

(iii) 1.0 cm3 of 2.5 M Nitric(V)acid (N =14.0.O =16.0, H=1.0)

Molar mass HNO3 = 63 g

Moles in 1 cm3 = Molarity x volume => 2.5 x 1 = 0.0025 / 2.5. x 10-3moles 1000 1000

Mass of HNO3 =Moles x molar mass => 2.5 x 10-3 x 40 = 0.1 g

  1. Calculate the volume required to dissolve :    (a)(i) 0.25moles of sodium hydroxide solution to form a 0.8M solution    Volume (in cm3) = moles x 1000 => 0.25 x 1000 = 312.5cm3    Molarity 0.8

   (ii) 100cm3 was added to the sodium hydroxide solution above. Calculate the concentration of the solution.

   C1 x V1 = C2 x V2 where:    C1 = molarity/concentration before diluting/adding water    C2 = molarity/concentration after diluting/adding water    V1 = volume before diluting/adding water    V2 = volume after diluting/adding water

   => 0.8M x 312.5cm3 = C2 x (312.5 + 100)

     C2  = 0.8M  x 312.5cm3 = 0.6061M
            412.5
   ```

(b)(ii) 0.01M solution  containing 0.01moles of  sodium hydroxide solution .
Volume (in cm3)   =   moles x 1000 =>  0.01  x 1000      = 1000 cm3
Molarity                0.01

(ii) Determine the quantity of water which must be added to the sodium hydroxide solution above to form a 0.008M solution. C1 x V1 = C2 x V2 where: C1 = molarity/concentration before diluting/adding water C2 = molarity/concentration after diluting/adding water V1 = volume before diluting/adding water V2 = volume after diluting/adding water

=> 0.01M x 1000 cm3 = 0.008 x V2 

 V2  = 0.01M  x 1000cm3 = 1250cm3
        0.008

```

Volume added = 1250 - 1000 = 250cm3