SOLUTION, COLLOIDS AND NANO-PARTICLES
2.1. SOLUTIONS INTRODUCTION
In this chapter we are going to study about true solutions and the different modes of expressing the concentrations of chemical solutions. Study of different units used to measure the concentration of solutions is essential, since in various estimations of chemicals in different fields these concentration units are employed.
Definition:
Solution is a homogeneous mixture of solute and solvent whose proportion varies within certain limits.
Solute:
Solute is the substance present in smaller quantities in a solution.
Solvent:
Solvent is the substance present in larger quantities in a solution.
Methods of expressing concentration of solutions:
1. Molarity 2. Molality 3. Normality 4. Mole-fraction
5. Percentage by mass Mole: n (See First lesson)
When the molecular mass of the substance expressed in grams it is known as gram molecular mass or one Mole.
Mass in grams Number of Moles (n) =
Molecular Mass Mole of the solvent is represented as n1 The mole of the solute is represented as n2
Molecular mass of water is 18
Gram molecular mass of water is =18 g or one mole.
One mole of water =18g Two mole of water is =36 g.
Molarity (M):
Molarity is the number of moles of solute present in 1000 ml or one litre of solution. The symbol used to represent molarity is ‘M’.
For example a 5.2 M solution of glucose contains 5.2 mole of glucose in every 1000 ml of solution.
The mass given in moles can be converted in to mass in grams provided we know the molar mass of a chemical.
Molar solution (Molarity =1M)
A Molar solution contains one mole of solute in one litre of solution.
Formula used to determine the molarity of a given solution
Molarity Molecular mass = Mass of solute present in 1000ml of solution.
Or
Mass of the solute x 1000
Molarity M =
Molecular mass of the solute x Volume of the solution Worked out Problems.
1. Calculate the molarity of a solution containing 4 grams of sodium hydroxide (NaOH) in 500 ml of the solution.(Molecular mass of sodium hydroxide=40) Mass of sodium hydroxide = 4 g
Molecular mass of sodium hydroxide=40 g (23X1+16X1+1X1) Mass of the solute x 1000
Molarity M =
Molecular mass of the solute x Volume of the solution 4 x 1000
Molarity M = 40 x 500
8
Molarity M = --- = 0.2M 40
2. Calculate the molarity of solution containing 40g of sugar (C12H22O11) in 200 ml of solution.(Molecular mass of sugar=342).
Formula
Molarity Molecular mass = Mass of solute present in 1000ml of solution.
Molecular mass of sugar = 1212 + 221 + 1611 = 342
Mass of sugar
Present in 200ml of solution = 40g Therefore
Mass of sugar present in 1000ml of solution = 200g Mass of the solute x 1000
Molarity M =
Molecular mass of the solute x Volume of the solution Substituting the values in the formula,
40 x 1000 Molarity M = 342 x 200
200
Molarity M = --- = 0.5848M 342
Therefore Molarity = 0.5848
3. Find the mass of urea CO (NH2)2 present in 4 litres of a 0.452M solution.(Molecular mass of urea=60)
Molecular mass of urea = 12+ 16 + 214 + 41 = 60 Formula
Mass of solute present in 1000ml or one litre of solution = Molarity Molecular mass
= 0.452M 60
= 27.12g. Therefore Mass of solute present in 4 litre of solution = 427.12 = 108.48g.
MOLALITY (M):
Molality is the number of moles of solute present in 1000g Or 1Kg of the solvent.
Molality is represented by the symbol ‘m’.
For example a 2.5 m solution of glucose (C6H12O6) contains 2.5 moles of glucose in 1000g of water.
Molal solution (Molality =1m)
A Molal solution contains one mole of solute in one Kg of solvent.
Formula used to determine the molality of the solution
Molality Molecular mass = Mass of solute in 1000g of solvent
Mass of solute x 1000
Molality m = Molecular mass of solute x Mass of solvent Worked out Example
1. Calculate the molality of a solution containing 4 grams of sodium hydroxide (NaOH) in 500 gms of water. (Molecular mass of sodium hydroxide=40).
Mass of sodium hydroxide = 4 g
Molecular mass of sodium hydroxide=40 g (23X1+16X1+1X1) Mass of the solute x 1000
Molality m = Molecular mass of the solute x mass of the solvent
4 x 1000 Molality m =
40 x 500 8
Molality m = --- = 0.2m 40
2. Calculate the molality of a solution containing 15g of methanol (CH3OH) in 300g of solvent. (Molecular mass of methanol=32)
Molecular mass of methanol = 12 + 16 + 41 = 32 Formula:
Mass of solute x 1000
Molality m = Molecular mass of solute x Mass of solvent
Mass of methanol present in 300g of water = 15g Therefore
15 x 1000 Molality m = 32 x 300
Molality32 = 50
Therefore Molality = 50/32 = 1.5625m
3. Calculate the molality of 5% solution of glucose (molecular mass of glucose=180)
In a solution the percentage of the solute glucose is= 5%
Hence the percentage of the solvent water is = 95%
Therefore
The mass of glucose is=5 g The mass of water is=95 g Molecular mass of glucose is=180
Mass of solute x 1000
Molality m = Molecular mass of solute x Mass of solvent
5 x 1000 5000
Molality m = ---=---=0.2926 m 180 x 95 17100
Mole-fraction of solvent(
x
1)Mole-fraction of the solvent is the ratio between the number of moles of solvent and the total number of moles present in solution. The mole of the solvent is denoted as n1and the mole of the solute is n2.
Number of Moles of solvent (n1) Mole-fraction of solvent =
Total number of Moles present in solution (n1+n2) n1
=
n1+n2
Mole-fraction of solute(x2)
Mole-fraction of the solute is the ratio between the number of moles of solute and the total number of moles present in solution.
Number of Moles of solute Mole-fraction of solute =
Total number of Moles present in solution
n2
=
---n1+n2
(Total number of moles present in the solution = number of moles of solute + number of moles of solvent)
In any solution, the sum of mole-fraction of solute and solvent is equal to one i.e. x1+x2=1.
Mass in gram
Note: Number of moles = Molecular Mass
In any solution, the sum of mole-fraction of solute and solvent is equal to one.
To prove that the sum of mole fractions of solute and solvent in a solution is equal to one
Let x and y be the number of moles of solute and solvent respectively x
Mole-fraction of solute = x + y
y Mole-fraction of solvent =
x + y
Mole-fraction of solute + Mole-fraction of solvent
X y x + y
= --- + --- = --- = 1 x + y x + y x + y
Hence it is proved.
Worked out example
1. Find the number of moles of solute and solvent in a solution containing 9.2g of ethyl alcohol (C2H5OH) in 180 g of water.
Molecular mass of ethyl alcohol = 212 + 61 + 16 = 46 Molecular mass of water = 18
Number of moles of ethyl alcohol = Mass / Molecular mass = 9.2/46 = 0.2
Number mole of water = 180/18 = 10 Total number of moles = 0.2 + 10 = 10.2
Mole fraction of ethyl alcohol = Number of moles of alcohol/total number of moles
= 0.2/10.2 = 0.0196
Mole fraction of water = Number of moles of water/total number of moles
= 10/10.2 = 0.9804
2. A solution is prepared by dissolving 4 g of sodium hydroxide in 18 g