Reactions in Solution
- A majority of reactions in the laboratories are carried out in solutions.
- The concentration of a solution or the amount of substance present in its given volume can be expressed in any of the following ways:-
- Mass per cent or weight percent (w/w% )
- Mole fraction
- Molarity
- Molality
Mass per cent or weight percent
- Mass percentage is one way of representing the concentration of an element in a compound or a component in a mixture.
- Mass percentage is calculated as the mass of a component divided by the total mass of the mixture, multiplied by 100%.
- For example: – Consider a compound of H2O which contains 2 moles of H2 and 1 mole of O2.
- Or it can be also stated as 2 x 1gram of H and 16grams of O.
- Therefore Mass % of H2=(2/18) x 100% and Mass% of O2=(16/18)x100%
- For example: – Consider a compound of H2O which contains 2 moles of H2 and 1 mole of O2.
Problem:-
A solution is prepared by adding 2 g of a substance A to 18 g of water. Calculate the mass per cent of the solute.
Answer:-
Mass of percent of A = (Mass of A)/ (Mass of solution) x100
= (2g)/ (2g of A + 18g of water) x100
= (2g/20g)/100
= 10 %
Mole Fraction
- It is the ratio of number of moles of a particular component to the total number of moles of the solution.
- For example: – Consider a substance ‘A’ dissolves in substance ‘B’ and their moles are nA and nB
- Mole fraction of A
- = (No. of moles of A)/(No. of moles of solution)
- (nA )/( nA + nB)
- Mole fraction of B
- =(No. of moles of B)/(No. of moles of solution)
- (nB )/( nA + nB)
Molarity
- Molarity is defined as the number of moles of the solute in 1 litre of the solution.
- It is widely used unit and is denoted by ‘M’.
- Molarity (M) = (No. of moles in solute)/(Volume of solution in litres)
Problem:-
A 4 g sugar cube (sucrose: C12H22O11) is dissolved in a 350 ml teacup filled with hot water. What is the molarity of the sugar solution?
Answer:-
Equation of molarity:-
M = (m/V)
Where M is molarity (mol/L)
m = number of moles of solute
V = volume of solvent (Litres)
For each of the atoms to get the total grams per mole:
C12H22O11 = (12) (12) + (1) (22) + (16) (11)
C12H22O11 = 144 + 22+ 176
C12H22O11 = 342 g/mol
To get the number of moles in a specific mass, divide the number of grams per mole into the size of the sample:
(4 g)/ (342 g/mol) = 0.0117 mol
Molality
- It is defined as the number of moles of solute present in 1 kg of solvent.
- It is denoted by m.
- Thus Molality(m) =
- (No. of moles of solute)/(Mass of solvent in kg)
Problem:-
The density of 3 M solution of NaCl is 1.25 g mL–1. Calculate molality of the solution.
Answer:-
M = 3 mol L–1
Mass of NaCl in 1 L solution = 3 × 58.5 = 175.5 g
Mass of 1L solution = 1000 × 1.25 = 1250 g
(Since density = 1.25 g mL–1)
Mass of water in solution = 1250 –175.5
= 1074.5g
Molality= (No. of moles of solute)/ (Mass of solvent in kg)
= (3 mol)/ (1.0745kg)
= 2.79 m.
Problem:-
How much copper can be obtained from 100 g of copper sulphate (CuSO4)?
Answer:-
1 mole of CuSO4 contains 1 mole of copper.
Molar mass of CuSO4 = (63.5) + (32.00) + 4(16.00)
= 63.5 + 32.00 + 64.00
= 159.5 g
159.5 g of CuSO4 contains 63.5 g of copper.
⇒ 100 g of CuSO4 will contain (63.5 x 100g)/ (159.5) of copper.
Therefore Amount of copper that can be obtained from 100 g CuSO4
= (63.5 x 100)/ (159.5) =39.81 g