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:-

1. Mass per cent or weight percent (w/w% )
2. Mole fraction
3. Molarity
4. 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%

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 n_A and n_B
  • Mole fraction of A
    • = (No. of moles of A)/(No. of moles of solution)
    • (n_A )/( n_A + n_B)
  • Mole fraction of B
    • =(No. of moles of B)/(No. of moles of solution)
    • (n_B )/( n_A + n_B)

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