Q. 1
What is the percentage composition of sodium and sulphur and oxygen in sodium sulphate?
• a. 30.4, 22, 45
• b. 31, 24, 46
• c. 31.4, 22.8, 45.7
• d. 30.1, 22.8, 46.7
• e. 30, 21.8, 46.7

• Explaination: Explanation: Method is as follows:
Molecular formula for sodium sulphate is Na2SO4
Molar mass of sodium sulphate is 44+32+64 = 140
Mass percent of sodium is (2*22) / (140) * 100 = 31.4
Mass percent of sulphur is (1*32) / (140) * 100 = 22.8
Mass percent of oxygen is (4*16) / (140) * 100 = 45.7
Q. 2
Identify the anode, cathode, oxidation half-cell and reduction half-cell in the following cell diagram. • a. A- anode, reduction half-cell; B- cathode, oxidation half-cell
• b. A- cathode, reduction half-cell; B- anode, oxidation half-cell
• c. A-anode, oxidation half-cell; B- cathode, reduction half-cell
• d. A- cathode, oxidation half-cell; B- anode, reduction half-cell
• e. A- anode, reduction half-cell; B- cathode, reduction half-cell

• Explaination: Explanation: The anodic reaction is always to be written on the left hand of the cell diagram and the cathodic reaction is to be written on the right hand side. An oxidation reaction always takes place at the anode and the reduction reaction always takes place at the cathode.
Q. 3
Which of the following is an example for an isolated system?
• a. A pot of boiling water
• b. Boiling a soup in an open sauce pan in a stove
• c. Cooking rice in a pressure cooker
• d. An open tank of water
• e. Hot water in a thermos flask

• Explaination: Explanation: Hot water in a thermos flask is an example for an isolated system where neither energy nor matter can enter or exit. Boiling a soup in an open sauce pan in a stove is an example for open system in which, it can freely exchange its energy and matter with its surroundings. Cooking rice in a pressure cooker is an example for a closed system where it can exchange only energy with its surroundings. A pot of boiling water and an open tank of water is an example for an open system as the matter gets exchanged with its surroundings.
Q. 4
The bond order for helium molecule is
• a. 1
• b. 2
• c. 2.5
• d. 3
• e. 0

• Explaination: Explanation: The electronic configuration of helium in the ground state is represented as (??1s)2 and in the excited state, it is represented as (??*1s)2.
So, number of electrons in bonding molecular orbital (Nb) is 2 and number of electrons in anti bonding molecular orbital (Na) is 2.
Bond order = (Nb-Na)/2 = (2-2)/2 = 0
The bond order for He2 is 0, so the molecule does not exist.
Q. 5
The order of ionisation energy is
• a. s
• b.
• c.
• d.
• e.