Question 1: A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.
Answer: Percentage of boron in given compound `= (0.096/0.24) xx 100 = 40%`
Percentage of oxygen in given compound `= (0.144/0.24) xx 100 = 60%`
Question 2: When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your answer?
Answer: We know that atoms of different elements combine in definite proportions to make a particular compound. Hence, 3 g of carbon will always give 11 g of carbon dioxide.
Question 3: What are polyatomic ions? Give examples.
Answer: An ion with more than one atom is called a polyatomic ion. Examples: Sulphate (SO42-), Nitrate (NO32-)
Question 4: Write the chemical formulae of the following.
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate.
Answer: (a) MgCl2, (b) CaO, (c) CuNO3, (d) AlCl3, (e) CaCO3
Question 5: Give the names of the elements present in the following compounds.
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate.
Answer: (a) Calcium, oxygen, (b) Hydrogen, bromine, (c) sodium, hydrogen, carbon, oxygen, (d) Potassium, sulphur, oxygen
Question 6: Calculate the molar mass of the following substances.
(a) Ethyne, C2H2
Answer: `2 xx 12 + 2 xx 1 = 26` u
(b) Sulphur molecule, S8
Answer: `8 xx 32 = 256` u
(c) Phosphorus molecule, P4 (Atomic mass of phosphorus = 31)
Answer: `4 xx 31 = 124`
(d) Hydrochloric acid, HCl
Answer: `1 xx 1 + 1 xx 35.5 = 36.5` u
(e) Nitric acid, HNO3
Answer: `1 xx 1 + 1 xx 14 + 3 xx 16 = 63` u
Question 7: What is the mass of:
(a) 1 mole of nitrogen atoms?
Answer: Atomic mass of nitrogen = 14 u
Hence, mass of 1 mole of nitrogen atoms = 14 g
(b) 4 moles of aluminium atoms (Atomic mass of aluminium = 27)?
Answer: Atomic mass of aluminium = 27 u
Hence, mass of 4 moles of aluminium atoms `= 4 xx 27 = 108`
(c) 10 moles of sodium sulphite (Na2SO3)?
Answer: Molecular mass of sodium sulphite
`= 2 xx 23 + 1 xx 32 + 3 xx 16 = 46 + 32 + 48 = 126 u`
Hence, mass of 10 mole of sodium sulphite `= 10 xx 126 = 1260 g`
Question 8: Convert into mole.
(a) 12 g of oxygen gas
Answer: Mass of 1 mole of oxygen = 32 g
Hence, number of mole in 12 g oxygen `= 12/32 = 0.375 M`
(b) 20 g of water
Answer: Molecular mass of water `= 2 xx 1 + 1 xx 16 = 18 u`
So, mass of 1 mole of water = 18 g
Hence, number of mole in 20 g of water `= 20/18 = 1.11 M`
(c) 22 g of carbon dioxide.
Answer: Molecular mass of carbon dioxide `= 1 xx 12 + 2 xx 16 = 44 u`
So, mass of 1 mole of carbon dioxide = 44 g
Hence, number of mole in 22 g of carbon dioxide `= 22/44 = 0.5 M`
Question 9: What is the mass of
(a) 0.2 mole of oxygen atoms?
Answer: Mass of 1 mole oxygen atoms = 16 g
Hence, mass of 0.2 mole of oxygen atoms `= 0.2 xx 16 = 3.2 g`
(b) 0.5 mole of water molecules?
Answer: Mass of 1 mole of water = 18 g
Hence, mass of 0.5 mole of water `= 0.5 xx 18 = 9 g`
Quesiton 10: Calculate the number of molecules of sulphur (S8) present in 16 g of solid sulphur.
Answer: Mass of 1 mole of sulphur `= 8 xx 32 = 256 g`
Hence, number of mole in 16 g of sulphur `= 16/256 = 0.625 M`
Question 11: Calculate the number of aluminium ions present in 0.051 g of aluminium oxide. (Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27 u)
Answer: Formula of aluminium oxide = Al2O3
Mass of aluminium atoms in aluminium oxide `= 2 xx 27 = 54 u`
Molecular mass of aluminium oxide `= 2 xx 27 + 3 xx 16 = 54 + 48 = 102 u`
Here; 2 mole aluminium ions are present in 102 g aluminium oxide
Existence of Atoms
In Text Solution
Hence, number of aluminium ions in 0.051 g aluminium oxide
`= (2/102) xx 0.051 = 2/2000` mole
`= (6.022 xx 10^23)/1000 = 6.022 xx 10^20`