Class 9 - Chemistry - Atoms and Molecules
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.
Mass of boron = 0.096g (Given)
Mass of oxygen = 0.144g (Given)
Mass of sample = 0.24g (Given)
Thus, percentage of boron by weight in the compound = (0.096/0.24)×100 % = 40%
Thus, percentage of oxygen by weight in the compound = (0.144/ 0.24) ×100 % = 60%
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?
The reaction of burning of carbon in oxygen may be written as:
Carbon + Oxygen à Carbon dioxide
C + O2 à CO2
3g of carbon reacts with 8 g of oxygen to produce 11g of carbon dioxide.
If 3g of carbon is burnt in 50g of oxygen, then 3g of carbon will react with 8 g of oxygen.
The remaining 42 g of oxygen will be left un-reactive. In this case also, only 11g of carbon dioxide will be formed.
The above answer is governed by the law of constant proportions.
What are polyatomic ions? Give examples.
A polyatomic ion is a group of atoms carrying a charge (positive or negative).
The ions which contain more than one atoms (same kind or may be of different kind) and behave as a single unit.
For example, ammonium ion (NH4 +), hydroxide ion (OH−), carbonate ion (CO32−), Sulphate ion (SO42−).
Write the chemical formulae of the following.
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate.
The chemical formulae of the following are as follows:-
(a) Magnesium chloride →MgCl2
(b) Calcium oxide →CaO
(c) Copper nitrate →Cu (NO3)2
(d) Aluminium chloride →AlCl3
(e) Calcium carbonate →CaCO3
Give the names of the elements present in the following compounds.
(a) Quick lime (b) Hydrogen bromide
(c) Baking powder (d) Potassium sulphate.
1. Quick lime
2. Hydrogen bromide
3. Baking powder
Sodium, hydrogen, carbon, oxygen
4. Potassium sulphate
Potassium, Sulphur, Oxygen
Calculate the molar mass of the following substances.
(a) Ethyne, C2H2
(b) Sulphur molecule, S8
(c) Phosphorus molecule, P4 (Atomic mass of phosphorus = 31)
(d) Hydrochloric acid, HCl
(e) Nitric acid, HNO3
(a) Molar mass of ethyne, C2H2 = (2 × 12) + (2 × 1) = 28g
(b) Molar mass of sulphur molecule, S8 = (8 × 32) = 256g
(c) Molar mass of phosphorus molecule, P4 = (4 × 31) = 124g
(d) Molar mass of hydrochloric acid, HCl = (1 + 35.5) = 36.5g
(e) Molar mass of nitric acid, HNO3 = (1 + 14) + (3 × 16) = 63g
What is the mass of—?
(a) 1 mole of nitrogen atoms?
(b) 4 moles of aluminium atoms (Atomic mass of aluminium = 27)?
(c) 10 moles of sodium sulphite (Na2SO3)?
Convert into mole.
(a) 12 g of oxygen gas
(b) 20 g of water
(c) 22 g of carbon dioxide
(a) 32 g of oxygen gas = 1 mole
Therefore, 12g of oxygen gas = (12/32) mole = 0.375 mole
(b) 18g of water = 1 mole
Therefore, 20 g of water = (20/18) mole = 1.11 moles (approx.)
(c) 44g of carbon dioxide = 1 mole
Therefore, 22g of carbon dioxide = (22/44) mole = 0.5 mole
What is the mass of?
(a) 0.2 mole of oxygen atoms?
(b) 0.5 mole of water molecules?
(a) Mass of one mole of oxygen atoms = 16g
Then, mass of 0.2 mole of oxygen atoms = (0.2 × 16g) = 3.2g
(b) Mass of one mole of water molecule = 18g
Then, mass of 0.5 mole of water molecules = (0.5 × 18g) = 9g
Calculate the number of molecules of sulphur (S8) present in 16 g of solid sulphur.
1 mole of solid sulphur (S8) = (8 × 32g) = 256g
i.e., 256g of solid sulphur contains = 6.022 × 1023 molecules
Then, 16g of solid sulphur contains ((6.022×1023)/ (256)) × 16 molecules
= 3.76 × 1022 molecules (approx.)
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)
Molar mass of aluminium oxide (Al2O3) = (2 × 27) + (3 × 16)
= 54 + 48 = 102g
=> 102g of Al2O3 will have 6.022 × 1023 molecules of Al2O3
Therefore, 0.051 g of Al2O3 contains
= ((6.022×1023)/ (102)) × 0.051 molecules
= 3.011 × 1020 molecules of Al2O3
The number of aluminium ions (Al3+) present in one molecules of aluminium oxide is 2.
Therefore, The number of aluminium ions (Al3+) present in
=3.011 × 1020 molecules (0.051g) of aluminium oxide (Al2O3) = (2 × 3.011 × 1020)
= 6.022 × 1020