Atoms And Molecules
Q71. Cinnabar (HgS) is a prominent ore of mercury. How many grams of mercury are present in 225 g of pure HgS? Molar mass of Hg and S are 200.6 g mol–1 and 32 g mol–1 respectively.
Ans. Molar mass of Hg = 200.6 g mol–1
Molar mass of S = 32 g mol–1
Molar mass of HgS = Molar mass of Hg + Molar Mass of S
= 200.6 + 32
= 232.6 g
232.6 g of HgS contains Hg = 200.6
225 g of HgS contains Hg = (200.6x225)/232.6
Q72. 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
Ans. (a) Ethyne, C2H2 = 2 x Atomic Mass of ‘C’ + 2 x Atomic Mass of ‘H’
= (2 x 12) + (2 x 1)
= 24 +2 = 26 g
(b) Sulphur molecule, S8 = 8 x Atomic Mass of ‘S’
= 8 x 32 = 256 g
(c) Phosphorus molecule, P4 = 4 x Atomic Mass of ‘P’
= 4 x 31 = 124 g
(d) Hydrochloric acid, HCl = Atomic Mass of ‘H’ + Atomic Mass of ‘Cl’
= 1 + 35.5 = 36.5 g
(e) Nitric acid, HNO3
= Atomic Mass of ‘H’ + Atomic Mass of ‘N’ + 3 x Atomic Mass of ‘O’
= 1 + 14 + (3 x 16)
= 63 g
Q73. Calculate the number of aluminium ions present in 0.051 g of aluminium oxide.
Ans. 1 mole of Al2O3 = Mass of Al2O3 in grams
= Mass of Al x 2 + Mass of O x 3
= 27 x 2 + 16 x 3
= 54 + 48
= 102 grams
Now, 1 mole aluminium oxide (Al2O3) = 6.022 x 1023
102 g = 6.022 x 1023
Therefore, 0.051 g aluminium oxide has = 6.022 x 1023 / 102 x 0.051
= 3.01 x 1020
1 molecule of aluminum oxide gives = 2Al+ ions
Hence, 0.051 g aluminium oxides gives = 2 x 3.01 x 1020
= 6.02 x 1020
Q74. Calculate the formula unit masses of ZnO, Na2O, K2CO3, given atomic masses of Zn = 65 u, Na = 23 u, K = 39 u, C = 12 u, and O = 16 u.
Ans. Formula unit mass of ZnO
= atomic mass of Zn + atomic mass of O
= 65 + 16
Formula unit mass of Na2O
= 2 x atomic mass of Na + atomic mass of O
= (2 x 23) + 16
Formula unit mass of K2CO3
= 2 x atomic mass of K + atomic mass of C + 3 x atomic mass O
= (2 x 39) + 12 + (3 x 16)
= 78 + 12 + 48
Q75. Which has more number of atoms?
100g of N2 or 100 g of NH3
Ans. Molar mass of N2 = 2x14 = 28g
28g of N2 has number of molecules = 6.022 x 1023
100g of N2 has number of molecules = (100 x 6.022 x 1023)/28
= 2.1 x 1024
Atoms in 100 g of N2 = 2.1 x 1024 x 2
= 4.2 x 1024
Molar mass of NH3 = 14+3x1 = 17g
17g of NH3 has number of molecules = 6.022 x 1023
100g of NH3 has number of molecules = (100 x 6.022 x 1023)/17
= 3.54 x 1024
Atoms in 100g of NH3 = 3.54 x 1024 x 4
= 1.416 x 1025
So, 100g of NH3 has more number of atoms.
Q76. What are the postulates of Dalton's atomic theory?
Ans. The postulates of this theory may be stated as follows:
(i) All matter is made of very tiny particles called atoms.
(ii) Atoms are indivisible particles, which cannot be created or destroyed in a chemical reaction.
(iii) Atoms of a given element are identical in mass and chemical properties.
(iv) Atoms of different elements have different masses and chemical properties.
(v) Atoms combine in the ratio of small whole numbers to form compounds.
(vi) The relative number and kinds of atoms are constant in a given compound.
Q77. State the law of constant proportions. Give one example to illustrate this law.
Explain giving a suitable example: Law of constant proportion.
How can you say the law of constant proportion is a valid justify with an example?
Ans. Lavoisier, along with other scientists, noted that many compounds were composed of two or more elements and each such compound had the same elements in the same proportions, irrespective of where the compound came from or who prepared it.
Example: In a compound such as water, the ratio of the mass of hydrogen to the mass of oxygen is always 1:8, whatever the source of water. Thus, if 9 g of water is decomposed, 1 g of hydrogen and 8 g of oxygen are always obtained.
This led to the law of constant proportions which is also known as the law of definite proportions. This law was stated by Proust as “In a chemical substance the elements are always present in definite proportions by mass”.
Q78. In a reaction, 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid. The products were 2.2 g of carbon dioxide, 0.9 g water and
8.2 g of sodium ethanoate. Show that these observations are in agreement with the law of conservation of mass.
sodium carbonate + ethanoic acid → sodium ethanoate + carbon dioxide + water
sodium carbonate (5.3g) + ethanoic acid (6g) → sodium ethanoate (8.2g) + carbon dioxide (2.2g) + water (0.9g)
As per the law of conservation of mass, the total mass of reactants must be equal to the total mass of products.
As per the reaction,
LHS (Total mass of reactants) = 11.3 g
RHS (Total mass of products) = 11.3 g
LHS = RHS
Hence, these observations are in agreement with the law of conservation of mass.
Q79. How do elements get their names and symbols?
Ans. Now-a-days, IUPAC (International Union of Pure and Applied Chemistry) approves names of elements. Many of the symbols are the first one or two letters of the element’s name in English. The first letter of a symbol is always written as a capital letter (uppercase) and the second letter as a small letter (lowercase).
(i) hydrogen, H
(ii) aluminium, Al and not AL
(iii) cobalt, Co and not CO.
Symbols of some elements are formed from the first letter of the name and a letter, appearing later in the name. Examples are: (i) chlorine, Cl, (ii) zinc, Zn etc.
Other symbols have been taken from the names of elements in Latin, German or Greek. For example, the symbol of iron is Fe from its Latin name ferrum, sodium is Na from natrium, potassium is K from kalium. Therefore, each element has a name and a unique chemical symbol.
Q80. Calculate the molecular masses of H2, O2, Cl2, CO2, CH4, C2H6,
C2H4, NH3, CH3OH.
Ans. Molecular mass of H2 = 2 x atomic mass of H = 2 x 1 = 2u
Molecular mass of O2 = 2 x atomic mass of O = 2 x 16 = 32u
Molecular mass of Cl2 = 2 x atomic mass of Cl = 2 x 35.5 = 71u
Molecular mass of CO2 = atomic mass of C + 2 x atomic mass of O
= 12 + (2x16) = 44u
Molecular mass of CH4 = atomic mass of C + 4 x atomic mass of H
= 12 + (4x1) = 16u
Molecular mass of C2H6 = 2 x atomic mass of C + 6 x atomic mass of H
= (2 x 12) + (6 x 1) = 30u
Molecular mass of C2H4 = 2 x atomic mass of C + 4 x atomic mass of H
= (2 x 12) + (4 x 1) = 28u
Molecular mass of NH3 = atomic mass of N + 3 x atomic mass of H
= 14 + (3 x 1) = 17u
Molecular mass of CH3OH = atomic mass of C + 3 x atomic mass of H + atomic mass of O + atomic mass of H
= 12 + (3 x 1) + 16 + 1 = 32u