1.    Mass of the given sample compound = 0.24g

Mass of boron in the given sample compound = 0.096g

Mass of oxygen in the given sample compound = 0.144g

% composition of compound = % of boron and % of oxygen

Therefore % of boron = mass of boron × 100/mass of the sample compound

= 0.096 × 100/0.24

= 40%

Therefore % of oxygen = mass of oxygen × 100/mass of the sample compound

= 0.144 × 100/0.24

= 60%.

2.    According to the law of chemical combination of constant proportions “in a chemical compound the elementary constituents always combine in constant proportions by weight/mass”. Therefore whether 3 g carbon is burnt in 8 g oxygen or 3g carbon is burnt in 50g oxygen in both cases only 11g carbon dioxide will be formed.

3.    When two or more atoms in a group is having a charge, such is called a polyatomic ion. For eg : NH4+, NO3– etc.

4.    Compound                                                             Chemical formula

Magnesium chloride                                                      MgCl2

Calcium oxide                                                                  CaO

Copper nitrate                                                                 CuNO3

Aluminium chloride                                                       AlCl3

Calcium carbonate                                                         CaCO3

5.    Compound                          Formula          Elements present

Quick lime                               CaO                     Calcium and oxygen

Hydrogen bromide                HBr                     Hydrogen and bromine

Baking powder                       NaHCO3             Sodium, hydrogen, carbon and oxygen

Potassium sulphate               K2SO4                 Potassium, sulphur and oxygen

6.    (a) Ethyene = C2H2 = 12 × 2 + 1 × 2 = 24 + 2 = 26 u = 26 g

(b) Sulphur molecular = S8 = 32 × 8 = 256 u = 256 g

(c) Phosphorus molecule = P4 = 31 × 4 = 124 u = 124 g

(d) Hydrochloric acid = HCl = 1 + 35.5 = 36.5 u = 36.5 g

(e) Nitric acid = HNO3 = 1 + 14 + (16 × 3) = 15 + 48 = 63 u = 63 g.

7.    (a) Atomic mass of nitrogen is 14 u.

       Therefore 1 mol of N = 14g

(b) Atomic mass of aluminium = 27u

       Therefore 1 mol of Al = 27g and so 4 mol of Al = 27 × 4 = 108g

(c) Molecular mass of Na2SO3 = 23 × 2 + 32 + 16 × 3 = 46 + 32 + 48 = 126 u

      therefore 1 mol of Na2SO3 has weight/mass 126g.

     hence, 10 mol of Na2SO3= 10 × 126 = 1260g

8.   (a) molecular mass of O2 = 32 u = 32g (1 mole)

since 32 g of O2 = 1mole then 12g of O2 = 1 × 12/32 =0.375 mole.

(b) molecular mass of H2O = 1 × 2 + 16 = 18 u= 18g (1 mole)

20g H2O = 1 × 20/18 = 1.11mole.

(c) molecular mass of CO2 = 12 + 16 × 2 = 12 + 32 = 44 u = 44g (1mole)

22g of CO2 = 1 × 22/44 = 0.5mole.

9.   (a) since 1 mole of O = atomic mass of O = 16u = 16g

then 0.2 mole of O = 0.2 × 16 = 3.2g

(b) 1mol of H2O = molecular mass of H2O = 1 × 2 + 16 =18 u =18g

then 0.5mol of H2O = 0.5 × 18 = 9g.

10.  1mol of S8 = molecular mass of S8 = 32 × 8 = 256u =256g

since 256g of S8 = 1mol = 6.022 × 1023 atoms (Avogadro number)

16g of S8 = 16 × 6.022 × 1023 /256 = 3.76 × 1022 molecules.

11.   1mol of Al2O3 = molecular mass of Al2O= 27 × 2 + 16 × 3 = 102u = 102g

aluminium ions present in Al2O3 = 2Al3+

102 g of Al2O3 contains aluminium ions = 2 × 6.022 × 1023

then 0.051 g Al2O3 contains aluminium ions = 2 × 6.022 × 1023 × 0.051/102= 6.022 × 1020

12.  According to law of conservation of mass :

mass of reactants = mass of products

Lets calculate and find out both results –

mass of reactants = mass of sodium carbonate +mass of ethanoic acid

= 5.3g + 6g

= 11.3g

mass of products = mass of sodium ethanoate + mass of carbon dioxide + mass of water

= 8.2g +2.2g + 0.9g

= 11.3g

Hence it is proved that these observations are in agreement with the law of conservation of mass.

13.  As per the given 1:8 ratio mass of oxygen gas required to react completely with 1g of hydrogen gas is 8g.

Therefore mass of oxygen gas required to react completely with 3g of hydrogen gas will be = 3 × 8 = 24g

14.  The postulate of Dalton’s atomic theory which is the result of the law of conservation of mass is mentioned as below :

Atoms are indivisible particles, which cannot be created or destroyed in a chemical reaction.

15.  The postulate of Dalton’s atomic theory which explains the law of definite proportions is

“Atoms combine in the ratio of small whole numbers to form compounds and the relative number and kinds of atoms are constant in a given compound.”

16.  According to the latest recommendations of International Union of Pure and Applied Chemistry (IUPAC) the atomic mass unit (amu) is abbreviated as u or unified mass. For chemical calculations the atomic masses of elements are expressed by taking the atomic mass of one atom of an element as the standard mass. Like the atomic mass of carbon is taken as 12 units and each unit is called as 1 a.m.u i.e.

1 amu = 1/12 of atomic masses of 6C12.

17.  An atom is an extremely minute particle and as such actual mass of an atom of hydrogen is considered to be 1.6 × 10-24 g. That is why it is not possible to see an atom with naked eyes.

18.  Compound                                                    Formula

Sodium oxide                                                      Na2O

Aluminium Chloride                                         AlCl3

Sodium Sulphide                                               Na2S

Magnesium Hydroxide                                     Mg(OH)2

19.  Formula                                                Compound

Al2(SO4)3                                                  Aluminium sulphate

CaCl2                                                          Calcium chloride

K2SO4                                                        Potassium sulphate

KNO3                                                         Potassium nitrate

CaCO3                                                       Calcium carbonate

20.  A chemical formula is the representation of elements present in a compound with the help of symbols and also the number of atoms of each element with those numbers only. for eg: A molecule of water (compound) contains 2 atoms of hydrogen and one atom of oxygen hence its chemical formula is H2O.

21.  (i)  2 atoms of hydrogen + 1 atom of sulphur = 3 atoms

(ii) 1 atom of phosphorus + 4 atoms of oxygen = 5 atoms

22.  Molecular mass of H2 = atomic mass of H × 2= 1 × 2 = 2u.

Molecular mass of O2 = atomic mass of O × 2 = 16 × 2 = 32u.

Molecular mass of Cl2 = atomic mass of Cl × 2 = 35.5 × 2 = 71u.

Molecular mass of CO2 = atomic mass of C + (atomic mass of O × 2)

= 12 + (16 × 2)

= (12 + 32)

= 44 u.

Molecular mass of CH4 = 12 + atomic mass of hydrogen × 4

= 12 + (1 × 4)

= 12 + 4

= 16 u.

Molecular mass of C2H6 = (12 × 2) + (1 × 6) = 24+6 = 30 u

Molecular mass of C2H4 = (12 × 2) + (1 × 4) = 24 + 4 = 28 u

Molecular mass of NH3 = 14 + (1 × 3) = 14 + 3= 17 u

Molecular mass of CH3OH = 12 + (1 × 3) + 16 + 1 = 12 + 3 + 16 + 1 = 32 u

23.  Formula unit mass of :

i) ZnO = Atomic mass of Zn + atomic mass of O = (65 + 16)u = 81 u

ii) Na2O = Atomic mass of Na + atomic mass of O = (23 × 2) + 16 = 46 + 16 = 62 u

iii) K2CO3 = (39 × 2) + 12 + (16 × 3) = 78 + 12 + 48 = 138 u.

24.  Weight of one mole of carbon = atomic mass of carbon (1 atom of carbon) = 12 u
Therefore one mole of carbon contains = 12 g = 6.022 x 1023 atoms (Avogadro number)

so 1 atom of carbon = 12/ g

or, 12 u = 12/6.022 × 1023 g

1 u = 12/6.022 × 1023 × 12 g

1 u = 1/6.022 × 1023 g

1 u = 0.1660577 × 10-23 g

or, 1 u = 1.660577 × 10-24 g

25.  We can find out the element with more number of atoms by calculating number of
moles of each of them :

Number of moles of sodium in 100g = m1/M1 = 100/23 = 4.34

Number of moles of iron in 100g = m2/M2 = 100/56 = 1.79

Therefore, the number of atoms is more for sodium as compared to iron.