It indicates that the element exist in the form of isotopes and its atomic mass is average atomic mass.

1 mole of CuS04 contains 1 mole (1 g atom) of Cu

Molar mass of CuS0₄= 63.5 + 32 + 4 x 16 = 159.5 g mol^-1

Thus, Cu that can be obtained from 159.5 g of CuS0₄ = 63.5 g

Therefore, Cu can be obtain from 159.5 g of CuSO₄ = 63.5 x 100 /159.5 

                                                                                    = 39.81g Answer

Mass of a molecule is that of a single molecule also known as its actual mass. But molecular mass is the mass of Avagadro’s number(6.022 x 10²³) of molecules.

In the text part we use the word mole while as a unit, we call it mol.

No, one gram mole of a gas occupies 22.4 L only under N.T.P or S.T.P conditions. i.e. at 273K temperature and under 760 mm pressure. If these conditions are not used, then the volume is not 22.4 L.

Empirical formula gives the simplest ration of the atoms of different elements in the molecules of a substance but the molecular formula gives their actual ratio.

This indicates that the compounds contains in it oxygen also and its percentage is (100 – 92) = 8.

A chemical equation has to be balanced in order to satisfy the law of conservation of mass. According to the law, there is no change in mass when the reactants change into products therefore, the chemical equations has to be balanced

Methane (CH₄) is regarded as the limiting reagent because air or oxygen is always present in excess. The amounts of C0₂ and H₂0 formed in the reaction depend upon the amount of methane only. Therefore, it is regarded as limiting reagent.

by definition, molarity (M) of a solution is the number of moles of the solute dissolved per litre of the solution and molality(m) is the number of moles of the solute dissolved per kilogram of the solvent. Therefore, in molarity, Volume of the solution is considered while in molality, mass of the solvent is taken into consideration.

Avagadro's number of molecules particles are present in one mole of a gas i.e. molecular mass of a gas expressed in grams. Any change in temperature and pressure has no influence on the number of particles.

No, there is a difference there are 0.5 mole of NaOH and 0.5 M(Molarity) means there are 0.5 moles present in Volume of solution (in Litre).

No, it is not possible. The fractional atomic mass of an element is it’s an average mass and not the actual mass. In the case, the element chlorine exists as two isotopes with atomic mass 35 u. and 37 u respectively in the ratio of 3: 1. The average come out to be fractional i.e. 35.5 u.

No, atomic masses of the elements are not actual masses. There are only relative because the actual masses are very small. For detail consult text part.

The coefficient of the reactant and product species involved in a chemical equation represent by the balanced form, are known as stiochiometric coefficient. For example.

N₂ (g) + 3H₂  ⇒ 2 NH3 (g)

Limiting reactant i.e. the reactant present in fixed amount is so named because it limits the participation of other reactant ever if present in excess in a particular reaction.

No, The actual masses are not fractional. It is the average of the atomic masses of the isotopes of an element which may be fractional. For example, the two isotopes of chlorine have atomic masses 35u and 37 u respectively the average atomic mass of chlorine 35.5 u.

Air which contains certain suspended particles such as dust particles is heterogeneous mixture and not a homogeneous mixture.

Since water (H₂O) is a compound It is chemically and physically different from hydrogen and oxygen, its properties are quite different from those of hydrogen and oxygen.

According to the Dalton atomic theory. An atom is the ultimate particles of matter which cannot be further divided into anything simpler than itself.

Molarity of a solution normally decreases with rise in temperature.

It means that 0.1 gram equivalent (4 g) of NaOH is dissolved per litre of the solution.

Yes, because HCL is a monobasic acid.  Its molecular mass and equivalent mass are both same. (36.5).

Methane (CH₄) is regarded as the limiting reactant in the combustion reaction because the other reactant i.e. oxygen is always present in excess. The extent of the combustion reaction will depend upon the amount of methane only. 

This illustrate the law of constant composition because sulphur and oxygen are combined in fixed ratio by mass n sulphur dioxide (S0₂) which is formed in this case.

(i) H = 1, 0 = 16 

Molar mass = 1 x 2 + 16 x 1 = 18 u or g/mol

(ii) C = 12  , 0 = 16 

Molar mass = 12 x 1 + 2 x 16 = 42 u or g/mol

(iii) C = 12, H = 1 

Moalr mass = 12 x 1 + 4 x 1 = 16 u or g/mol  

Na = 23u , S = 32u , 0 = 16u

Na₂SO₄ = 2 x 23 + 32 + 16 x 4 = 142 u 

% of Na in Na₂SO_{4 =}     = 32.99 % 

% of S in Na₂SO_{4 =}  

% of 0 in Na₂SO_{4 =}  

Iron = 69.9% , dioxide = 30.1%

Atomic Mass of Fe = 55.85

Atomic mass of O = 16

Mole of Oxygen =   

Mole of fe  = 

Molecular ratio in ratio simplest form = 2 : 3