## Q1. Explain the law of conservation of mass.

Ans. The law of conservation of mass states that mass can neither be created nor destroyed in a chemical reaction. For example, in the reaction $$C+{ O }_{ 2 }\longrightarrow C{ O }_{ 2 }$$, total mass of carbon dioxide produced is equal to the total mass of carbon and oxygen gas taking part in the reaction.

## Q2. What is the law of constant proportions?

Ans. The law of constant proportions, also known as the law of definite proportions, was established by Proust. According to this law, in a chemical substance the elements are always present in definite proportion by mass. All pure samples of a compound contain the same elements combined together in the same proportion by mass. For example, a sample of water would always contain hydrogen and oxygen in the ratio of 1:8 by mass irrespective of the source of water.

## Q3. What is the full form of IUPAC? What is the present accepted norm of IUPAC for naming symbols of elements?

Ans.

The full name of IUPAC is International Union of Pure and Applied Chemistry.

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 (upper case) and the second letter is always a small letter (lower case). For example, hydrogen is written as H and aluminium is written as Al. Some symbols are formed from the first letter of the name and a letter, appearing later in the name. For example, chlorine as Cl and zinc as Zn. Some other symbols are taken from the names of elements in Latin, German or Greek. For example, the symbol of iron is Fe from Latin name ferrum.

## Q4. Find the ratio by number of calcium and oxygen in the compound calcium oxide. (Given ratio by mass of calcium and oxygen in the compound is 5:2. Atomic mass of Ca = 40 u and O = 16 u)

Ans.

 Element Mass ratio (given) Atomic mass (u) Number of atoms (Mass ratio/atomic mass) Simplest ratio Ca 5 40 0.125 1 O 2 16 0.125 1

Thus, the formula of calcium oxide should be CaO as the simplest ratio of the elements is 1:1.

## (ii). 6.022 x $${ 10 }^{ 23 }$$numbers of $${ O }_{ 2 }$$ atoms (atomic mass of oxygen = 16 u)

Ans.

(i) The number of moles = $$\frac { Number\quad of\quad molecules }{ Mass\quad Avagadro’s\quad constant }$$

= $$\frac { 6.022\times { 10 }^{ 23 } }{ 6.022\times { 10 }^{ 23 } }$$ = 1

Now, Mass = Mole x Molecular mass = 1 x 32 = 32 g

(ii). The number of moles = $$\frac { Number\quad of\quad molecules }{ Mass\quad Avagadro’s\quad constant }$$

= $$\frac { 6.022\times { 10 }^{ 23 } }{ 6.022\times { 10 }^{ 23 } }$$ = 1

Now, Mass = Mole x Molecular mass = 1 x 16 = 16 g