Atoms and Molecuels

Question: (1) In a reaction, 5.3 g 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

Solution:

Given, reaction

Sodium carbonate +ethanoic acid ⇒ sodium ethanoate + carbon dioxide + water

In this reaction,

Reactant = Sodium carbonate +ethanoic acid

And product = sodium ethanoate + carbon dioxide + water

Total mass of reactant = Mass of sodium carbonate + Mass of ethanoic acid

= 5.3 g + 6 g = 11.3 g

Thus, total mass of reactant = 11.3 g

Now, total mass of product = mass of sodium ethanoate + mass carbon dioxide + mass of water

= 8.2 g + 2.2 g + 0.9 g = 11.3 g

Thus, mass of product = 11.3 g

Here, since mass of reactant = mass of product.

Thus, these observations are in agreement with the Law of Conservation of Mass. As Law of Conservation of Mass says that mass can neither be created nor destroyed in a chemical reaction.

Question: (2) Hydrogen and oxygen combine in the ratio of 1:8 by mass to form water. What mass of oxygen gas would be required to react completely with 3 g of hydrogen gas?

Solution :

∵ To react completely, for 1 g of hydrogen gas mass of oxygen required = 8 g

∴ To react completely, for 3 g of hydrogen gas mass of oxygen required = 8 × 3 g = 24 g

Thus, for 3g of hydrogen gas to react completely mass of oxygen required = 24 g Answer

Alternate Method

Ratio of hydrogen to oxygen required to form water by mass = 1: 8

i.e. ratio `=1/8`

Thus, when mass of hydrogen = 3 g

Then required ratio `=(1xx3)/(8xx3)=1/24`

Thus, required mass of oxygen = 24 g Answer

Question: (3) Which postulate of Dalton's atomic theory is the result of the law of conservation of mass?

Solution:

'Atoms are indivisible particles, which cannot be created or destroyed in a chemical reaction' is one of the postulates of Dalton's atomic theory and is the result of the law of conservation of mass.

Question: (4) Which postulate of Dalton's atomic theory can explain the law of definite proportions?

Solution:

'The relative number and kinds of atoms are constant in a given compound' is one of the postulates of Dalton's atomic theory and can explain the law of definite proportions.

Question: (5) Define the atomic mass unit.

Solution:

Atomic mass unit is the unit of relative atomic mass. It is denoted by letter 'u'.

One Atomic Mass Unit is a mass unit equal to exactly one–12th (`1/12`th) the mass of one atom of carbon–12.

The relative atomic masses of all the element have found with respect to an atom of carbon–12.

Example:

The relative atomic mass of Hydrogen = 1 u

This means that hydrogen atom is 1 times heavier than `1/12^(th)` of the carbon-12 atom.

Question: (6) Why is it not possible to see an atom with naked eyes?

Solution:

Atoms are very small in size. Actually atoms are smaller than anything that we can imagine or compare with.

Size of atoms are measured in nanometer. 1 nm `=10^(-9)` m.

And size of hydrogen atom is equal `10^(-10)` m.

Thus, because of very very small size it is not possible to see an atom with naked eyes.

Question: (7) Write down the formulae of

(i) sodium oxide

(ii) aluminium chloride

(iii) sodium sulphide

(iv) magnesium hydroxide

Solution:

(i) sodium oxide

Symbol of sodium : Na

Symbol of oxide : O

Valency of Sodium (Na) = +1

Valency of oxide (O) = –2

Thus, Formula of sodium oxide is Na₂O

(ii) aluminium chloride

Symbol of Aluminium : Al

Symbol of chloride: Cl

Valency of aluminium (Al) = 3

Valency of chloride (Cl) = 1

Thus, formula of Aluminium chloride is AlCl₃

(iii) sodium sulphide

Symbol of sodium : Na

Symbol of sulphide: S

Valency of sodium: 1

Valency of sulphide: 2

Thus, Formula of sodium sulphide is Na₂S

(iv) magnesium hydroxide

Symbol of Magnesium : Mg

Symbol of hydroxide: OH

Valency of magnesium: 2

Valency hydroxide (OH): –1

After cross multiplication of symbol and valency of magnesium and hydroxide

Thus, formula of Magnesium hydroxide is Mg(OH)₂

Question: (8) Write down the names of compounds represented by the following formulae:

(i) Al₂(SO₄)₃

(ii) CaCl₂

(iii) K₂SO₄

(iv) KNO₃

(v) CaCO₃

Solution:

(i) Al₂(SO₄)₃

Here, Al is the symbol for Aluminium

And SO₄ is the symbol used for sulphate.

Thus, name of given compound is Aluminium sulphate

(ii) CaCl₂

Here, Ca is the symbol for Calcium

And Cl is the symbol for chloride

Thus, name of the given compound is Calcium chloride

(iii) K₂SO₄

Here, K is the symbol used for Potassium

And SO₄ is the symbol used for sulphate

Thus, name of the given compound is Potassium sulphate

(iv) KNO₃

Here, K is the symbol used for Potassium

And symbol NO₃ is used to denote nitrate

Thus, name of the given compound is Potassium nitrate

(v) CaCO₃

Here, Ca is a symbol used for Calcium

And symbol CO₃ is used for carbonate

Thus, name of the given compound is Calcium carbonate

Question: (9) What is meant by the term chemical formula?

Solution:

The chemical formula of a compound is a symbolic representation of its composition. A chemical formula gives more information about a compound in comparison to its name in words.

Example: Chemical formula of water is H₂O

Chemical formula of sodium chloride is NaCl

Chemical formula of carbon dioxide is CO₂, etc.

Question: (10) How many atoms are present in a

(i) H₂S molecule and

(ii)PO₄^3– ion?

Solution:

(i) H₂S molecule and

In the given compound,

Number of hydrogen atoms = 2

And number of sulphide atoms = 1

∴ Total number of atoms present = 2+1 = 3

Thus, total number of atoms present in the given molecule = 3

And the name of given molecule is hydrogen sulphide

(ii) PO₄^3– ion?

Number of phosphorous atom = 1

Number of oxygen atoms = 4

∴ total number of atoms = 1 + 4 = 5

Thus, total number of atoms present in the given ion = 5

And name of the given ion is phosphate ion.

Question: (11) Calculate the molecular masses of H₂, O₂, Cl₂, CH₄, NH₃, CH₃OH.

Solution:

(a) H₂

Atomic mass of hydrogen (H) = 1 u

Thus, molecular mass of H₂ = 1 × 2 = 2 u

(b) O₂

Atomic mass of Oxygen (O) = 16 u

Thus, molecular mass of O₂ = 16 × 2 = 32 u

(c) Cl₂

Atomic mass of Chlorine (Cl) = 35.5 u

Thus, molecular mass of Cl₂

= 35.5 u × 2 = 71 u

(d) CH₄

Atomic mass of carbon (C) = 12 u

Atomic mass of hydrogen (H) = 1 u

Thus, molecular mass of CH₄

= 12 u + (1 u × 4) = 12+4=16 u

(e) NH₃

Atomic mass of nitrogen (N) = 14 u

Atomic mass of hydrogen (H) = 1 u

Molecular mass of NH₃

= 14 u + (1 u × 3) = 14 + 3 = 17 u

(f) CH₃OH

Atomic mass of carbon (C) = 12 u

Atomic mass of hydrogen (H) = 1 u

Atomic mass of oxygen (O) = 16 u

Thus, Molecular mass of CH₃OH

= 12 u + (1 u × 3) + 16 u + 1 u

= 12 u + 3 u + 16 u + 1 u

= 32 u

Question: (12) 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 = 16u.

Solution:

(a) ZnO

Atomic mass of Zinc (Zn) = 65 u

Atomic mass of Oxygen (O) = 16 u

Thus, formula unit mass of ZnO

= 65 u + 16 u = 81 u

(b) Na₂O

Atomic mass of sodium (Na) = 23 u

Atomic mass of oxygen (O) = 16 u

Thus, formula unit mass of Na₂O

= (23 u × 2) + 16 u

= 46 u + 16 u = 62 u

(c) K₂CO₃

Atomic mass of potassium (K) = 39 u

Atomic mass of carbon (C) = 12 u

Atomic mass of oxygen (O) = 16 u

Thus, formula unit mass of K₂CO₃

= (39 u × 2) + 12 u + (16 u × 3)

= 78 u + 12 u + 48 u

= 138 u

Question: (13) If one mole of carbon atoms weighs 12 grams, what is the mass (in grams) of 1 atom of carbon?

Solution:

Given, weight (mass) of 1 mole of carbon atoms = 12 gram

Since, 1 mole of carbon = 6.022 × 10²³

&beaus; mass of 6.022 × 10²³ carbon atoms = 12 g

∴ mass of 1 atom of carbon `=(12\ g)/(6.022xx10^23)`

= 1.9926 × 10^–23 g Answer

Question: (14) Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given, atomic mass of Na = 23 u, Fe= 56 u)?

Solution:

Given,

Mass of sodium = 100 g

Atomic mass of sodium = 23 u

Thus, gram atomic mass of sodium (Na) = 23 g

∵ Number of atoms present in 23 g of sodium = 6.022 × 10²³

∴ Number of atoms present in 1 g of sodium `=(6.022xx10^23)/23`

∴ Number of atoms present in 100 gram of sodium `=(6.022xx10^23)/23xx100`

= 2.6182 × 10²⁴

Again, given Mass of iron (Fe) = 100 gram

Atomic mass of iron (Fe) = 56 u

Thus, gram atomic mass of iron (Fe) = 56 g

∵ Number of atoms present in 56 g of iron = 6.022 × 10²³

∴ Number of atoms present in 1 g of iron `=(6.022xx10^23)/56`

∴ Number of atoms present in 100 gram of iron `=(6.022xx10^23)/56xx100`

= 1.0753 × 10²⁴

Thus, 100 gram of sodium contain more number of atoms than 100 gram of iron.