NIOS Class 10 Science and Technology Chapter 3 Atoms and Molecules

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NIOS Class 10 Science and Technology Chapter 3 Atoms and Molecules

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Atoms and Molecules

Chapter: 3

INTEXT QUESTIONS 3.1

1. Name the scientists who proposed the law of conservation of mass and law of constant proportions.

Ans: Lavoisier proposed the law of conservation of mass and Proust proposed the law of constant proportions.

2. 12 g of magnesium powder was ignited in a container having 20 g of pure oxygen. After the reaction was over, it was found that 12 g of oxygen was left unreacted. Show that it is according to the law of constant proportions.

2Mg + O2 ⎯→ 2MgO.

Ans: In the container, 12g of Oxygen was left unreacted. Therefore, amount of unreacted Oxygen = (20 – 12)g = 08g. Thus 12g of magnesium reacted with 8g of oxygen in the ratio 12:8. This is what we expected for MgO i.e 24g of Mg reacted with 16g of Oxygen or 12g of Mg will react with 8g of Oxygen.

INTEXT QUESTIONS 3.2

1. Nitrogen forms three oxides: NO, NO2 and N2O3. Show that it obeys laws of multiple proportions.

Ans: Atomic mass of nitrogen is 14u and that of oxygen is 16u.

In NO, 14g of nitrogen reacted with 16g of oxygen 

In NO2, 14g of nitrogen reacted with 32g of oxygen

In N2O3, 28g of nitrogen reacted with 48g of oxygen.

2. Atomic number of silicon is 14. If there are three isotopes of silicon having 14, 15 and 16 neutrons in their nuclei, what would be the symbol of the isotope?

Ans: Atomic number of Si is 14 Mass number of silicon atoms having 14,15 and 16 neutrons will be 28,29 and 30 respectively and therefore symbols of isotopes of silicon will be 

3. Calculate molecular mass of the compounds whose formulas are provided below: 

C2H4, CH2O and CH3OH.

Ans: Molecular mass of C2H4 = mass of two atom of carbon + mass of 4 atom of hydrogen

= 2 X lu +1 + 16u = 18u.

Molecular mass of H2O = mass of two atoms of hydrogen + mass of one atom of oxygen.

= 2 x 1u + x 16u=18u.

Molecular mass of CH3OH = mass of one atom of carbon +mass of 4 atoms of hydrogen + mass of one atom of oxygen

= 1 x 2u + 4 + 1u x  16u = 32u.

INTEXT QUESTIONS 3.3

1. Work out a relationship between the number of molecules and mole.

Ans: 1 mole of a substance contains 6.023 × 1023 molecules of that substance i.e 1 mole of a substance = 6.023 × 1023 molecules of that substance.

2. What is molecular mass? In what way it is different from the molar mass?

Ans: Molecular mass is the sum of atomic masses of all the atoms present in that molecule.

Molecular mass is the mass of one molecule whereas molar mass is the mass of 1 mol or 6.023 × 1023 elementary entities (atoms, molecules, ions).

3. Consider the reaction.

C (s) + O2 (g) ⎯→ CO2 (g)

18 g of carbon was burnt in oxygen. How many moles of CO2 are produced?

Ans: 

4. What is the molar mass of NaCl?

Ans: Molar mass of NaCl = ( 23.0 + 35.5 )g mol-1 

                                         = 58.5 g mol-1.

INTEXT QUESTIONS 3.4

1. Write the name of the expected compound formed between.

(i) Hydrogen and sulphur.

Ans: H2S.

(ii) Nitrogen and hydrogen.

Ans: NH3.

(iii) magnesium and oxygen.

Ans: MgO.

2. Propose the formulas and names of the compounds formed between.

(i) potassium and iodide ions.

Ans: KI, Potassium iodide.

(ii) sodium and sulphate ions.

Ans: Na2SO4, Sodium Sulphate.

(iii) Aluminium and chloride ions.

Ans: AlCl3, Aluminium Chloride. 

3. Write the formula of the compounds formed between.

(i) Hg2+ and Cl

Ans: HgCl2.

(ii) Pb2+ and 3PO4

Ans: Pb3(PO4)2.

(iii) Ba2+ and 2SO 4

Ans: BaSO4.

TERMINAL EXERCISE

1. Describe the following:

(a) Law of conservation of mass?

Ans: The law of conservation of mass states that. “The mass in an isolated system can neither be created nor be destroyed but can be transformed from one form to another”. According to the law of conservation of mass, the mass of the reactants must be equal to the mass of the products for a low energy thermodynamic process.

(b) Law of constant proportions.

Ans: The law of constant proportions states that chemical compounds are made up of elements that are present in a fixed ratio by mass. This implies that any pure sample of a compound, no matter the source, will always consist of the same elements that are present in the same ratio by mass.

(c) Law of multiple proportions.

Ans: Law of multiple proportions, statement that when two elements combine with each other to form more than one compound, the weights of one element that combine with a fixed weight of the other are in a ratio of small whole numbers.

2. What is the atomic theory proposed by John Dalton? What changes have taken place in the theory during the last two centuries?

Ans: The English scientist John Dalton was by no means the first person to propose the existence of atoms, such ideas date back to classical times. Dalton’s major contribution was to arrange those ideas in proper order and give evidence for the existence of atoms. He showed that the mass relationship expressed by Lavoisier and Proust (in the form of law of conservation of mass and law of constant proportions) could be interpreted most suitably by postulating the existence of atoms of the various elements.

In 1803, Dalton published a new system of chemical philosophy in which the following statements comprise the atomic theory of matter:

(a) Matter consists of indivisible atoms.

(b) All the atoms of a given chemical element are identical in mass and in all other properties.

(c) Different chemical elements have different kinds of atoms and in particular such atoms have different masses.

(d) Atoms are indestructible and retain their identity in chemical reactions.

(e) The formation of a compound from its elements occurs through the combination of atoms of unlike elements in a small whole number ratio. 

3. Write the number of protons, neutrons and electrons in each of the following isotopes.

Ans: Do Yourself.

4. Boron has two isotopes with masses 10.13 u and 11.01 u and abundance of 19.77% and 80.23% respectively. What is the average atomic mass of boron?

Ans: To find atomic mass of boron of the ratio of its isotopes is determined:

= 10:130 – 19.77 %

= 11.01u – 80.23 %

∴ 1: 4

average atomic mass of boron = (10.13u x 1 + 11.01u x 4)/5 = (54.17)/5 = 10.834u.

5. Give symbol for each of the following isotopes:

(a) Atomic number 19, mass number 40. 

Ans: Atomic number 19, mass number 40 – 40 / 19 k.

(b) Atomic number 7, mass number 15.

Ans: Atomic number 7, mass number 15 – 15/7 N.

(c) Atomic number 18, mass number 40.

Ans: Atomic number 18, mass number 40 – 40/80 AR.

(d) Atomic number 17, mass number 37.

Ans: Atomic number 17, mass number 37 – 37/17 Cl. 

6. How does an element differ from a compound? Explain with suitable examples.

Ans: Elements are formed with the same types of atoms. For example, we are considering oxygen (O2). Oxygen is formed with two oxygen (O) atoms of the same kind, atomic number, and properties. Compounds are constructed with more than one type of atom.

7. Charge of one electron is 1.6022×10–19 coulomb. What is the total charge on 1 mol of electrons?

Ans: Charge of one electron is 1.6022 x 10-9

Number of electrons in 1 mole  = avogadro number = 6.022 x 10 x 10 23

∴ charge on 1 mole coulomb 

= 1.6022 x 10 -19 coulomb  x 6.22 x 10 23

= 96485 x 10 4 coulomb 

= 96485 coulomb. 

8. How many molecules of O2 are in 8.0 g of oxygen? If the O2 molecules were completely split into O (Oxygen atoms), how many mole of atoms of oxygen would be obtained?

Ans: There are 6.022 x 1023 Atoms in a mole or 16 grams of oxygen There will be 6.022 x 10 23 molecules in 1 mole of oxygen  i.e . 32 grams of oxygen  8 grams of  oxygen is ¼ of 32 grams

∴ Numbers of molecules in 8 grams of oxygen is = 6.022 x 10 22 /4 = 1.51 x 1023 

Hence the numbers of molecules from 1 molecule. 

2 atoms of oxygen together from 1 molecule Numbers of molecules in 8 grams of oxygen is

=2 x 1.51 x 10 23 = 3.02 x 10 23   

Numbers of moles of atoms in mole oxygen atom i.e 16 grams of oxygen = 6.022 x 10 23 

∴ 3.02 x 1023  moles in atoms = 3.02 x 10 23 / 6.022 x 1023 = ½ = 0.5 moles 

Hence the number of molecules in 8 grams of oxygen is 3.02  x 10 23 and 0.5 moles.

9. Assume that the human body is 80% water. Calculate the number of molecules of water that are present in the body of a person whose weight is 65 kg.

Ans: The mass of water is (80/100×65) =52kg.

So, the no of molecules present in human body is = (52/18 × 6.023×10^23) = 1.73×10^22.

10. Refer to atomic masses given in the Table (3.2) of this chapter. Calculate the molar masses of each of the following compounds:

HCl, NH3, CH4, CO and NaCl

Ans: Do yourself.

11. Average atomic mass of carbon is 12.01 u. Find the number of moles of carbon in 

(a) 2.0 g of carbon.

 Ans: 2.0 g of carbon has the number of moles is (2/12.01= 0.167.

(b) 8.0 g of carbon.

Ans: 8.0 g of carbon has the number of moles is (8/12.01)=0.667.

12. Classify the following molecules as di, tri, tetra, penta and hexa atomic molecules: H2, P4, SF4, SO2, PCl3, CH3OH, PCl5, HCl.

Ans: To classify the molecules based on the number of atoms they contain:

(i) H2 (Hydrogen gas) – Diatomic (contains 2 atoms).

(ii) P4 (Phosphorus) – Tetraatomic (contains 4 atoms).

(iii) SF4 (Sulphur tetrafluoride) – Polyatomic (contains 5 atoms).

(iv) SO2 (Sulphur dioxide) – Triatomic (contains 3 atoms).

(v) PCl3 (Phosphorus trichloride) – Tetraatomic (contains 4 atoms).

(vi) CH3OH (Methanol) – Tetraatomic (contains 4 atoms).

(vii) PCl5 (Phosphorus pentachloride) – Polyatomic (contains 5 atoms).

(viii) HCl (Hydrogen chloride) – Diatomic (contains 2 atoms)

So, the classification is:

(a) Diatomic: H2, HCl.

(b) Triatomic: SO2.

(c) Tetraatomic: P4, PCl3, CH3OH.

(d) Polyatomic: SF4, PCl5.

13. What is the mass of.

(a) 6.02×1023 atoms of oxygen.

Ans: Molar mass of oxygen (O) = 16.00 g/mol

The mass of 1 mole of oxygen atoms = 16.00 g

So, the mass of 6.02×1023 atoms of oxygen = (16.00 g/mol) * (6.02×1023 atoms) =96.32 g. 

(b) 6.02×1023 molecules of P4.

Ans: The molar mass of 𝑃4 (phosphorus) is approximately 123.89/g  mol Therefore, the mass of 6.02×1023 molecules of 𝑃4 would be:

6.02 x 10 23   molecules x 1 mol /6.02 10 23      molecules x 123.89g / mol = 123.89 grams.

(c) 3.01×1023 molecules of O2.

Ans: Oxygen gas  (O2) has a molar mass of approximately 32g/mole  therefore , the mass of 3.01 x 10 23  molecules  x 1 mol/6.02 x 1023    molecules 32g/mol

 = 16 grams.  

14. How many atoms are present in:

(a) 0.1 mol of sulphur. 

Ans: 1 mole of sulphur has 6.023×10^23 no of molecules.

So, 0.1 mole has 6.023×10^22.

(b) 18.g of water (H2O).

Ans: 18gm of water means 1 mole of water. So, as we all know in one mole of water has 6.023×10^23.

(c) 0.44 g of carbon dioxide (CO2).

Ans: 44g of CO2 has 6.023×10^23.

So, 0.44g of CO2 has 6.023×10^21 no of molecules.

​15. Write various postulates of Dalton’s atomic theory?

Ans: The English scientist John Dalton was by no means the first person to propose the existence of atoms, as we have seen in the previous section, such ideas date back to classical times. Dalton’s major contribution was to arrange those ideas in proper order and give evidence for the existence of atoms. He showed that the mass relationship expressed by Lavoisier and Proust (in the form of law of conservation of mass and law of constant proportions) could be interpreted most suitably by postulating the existence of atoms of the various elements. In 1803, Dalton published a new system of chemical philosophy in which the following statements comprise the atomic theory of matter:

(i) Matter consists of indivisible atoms.

(ii) All the atoms of a given chemical element are identical in mass and in all other properties. 

(iii) Different chemical elements have different kinds of atoms and in particular such atoms have different masses.

(iv) Atoms are indestructible and retain their identity in chemical reactions.

(v) The formation of a compound from its elements occurs through the combination of atoms of unlike elements in a small whole number ratio.

Dalton’s fourth postulate is clearly related to the law of conservation of mass. Every atom of an element has a definite mass. Also in a chemical reaction there is rearrangement of atoms. Therefore after the reaction, the mass of the product should remain the same. The fifth postulate is an attempt to explain the law of definite proportions. A compound is a type of matter containing the atoms of two or more elements in a small whole number ratio. Because the atoms have definite mass, the compound must have the elements in definite proportions by mass.Dalton’s atomic theory not only explained the laws of conservation of mass and law of constant proportions but also predicted the new ones. He deduced the law of multiple proportions on the basis of his theory. The law states that when two elements form more than one compound, the masses of one element in these compounds for a fixed mass of the other element are in the ratio of small whole numbers. 

For example: carbon and oxygen form two compounds: Carbon monoxide and carbon dioxide. Carbon monoxide contains 1.3321 g of oxygen for each 1.000g of carbon, whereas carbon dioxide contains 2.6642 g of oxygen for 1.0000 g of carbon. In other words, carbon dioxide contains twice the mass of oxygen as is contained in carbon monoxide (2.6642 g = 2 × 1.3321 g) for a given mass of carbon. Atomic theory explains this by saying that carbon dioxide contains twice as many oxygen atoms for a given number of carbon atoms as does carbon monoxide. The deduction of law of multiple proportions from atomic theory was important in convincing chemists of the validity of the theory.

16. Convert into mole: 

(a) 16 g of oxygen gas (O2). 

Ans: First, we find the molar mass of oxygen gas, which is 2×16.00 g/mol=32.00 g/mol 2×16.00g/mol=32.00g/mol since oxygen exists as O2 molecules.

Now, to find the number of moles  

Number of moles = Molar mass (g/mol)/ Mass (g)

Number of moles = 16g / 32.00g/mol

Number of moles=0.5mol

So, 16 g of oxygen gas (O2) is equal to 0.5 moles.

(b) 36 g of water (H2O).

Ans: The molar mass of water (H2O) is 2(1.008 g/mol) +1 (15.999 g/mol) =18.015 g/mol

Number of moles = Mass (g)/Molar mass (g/mol) 

Number of moles = 36g/18.015g/mol

Number of moles = 1.998mol

So, 36 g of water (H2O) is approximately 1.998 moles.

(c) 22 g of carbon dioxide (CO2).

Ans: The molar mass of carbon dioxide (CO2) is 12.011  g/ mol + 2× 15. 999  g/ mol =44.01 g/mol

Number of moles = Mass (g)/Molar mass (g/mol) 

Number of moles = 22g/ 44.01g/mol

Number of moles = 0.4995mol 

So, 22 g of carbon dioxide (CO2) is approximately 0.4995 moles.

17. What does a chemical formula of a compound represent?

Ans: A compound is a substance made up of a definite proportion of two or more elements. A chemical formula tells us the number of atoms of each element in a compound. It contains the symbols of the atoms of the elements present in the compound as well as how many there are for each element in the form of subscripts. 

18. Write chemical formulas of the following compounds:

(a) Copper (II) sulphate.

(b) Calcium fluoride.

(c) Aluminium bromide.

(d) Zinc sulphate. 

(e) Ammonium sulphate. 

Ans: (a) Copper (II) sulphate has the chemical formula CuSO₄.

(b) (b) Calcium fluoride: CaF₂.

(c) Aluminium bromide: AlBr₃.

(d) Zinc sulphate: ZnSO₄.

(e) Ammonium sulphate: (NH₄)₂SO₄.

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