NIOS Class 10 Science & Technology Chapter 3 Solution – Atoms And Molecules
NIOS Class 10 Science & Technology Solution Chapter 3 Atoms And Molecules. NIOS Class 10 Science & Technology Chapter 3 Question Answers Download PDF. NIOS Class 10 Science Notes.
NIOS Class 10 Science & Technology Chapter 3 Solution
| Board | NIOS |
| Class | 10th (Secondary) |
| Subject | Science and Technology (212) |
| Topic | Question Answer, Solution, Notes |
CHAPTER: – 3 (ATOMS AND MOLECULES)
INTEXT QUESTIONS 3.1
1) Name the scientists who proposed the law of conservation of mass and law of constant proportions.
Ans: – The law of conservation of masses and law of constant property is given by the scientist Lavoisier.
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 law of constant proportions.
2Mg + O2 ⎯→ 2MgO
Ans: – As from the explanation it is clear that the oxygen which is reacted is (20-12=8) 8gm. So we can see that 12gm of Mg is reacting with 8gm of O2 and in MgO also the 24 gm magnesium is reacting with 16 gm of oxygen or in the same ratio.
INTEXT QUESTIONS 3.2
1) Nitrogen forms three oxides: NO, NO2 and N2O3. Show that it obeys law of multiple proportions.
Ans: – As we know in a multiple property there are always same elements is reacting with a certain ratio in NO, we can see that nitrogen and oxygen are in ratio of 14:16 or 7:8 and in NO2 this ratio is 14:32 or 7:16, in N2O3 the ratio between the nitrogen and oxygen is 28:48 or 7:12. So here we see that all are bonded in a specific ratio.
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: –
3) Calculate molecular mass of the compounds whose formulas are provided below:C2H4, H2O and CH3OH.
Ans: –
INTEXT QUESTIONS 3.3
1) Work out a relationship between number of molecules and mole.
Ans: – The molecules of any element of 1 mole is present in same no which we came to know from the Avogadro’s principal. And as a reason for this in 1 mole of oxygen or 1 mole of nitrogen always same no of molecules of 6.023×10^23 present.
2) What is molecular mass? In what way it is different from the molar mass?
Ans: – The molecular mass and molar mass are different things as we know the molecular mass denotes that sum of total mass of molecules present in it where as the molar mass is the mass of 1 mole of that element.
3) Consider the reaction18 g of carbon was burnt in oxygen. How many moles of CO2 is produced?
C (s) + O2 (g) ⎯→ CO2 (g).
Ans: – The mass of CO2 is 44 gm so, we can say that in the upper reaction the 12g of carbon or 32 gm of oxygen gave the 1 mole of CO2 or 44gm of carbon dioxide. So we clearly see that 18 gm of carbon will gave 1.5 mole of CO2.
4) What is the molar mass of NaCl ?
Ans: – Molar mass of NaCl is = (molar mass of Na and Cl)
=(23 + 35.5)
=58.5.
INTEXT QUESTIONS 3.4
1) Write the name of the expected compound formed between
(i) hydrogen and sulphur
(ii) nitrogen and hydrogen
(iii) magnesium and oxygen.
Ans: – (i) The compound will from after the reaction between hydrogen and sulphur is hydrogen sulphide.
(ii) Ammonia will be produced after the reaction between the nitrogen and hydrogen.
(iii) MgO or magnesium oxide will be the product of the reactant of magnesium and oxygen.
2) Propose the formulas and names of the compounds formed between
(i) potassium and iodide ions
(ii) sodium and sulphate ions
(iii) aluminium and chloride ions.
Ans: – (i) I2 + 2K —> 2KI.
(ii) 2Na + SO4- —> Na2SO4.
(iii) 3Cl + Al —> AlCl3.
3)Write the formula of the compounds formed between
(i) Hg2+ and Cl–
(ii) Pb2+ and PO4(3-)
(iii) Ba2+ and SO4(2-)
Ans: –(i) Hg2+ + Cl- —> HgCl2.
(ii) Pb2+. + PO4(3-) —> Pb3(SO4)2.
(iii) Ba2+ + SO4(2-) —> BaSO4.
TERMINAL EXERCISE
1)Describe the following:
(a) Law of conservation of mass
(b) Law of constant proportions
(c) Law of multiple proportions.
Ans: – (i) Law of conservation of mass —> According to the law of conservation of mass it say that the mass of the reactant and the product will be same after the reaction.
(ii) In laws of constant property, the compound is bonded or formed by a certain ratio like in the water the oxygen and hydrogen are mixed with the ratio of 1:2.
(iii) In the multiple proportion laws States that the same element from different compound by mixing with the same elements but in different proportion like in NO or N2O or N2O3 all have the same oxygen and nitrogen but in different ratio of their amount.
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 Dalton atomic theory is the first to explain the formation of the different atom present in the universe where he says that the atom is consist of the positive and negative ion inside it. But after the various study for decades, it comes that some of the principals are not same which he states like the atomic structure are some different.
3) Write the number of protons, neutrons and electron in each of the following isotopes:
2H1 , 18O8, 19F9 , 40Ca20.
Ans: – In hydrogen proton is 1, neutron is 1 and electron is 1.
In this oxygen proton is 8, neutron is 10 and electron is 8.
In fullerene proton and electron is 9 and neutron is 10.
In this calcium proton and electron is 20 and neutron is 20.
4) Boron has two isotopes with masses 10.13 u and 11.01 u and abundance of19.77% and 80.23% respectively. What is the average atomic mass of boron?
Ans: – The average atomic mass of boron is = (mass of A × percentage + mass of B × percentage)
= (10.13×19.77/100 + 11.01×80.23/100)
=10.81u.
5) Give symbol for each of the following isotopes:
(a) Atomic number 19, mass number 40
(b) Atomic number 7, mass number 15
(c) Atomic number 18, mass number 40
(d) Atomic number 17, mass number 37.
Ans: –
6) How does an element differ from a compound? Explain with suitable examples.
Ans: – An element is different from the compound like oxygen is an element but the carbon dioxide is a compound. So, the elements are consisted of same no to atoms but the compound are consisted of different no of element.
7) Charge of one electron is 1.6022×10–19 coulomb. What is the total charge on1 mole of electrons?
Ans: –The charge on 1 mole of electron will ne same as the charge of one electron of 1.6022×10^-19.
8) How many molecules of O2 are in 8.0 g of oxygen? If the O2 molecules werecompletely split into O (Oxygen atoms), how many moles of atoms of oxygenwould be obtained?
Ans: – As we all know 16 gm of O2 is contain of 1 mole of total value which have 6.023×10^23 no of molecules. So, in 8gm of O2 there are 3.011×10^23 no of molecules are present.
9) Assume that human body is 80% water. Calculate the number of molecules ofwater 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 of 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 themolar masses of each of the following compounds :
HCl, NH3, CH4, CO and NaCl.
Ans: –
11) Average atomic mass of carbon is 12.01 u. Find the number of moles of carbon in (a) 2.0 g of carbon. (b) 8.0 g of carbon.
Ans: – (a) 2.0 g of carbon has the number of moles is (2/12.01=0.167.
(b)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: – Here di or diatomic molecules is H2, HCl; tri or triatomic is SO2, tetratomic is PCl3 and P4, and penta is SF4 and hexa atomic is CH3OH, PCl5.
13) What is the mass of
(a) 6.02×1023 atoms of oxygen.
Ans: – The masses of oxygen are 16gm as 1 mole oxygen has this no of atoms
(b) 6.02×1023 molecules of P4.
Ans: – 31g phosphorus has this no of molecules so the total weight of P4 is 31×4=124.
(c) 3.01×1023 molecules of O2.
Ans: – 1 mole O2 has 16g of oxygen so this no of molecules has 0.5 mole or 8gm of molecules.
14) How many atoms are present in:
(a) 0.1 mole 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 postulate of Dalton’s atomic theory is,
- The atoms are formed by the various indivisible particles.
- Atoms are neither be created or destroyed.
- The property of different atoms are same.
16) Convert into mole:
(a) 16 g of oxygen gas (O2)
(b) 36 g of water (H2O)
(c) 22 g of carbon dioxide (CO2).
Ans: – (a) 16gm of oxygen gas 1 mole of O2.
(b) 18gm of water has 1 mole.
So, 36gm of water has 2 moles.
(c)44g of carbon dioxide has 1 mole.
So, 22g of carbon has o.5 mole of carbon dioxide.
17) What does a chemical formula of a compound represents?
Ans: – The chemical formula of any compound represents the different no of element that present in that. This formula tells us about the various characteristics of that compound. The different property of the different particles presents in that we came to know.
18) Write chemical formulas of the following compounds:
(a) Copper (II) sulphate.
Ans: – CuSO4.
(b) Calcium fluoride.
Ans: – CaF2.
(c) Aluminium bromide.
Ans: – AlBr3.
(d) Zinc sulphate.
Ans: – ZnSO4.
(e) Ammonium sulphate.
Ans: – (NH4)2SO4.
