1. Which of the following correctly represents 360 g of water?

(i) 2 moles of H2O

(ii) 20 moles of water

(iii) 6.022 × 1023 molecules of water

(iv) 1.2044 × 1025 molecules of water

Options:

(a) (i)

(b) (i) and (iv)

(c) (ii) and (iii)

(d) (ii) and (iv)

Ans: The two moles of water molecules (H2O) is 36 grams. Therefore the 20 moles of water molecule (H2O) will be 360 grams. Now the mass of 6.022 × 1023 molecules of water molecule is 18g. Also the mass of 1.2044 × 1025 of water molecules is 20 x 18g = 360g. 

Therefore the options representing 360 grams of water is option (ii) and option (iv). Hence the correct answer is option d.

2. Which of the following statements is not true about an atom?

(a) Atoms are not able to exist independently.

(b) Atoms are the basic units from which molecules and ions are formed.

(c) Atoms are always neutral in nature.

(d) Atoms aggregate in large numbers to form the matter that we can see, feel or touch.

Ans: Only the atoms of inert gasses can exist independently. The atoms are the basic units from which the molecules and ions are formed. The atoms are always neutral in nature. The atoms aggregate in a large number to form the matter that we can see, feel or touch. Thus rest all the given statements about an atom are true. Hence the correct answer is option a.

3. The chemical symbol for nitrogen gas is:

(a) N

(b) N2

(c) N+

(d) N

Ans: The chemical symbol for nitrogen gas is N2 since the molecules of the nitrogen gas are diatomic in nature. Hence correct answer is option b.

4. The chemical symbol for sodium is:

(a) So

(b) Sd

(c) NA

(d) Na

Ans: The chemical symbol for sodium is Na as the name "sodium" has been derived from the Latin word Natrium. Hence the correct answer is option d.

5. Which of the following would weigh the highest?

(a) 0.2 mole of sucrose (C12H22O11)

(b) 2 moles of CO2

(c) 2 moles of CaCO3

(d) 10 moles of H2O

Ans: 

 (a) Chemical formula of Sucrose is C12H22O11.

Molar Mass of Sucrose = (12 x 12) + (22 x 1) + (11 x 16) 

= 144 + 22 + 176 

= 342 g

Or 0.2 M Sucrose= 0.2 x 342

0.2 M Sucrose =68.4 g

(b) 1 M carbon dioxide = (1 x 12) + (2 x 16) 

= 44 g

Or 2 M CO2 = 44 x 2

                           2 M CO2= 88 g

(c)                 1 M CaCO3 = (1 x 40) + (1 x 12) + (3 x 16) 

= 40 + 12 + 48 

= 100 g

                Or 2 M CaCO3 = 100 x 2

                                         = 200 g

(d)                   1 M H2O = (2 x 1) + (1 x 16) 

                                        = 18 g

             Or 10 M water = 18 x 10

                                        =180 g

 Therefore the 2 moles of CaCO3 weigh the highest. Hence the correct answer is option c. 

6. Which of the following has maximum number of atoms?

(a) 18 g of H2O

(b) 18 g of O2

(c) 18 g of CO2

(d) 18 g of CH4

Ans: 

The number of atoms 

(a) For 18 g of H2O = 18g x 3 x NA = 3NA

      18 g

(b) For 18 g of O2 = 18g x 2 x NA = 1.125NA

      32 g

(c) For 18 g of CO2 = 18g x 3 x NA = 1.23NA

      44 g

(d) For 18 g of CH4 = 18g x 5 x NA = 5.625NA

      16 g

Therefore 18 grams of CH4 has a maximum number of atoms. Hence the correct answer is option d.

7. Which of the following contains the maximum number of molecules?

(a) 1 g CO2

(b) 1 g N2

(c) 1 g H2

(d) 1 g CH4

Ans:

Number of molecules = Mass of substance x NA

Molar mass

For 1 g of Hydrogen = [1g / 2g] x NA

= 0.5 g NA 

= 0.5 x 6.022 x 1023

=3.011 x 1023

Therefore the molar mass of other molecules is very much higher than that of given mass thus the number of molecules in them will be less than that in hydrogen.

Hence the correct answer is option c.

8. Mass of one atom of oxygen is:

(a)         16       g    

    6.023 x 1023   

(b)         32      g    

    6.023 x 1023   

(c)         1        g    

    6.023 x 1023   

(d) 8u

Ans: Mass of one atom of oxygen = Atomic Mass 

 NA

=         16          g    

    6.023 x 1023   

Hence the correct answer is option a.

9. 3.42 g of sucrose are dissolved in 18 g of water in a beaker. The number of oxygen atoms in the solution is:

(a) 6.68 x 10 23

(b) 6.09 x 1022

(c) 6.022 x 1023

(d) 6.022 x 1021

Ans: 

 The number of moles = (Mass of substance) 

(Molar mass)

The molar mass of sucrose = 342 g

No. of moles for 3.42 g sucrose = 3.42g

342g

= 0.01 M

1 M sucrose (C11H22O11) contains 11 x NA atoms of oxygen.

Therefore the number of oxygen atoms in 0.01 M sucrose= 0.11 NA

No. of moles for 18 g water= 18g = 1M

18g

1 M water contains 1 NA oxygen atoms

Therefore total number of oxygen atoms in given solution,

= 0.11 NA + 1.0 NA

= 1.11NA

=1.11 x 6.022 x 1023

=6.68 x 1023

Hence the correct answer is option a.

10. A change in the physical state can be brought about:

(a) Only when energy is given to the system.

(b) Only when energy is taken out from the system.

(c) When energy is either given to, or taken out from the system.

(d) Without any energy change.

Ans:  A change in the physical state can be brought when the energy is either given to or taken out from the system. When a solid changes into liquid, it absorbs the energy. When a liquid changes into solid, it releases the energy. Hence the correct answer is option c.

11. Which of the following represent the correct chemical formula? Name it.

(a) CaCl

(b) BiPO4

(c) NaSO4

(d) NaS

Ans: In the option (b), Bismuth (Bi) is trivalent and therefore this depicts the correct chemical formula. In the option (a), Ca is bivalent while the chlorine is monovalent thus the correct chemical formula should be as CaCl2. In the options (c), Na is monovalent while the sulfate radical is bivalent thus the correct chemical formula should be as Na2SO4. In the options (d), Na is monovalent while the sulfur is bivalent thus the correct chemical formula should be as Na2S. Therefore the BiPO4 represents the correct chemical formula. Hence the correct answer is option b.

12. Write the molecular formulae for the following compounds:

(a) Copper (II) bromide

Ans: The valency of copper (metal) is +2. The valency of bromine (non metal) is -1.

Therefore the molecular formula of Copper (II) bromide is CuBr2.

 (b) Aluminum (III) nitrate

Ans: The valency of aluminum is +3 and the valency of nitrate radical is -1.

Therefore the molecular formula of Aluminum (III) nitrate is Al (NO3)3.

(c) Calcium (II) phosphate

Ans: The valency of calcium is +2 and the valency of phosphate radical is -3.

Therefore the molecular formula of Calcium (II) phosphate is Ca3 (PO4)2.

(d) Iron (III) sulfide

Ans: The valency of iron is +3 and the valency of sulfur is -2.

Therefore the molecular formula of Iron (III) sulfide is Fe2S3.

(e) Mercury (II) chloride

Ans: The valency of mercury is +2 and the valency of chlorine is -1.

Therefore the molecular formula of Mercury (II) chloride is HgCl2.

(f) Magnesium (II) acetate

Ans: The valency of magnesium is +2 and the valency of acetate radical is -1.

Therefore the molecular formula of Magnesium (II) acetate is Mg (CH3COO) 2.

13. Write the molecular formulae of all the compounds that can be formed by the combination of following ions:

Ca2+, Na+, Fe3+, Cl- , SO42- , PO43-

Ans: Following is the molecular formulae of all the compounds that can be formed by the combination of given ions:

• Fe2(SO4)3

• FeCl3

• Na2SO4

• CuCl2

• Cu3(PO4)2

• Na3PO4

• FePO4

• CuSO4 

• NaCl

14. Write the cations and anions present (if any) in the following compounds:

(a) CH3COONa

(b) NaCl

(c) H2

(d) NH4NO3

Ans:

15. Give the formulae of the compounds formed from the following sets of elements:

(a) Calcium and fluorine

Ans: The valency of calcium is +2 and the valency of fluorine is -1.

The formula of compounds formed from Calcium and fluorine is CaF2.

(b) Hydrogen and sulfur

Ans: The valency of hydrogen is +1 and the valency of sulfur is -2.

The formula of compounds formed from Hydrogen and sulfur is H2S.

(c) Nitrogen and hydrogen

Ans: The valency of nitrogen is -3. The valency of hydrogen is +1.

The formula of compounds formed from Nitrogen and hydrogen is NH3.

(d) Carbon and chlorine

Ans: The valency of carbon is +4 and the valency of chlorine is -1.

The formula of compound formed from Carbon and chlorine is CCI4.

(e) Sodium and oxygen

Ans: The valency of sodium is +1 and the valency of oxygen is -2.

The compound formed from Sodium and oxygen is Na2O.

(f) Carbon and oxygen

Ans: The valency of carbon is +4 and the valency of oxygen is -2.

The formula of compounds formed from Carbon and oxygen are CO and CO2.

16. Which of the following symbols of elements are incorrect? Give their correct symbols.

(a) Cobalt-CO

Ans: The given symbol for cobalt is incorrect; the correct symbol of cobalt is Co. 

(b) Carbon-c

Ans: The given symbol of Carbon is incorrect, the correct symbol of carbon is C. 

(c) Aluminum-AL

Ans: The given symbol of aluminum is incorrect; the correct symbol of aluminum is Al.

(d) Helium-He

Ans: The given symbol for helium is correct. The correct symbol for Helium is He. 

(e) Sodium-So

Ans: The given symbol of sodium is incorrect; the correct symbol of sodium is Na. 

17. Give the chemical formula for the following compounds and compute the ratio by mass of the combining elements in each one of them.

(a) Ammonia

Ans: The chemical formula of ammonia is NH3.

The mass of N is 14 and the mass of H is 3.Therefore their mass ration will be 14:3.

(b) Carbon monoxide

Ans: The chemical formula of carbon monoxide is CO.

The mass of C is 3 and the mass is 4. Therefore their mass ration will be 3:4.

(c) Hydrogen chloride

Ans:  The chemical formula of hydrogen chloride is HCl.

The mass of H is 1 and the mass of Cl is 35.5. Therefore their mass ration will be 1:35.5.

(d) Aluminum fluoride

Ans: The chemical formula of aluminum fluoride is AlF3.

The mass of Al is 9 and the mass of F is 19. Therefore their mass ration will be 9: 19.

(e) Magnesium sulfide

Ans: The chemical formula of magnesium sulfide is MgS.

The mass of Mg is 3 and the mass of S is 4. Therefore their mass ration will be 3: 4.

18. State the number of atoms present in each of the following chemical species:

(a) CO32-

Ans: In the given compound there is 1 atom of carbon and 3 atoms of oxygen. Hence total 4 numbers of atoms present.

(b) PO43-

Ans: In the given compound there is 1 atom of phosphorus and 4 atoms of oxygen. Hence total 5 numbers of atoms present.

(c) P2O5

Ans: In the given compound there are 2 atoms of phosphorus and 5 atoms of oxygen. Hence total 7 numbers of atoms present.

(d) CO

Ans: In the given compound there is 1 atom of carbon and 1 atom of oxygen. Hence total 2 number of atoms present.

19. What is the fraction of the mass of water due to neutrons?

Ans: By Avogadro number the mass of one mole of neutrons is nearly 1 g.

Therefore the mass of one neutron= 1 g

NA

The mass of one molecule of water= Molar mass

NA 

= 18 g

NA

There are total 8 neutrons present in one atom of oxygen.

Therefore the mass of 8 neutrons= 8 g

NA

Hence the fraction of the mass of water due to neutrons is nearly equal to 8 g.

18

20. Does the solubility of a substance change with temperature? Explain with the help of an example.

Ans: Yes the solubility of a substance depends on the temperature. Generally the solubility increases with the increase in temperature. For example: The solubility of sugar is less in cold water as that of in hot water. 

21. Classify each of the following on the basis of their atomicity:

(a) F2

Ans: The atomicity of F2 is 2 atoms.

(b) NO2

Ans: The atomicity of NO2 is 3 atoms.

(c) N2O

Ans: The atomicity of N2O is 3 atoms.

(d) C2H6

Ans: The atomicity of C2H6 is 8 atoms.

(e) P4

Ans: The atomicity of P4 is 4 atoms.

(f) H2O2

Ans: The atomicity of H2O2 is 4 atoms.

(g) P4O10

Ans: The atomicity of P4O10 is 14 atoms.

(h) O3

Ans: The atomicity of O3 is 3 atoms.

(i) HCl

Ans: The atomicity of HCl is 2 atoms.

(j) CH4

Ans: The atomicity of CH4 is 5 atoms.

(k) He

Ans: The atomicity of He is 1 atom.

(l) Ag

Ans: The polyatomic. 

22. You are provided with a fine white colored powder which is either sugar or salt. How would you identify it without tasting?

Ans: If the provided powder solution is conducting electricity then it is a salt solution and the white colored powder is of a salt. And if the given powder solution is not conducting the electricity then it is of a sugar solution and the white colored powder is a sugar. Also if the given powder is charged when heated then it will be of sugar. This is how we can check whether the solution is of salt or sugar without tasting.

23. Calculate the number of moles of magnesium present in a magnesium ribbon weighing 12 g. Molar atomic mass of magnesium is 24 g mol–1.

Ans: We know that,

The number of moles = Mass of magnesium ribbon 

Molar mass 

= 12 g 

24 g 

= 0.5 mol

Therefore the number of moles of the magnesium present in a magnesium ribbon is 0.5 moles.

24. Verify by calculating that:

(a) 5 moles of CO2 and 5 moles of H2O do not have the same mass.

Ans: We know that,

The molar mass of CO2 = 44 g mol–1

Therefore the molar mass of 5 moles of CO2 = 5 × 44 g 

= 220 g

Also the molar mass of H2O = 18 g mol–1

Therefore the molar mass of 5 moles of H2O = 5 × 18 g 

= 90 g

By the above solution we conclude that the 5 moles of CO2 have a mass of 220 g and 5 moles of H2O have a mass of 90 g

Therefore they do not have the similar mass.

(b) 240 g of calcium and 240 g magnesium elements have a mole ratio of 3 : 5.

Ans: We have,

The number of moles in 240 g Ca metal=240

40

=6

Similarly the number of moles in 240 g of Mg metal=240

24

=10

Therefore the mole ratio = 6: 10 

= 3: 5

Therefore the mole ration is 3: 5.

25. Find the ratio by mass of the combining elements in the following compounds:

(a) CaCO3

Ans: The mass of Ca=40

The mass of C=12

Also the mass of O3=16x3=48

Therefore their ratio by mass will be 40: 12: 48, i.e. 10:3:12.

(b) MgCl2

Ans: The mass of Mg=24

And the mass of Cl2=35.5x2=71

Therefore their ratio by mass will be 24: 71.

(c) H2SO4

Ans: The mass of H2=1x2=2

The mass of S =32

Also the mass of O4=16x4=64

Therefore their ratio by mass will be 2: 32: 64 i.e. 1: 16: 32.

 (d) C2H5OH

Ans: The mass of C2=12x2=24

The mass of H6=1x6=6

Also the mass of O=16

Therefore their ratio by mass will be 24: 6: 16 i.e. 12: 3: 8.

(e) NH3

Ans: The mass of N=14

And the mass of H3=1x3=3

Therefore their ratio by mass will be 14: 3.

(f) Ca (OH) 2

Ans: The mass of Ca =40

The mass of O2=16x2=32

Also the mass of H2=1x2=2

Therefore their ratio by mass will be 40: 32: 2 i.e. 20: 16: 1.

26. Calcium chloride when dissolved in water dissociates into its ions according to the following equation: 

CaCl 2(aq) Cl2+ (aq) + 2Cl- (aq)

Calculate the number of ions obtained from CaCl2 when 222 g of it is dissolved in water.

Ans: We know that,

1 mole of calcium chloride = 111 g

Therefore 2 moles of CaCl2 is equivalent to 222 g of CaCl2.

We get 3 ions from 1 formula unit of CaCl2 thus 1 mol of CaCl2 will give 3 moles of ions.

Also by this 2 moles of CaCl2 will give 6 moles of ions.

Hence the no. of ions = no. of moles of ions × Avogadro number

= 6 × 6.022 × 1023 

= 36.132 × 1023 

= 3.6132 × 1024 ions

Hence the number of ions obtained from CaCl2 when 222 g of it is dissolved in water is 3.6132 × 1024 ions.

27. The difference in the mass of 100 moles each of sodium atoms and sodium ions is

5.48002 g. Compute the mass of an electron.

Ans: We have the difference between sodium atom and ion as one electron. 

Therefore for the 100 moles each of sodium atoms and ions there will be a difference of 100 moles of electrons.

Therefore the mass of 100 moles of electrons = 5.48002 g

Therefore the mass of 1 mole of electron=5.48002 g

100

Also the mass of one electron= (5.48002)         g

100x6.022x1023

=9.1x10-28 g

=9.1x10-31 Kg

Therefore the mass of an electron is 9.1x10-31 Kg.

28. Cinnabar (HgS) is a prominent ore of mercury. How many grams of mercury are present in 225 g of pure HgS? Molar mass of Hg and S are 200.6 g mol-1 and 32 g mol- 1 respectively.

Ans: We have,

The molar mass of HgS = 200.6 + 32 

= 232.6 g mol-1

Also the mass of Hg in 232.6 g of HgS = 200.6 g

Therefore the mass of Hg in 225 g of HgS=200.6 x 225

232.6

=194.04 g

Hence 194.04 grams of mercury are present in 225 g of pure HgS.

29. The mass of one steel screw is 4.11 g. Find the mass of one mole of these steel screws.

Compare this value with the mass of the Earth (5.98 x 1024 kg). Which one of the two is heavier and by how many times?

Ans: We have,

The one mole of screws weigh = 4.11 × (6.022 × 1023) 

= 2.475 × 1024 g 

= 2.475 × 1021 kg

Therefore, (The mass of earth) =5.98x 1024 Kg

One mole of screw     2.475 × 1021

=2.4x 103

Hence the mass of the Earth is 2.4x 103 times to that of the mass of one mole of steel screws.

Therefore the earth is 2400 times heavier than that of one mole of steel screws.

30. A sample of vitamin C is known to contain oxygen atoms. How many moles of oxygen atoms are present in the sample?

Ans: We have,

The 1 mole of oxygen atoms = 6.022 × 1023 atoms

Therefore the number of moles of oxygen atoms= (2.58x1024)

 6.022 × 1023

=4.28 moles

Therefore 4.28 moles of oxygen atoms are present in the given sample of vitamin C.

31. Raunak took 5 moles of carbon atoms in a container and Krish also took 5 moles of sodium atoms in another container of the same weight.

(a) Whose container is heavier?

Ans: The mass of sodium atoms carried by Krish will be (5 x 23) g = 115g

Also the mass of carbon atom carried by Raunak will be (5 x 12) g = 60g

Hence the container of Krish is heavier than that of the Raunak's container.

(b) Whose container has more atoms?

Ans: Since they have the same number of moles of atoms therefore the both containers have the same number of atoms.

32. Fill in the missing data in Table given below:

Ans: The number of particles of H2O is 2 × 6.022 × 1023 = 1.2044 × 1024

Also the no. of moles of CO2 = (3.011 × 1023) 

 (6.022 × 1023) 

= 0.5 mol

The molar mass of CO2 = the atomic mass of C + (2 × atomic mass O) 

= 12 + (2 × 16) 

= 44 g

The mass of 0.5 moles of CO2 = 0.5 × 44 g 

= 22 g

The number of moles of sodium (Na) atom = 115 

23

 = 5 mol

The number of particles of sodium (Na) atom = 5 × 6.022 × 1023 

= 3.011 × 1024

The molar mass of MgCl2 = atomic mass of Mg + (2 × atomic mass of Cl) 

= 24 + (2 × 35.5) 

= 95 g

The mass of 0.5 mol of MgCl2 = 0.5 × 95 g 

= 47.5 g

The number of particles of MgCl2 = 0.5 × 6.022 × 1023 

= 3.011 × 1023

 Therefore the complete table is as follows:

33. The visible universe is estimated to contain 1022 stars. How many moles of stars are present in the visible Universe?

Ans: The number of moles of stars in the visible Universe will be calculated as:

= 10/ 6.023x1023

= 0.0166 moles

Therefore the number of moles of stars present in the visible Universe is 0.0166 moles.

34. What is the SI prefix for each of the following multiples and submultiples of a unit?

(a) 103

Ans: The SI prefix for the above unit is kilo.

 (b) 10-1 

Ans: The SI prefix for the above unit is deci.

(c) 10-2 

Ans: The SI prefix for the above unit is centi.

(d) 10-6

Ans: The SI prefix for the above unit is micro.

(e) 10-9 

Ans: The SI prefix for the above unit is nano.

(f) 10-12

Ans: The SI prefix for the above unit is pico.

35. Express each of the following in kilograms:

(a) 5.84 × 10-3 mg

Ans: 5.84 × 10-3 mg = 5.84x10-9 Kg

(b) 58.34 g

Ans: 58.34 g= 5.834x10-2 Kg

(c) 0.584 g

Ans: 0.584 g=5.84x10-4 Kg

(d) 5.873 × 10-21 g

Ans: 5.873 × 10-21 g = 5.837x10-24

36. Compute the difference in masses of 103 moles each of magnesium atoms and magnesium ions. (Mass of an electron = 9.1 × 10-31 kg)

Ans: There is a difference of two electrons between the Mg2+ ion and Mg atom.

Therefore there will be 2x103 moles of electrons difference between 103 moles of Mg2+ and Mg atoms.

Hence the mass of 2 ×103 moles of electrons = (2x103x6.023x1023x9.1x10-31) Kg

= 2x6.022x9.1x10-3 Kg

=1.096x10-3 Kg

Hence the difference in masses of 103 moles each of magnesium atoms and magnesium ions is 1.096x10-3 Kg.

37. Which has more number of atoms?

100 g of N2, 100 g of NH3

Ans: The 100 g of N2 = 100 moles

28

Therefore the number of atoms of N2 = 2 x 100 x 0.622 x 1023

28

=43.01 x 1023

The 100 g of NH3 = 100 moles

                                   17

Therefore the number of atoms of NH3 = 4 x 100 x 6.022 x 1023

17

=141.69 x 1023

Hence NH3 has more number of atoms.

38. Compute the number of ions present in 5.85 g of sodium chloride.

Ans: The 5.85 g of NaCl= (5.58/5.58) =0.1 moles

Or we can say 0.1 moles of NaCl particle

Therefore each NaCl particle is equivalent to one Na+ and one CI-

For 2 ions

Therefore the total moles of ions = 0.1 × 2

For 2 moles

No. of ions = 0.2x6.022x 1023 = 1.2042x1023 ions

Therefore the number of ions present in 5.85 g of sodium chloride is 1.2042x1023.

39. A gold sample contains 90% of gold and the rest copper. How many atoms of gold are present in one gram of this sample of gold?

Ans: we know that the one gram of gold sample will contain 0.9 g of gold.

Therefore the number of moles of gold=   (mass of the gold)          .

Atomic mass of the gold

=0.9

  197

=0.0046

Therefore one moles of gold contains NA atoms = 6.022 x 1023

Hence 0.0046 mole of gold will contain = 0.0046x 6.022 1023

=2.77x1021

Therefore the atoms of gold present in one gram of a given sample of gold is 2.77x1021.

40. What are ionic and molecular compounds? Give examples.

Ans: 

• The compounds that contain charged species of metals as well as nonmetals are called the ionic compounds. The charged species present are called the ions. An ion is a charged particle and can be charged negatively or positively. A negatively charged ion is called an anion and the positively charged ion is called cation. The ionic compounds are formed when the ionic bonds are formed between the different elements through the transfer of electrons. The examples of ionic compounds are sodium chloride, calcium oxide etc. 

• The molecular compounds or the covalent compounds are those compounds in which the elements share electrons through the covalent bonds. The examples are water, ammonia, and carbon dioxide etc.

41. Compute the difference in masses of one mole each of aluminium atoms and one mole of its ions. (Mass of an electron is 9.1 x 10-28 g). Which one is heavier?

Ans: The mass of 1 mole of aluminum atom is equal to the molar mass of aluminum which is equal to 27 g mol-1.

The aluminum atom needs to lose their three electrons to become an ion AI3+

For the one mole of AI3+ ion there will be lose of three moles of electrons.

Therefore the mass of the three moles of electrons= 3 x (9.1x10-28) x 6.022x1023 g

= 64.400x10-5 g

=0.00164 g

Therefore the molar mass of Al3+ = (27-0.00164) gmol-1

= 26.998  

Hence the difference in mass will be as 27 - 26.9964 =0.0016 g

42. A silver ornament of mass ‘m’ gram is polished with gold equivalent to 1% of the mass of silver. Compute the ratio of the number of atoms of gold and silver in the ornament.

Ans: We have the mass of silver = m g

Also the mass of gold= m/100 g

The number of atoms of silver=mass x N A

Atomic mass

= m /108 x NA

Therefore the number of atoms of gold= m/ (100x197)x NA

Therefore the ratio of number of atoms of gold to that of silver is

= Number of atoms of Au: Number of atoms of Ag

= m/(100x197)x NA :  m /108 x NA

108: 100x 197=1:182.41

43. A sample of ethane (C2H6) gas has the same mass as 1.5x1020 molecules of methane

(CH4). How many C2H6 molecules does the sample of gas contain?

Ans: Mass of 1 molecule of methane (CH4)=16g/NA

The mass of 1.5x 1020 molecules of methane=1.5x1020 x 16/NA

Mass of C2H6 molecules=1.5x1020 x 16/NA

But mass of 1 molecule of C2H6= 30/NA

Therefore the number of molecules of ethane in the sample of gas

= Mass of C2H6 molecules in the sample / Mass of 1 molecule of C2H6

=0.8x1020

44. Fill in the blanks:-

(a) In a chemical reaction, the sum of the masses of the reactants and products remains unchanged. This is called ______.

Ans: In a chemical reaction, the sum of the masses of the reactants and products remains unchanged. This is called law of conservation of mass.

(b) A group of atoms carrying a fixed charge on them is called _______.

Ans: A group of atoms carrying a fixed charge on them is called polyatomic ion.

(c) The formula unit mass of Ca3 (PO4)2 is _______.

Ans: The formula unit mass of Ca3 (PO4)2 is (3 × atomic mass of Ca) + (2 × atomic mass of phosphorus) + (8 × atomic mass of oxygen) = 310

(d) Formula of sodium carbonate is ______ and that of ammonium sulphate is _______.

Ans: Formula of sodium carbonate is Na2CO3 and that of ammonium sulphate is (NH4)2SO4.

45. Complete the following crossword puzzle (Fig. 3.1) by using the name of the chemical elements. Use the data given in Table 3.2.

Ans:

1. Silver (2) Gold (3) iron (4) Copper (5) Phosphorus (6) Mercury (7) Hydrogen (8) Lead

46. (a) In this crossword puzzle (Fig 3.2), names of 11 elements are hidden symbols of

these are given below. Complete the puzzle. (1.) Cl (2.) H (3.) Ar (4.) O (5.) Xe (6.)  (7.) 

He (8.) F (9.) Kr (10.) Rn (11.) Ne

Ans.

(b) Identify the total number of inert gasses, their names and symbols from this cross word puzzle.

Ans: Insert Gasses: There are six insert gasses in this crossword which are Helium (He), Argon (Ar), Xenon (Xe), Krypton (Kr), Radon (Rn) and Neon (Ne).

47. Write the formula for the following and calculate the molecular mass for each one of them. (a) Caustic potash (b) Baking powder (c) Limestone (d) Caustic soda (e) Ethanol (f) Common salt

Ans:

The formulae for the given and calculate the molecular mass for each one of them as is follows:

48. In photosynthesis, 6 molecules of carbon dioxide combine with an equal number of water molecules through a complex series of reactions to give a molecule of glucose having a molecular formula C6H12O6. How many grams of water would be required to produce 18 g of glucose? Compute the volume of water so consumed assuming the density of water to be 1 g cm-3.

Ans: The equation for photosynthesis is as follows:

6CO2 + 6 H2 O (in the presence of Chlorophyll /Sunlight) → C6 H12 O6 + 6O2

Therefore 1 mole of glucose needs 6 moles of water and the 180 g of glucose needs (6×18) g of water and 1 g of glucose will need 108/ 180 g of water.

Thus 18 g of glucose will need (108 /180) × 18 g of water i.e. 10.8 g

Now the volume of water used = Mass / Density

= 10.8 g/ 1g cm-3

= 10.8 cm3

Hence the volume of water is 10.8 cm3.

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