Class 9 Science Exemplar Solution Chapter-3 Atoms and Molecules

EXEMPLAR
SOLUTIONS
SOLUTIONS
SOLUTIONS
SCIENCE
For Revised Syllabus Session 2024-25
Artham
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Chapter :Atoms and Molecules

Chapter 3
toms
A and Molecules
Multiple Choice Questions
1.Which of the following correctly represents 360 g of water?
(i) 2 moles of H20
(ii) 20 moles of water
(iii) 6.022 × 1023 molecules of water
(iv) 1.2044×1025 molecules of water
(a) (i)
(b) (i) and (iv)
(c) (ii) and (iii)
(d) (ii) and (iv)
Soln:
Answer is (d) (ii) and (iv)
Explanation:
Number of moles = Mass of water
Molar mass of water
Number of moles = 360g
12g/mol
Number of moles = 20
Number of molecules = 20 x 6.022 x 1023
= 1.2044 x 1025
molecules of water
Thus, option (d) is correct.
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
Soln:
Answer is d) Atoms aggregate in large numbers to form the matter that we can see, feel or touch
Explanation:
Atoms aggregate in large numbers to form the matter But we cannot see the matter with our naked eyes.
3. The chemical symbol for nitrogen gas is
(a) Ni
(b) N2
(c) N+
(d) N

Soln:
Answer is (b) N2
Explanation:
Chemical formula of Nitrogen is N but Nitrogen exist in molecule of two ions hence chemical symbol of Nitrogen
is written as N2.
4. The chemical symbol for sodium is
(a) So
(b) Sd
(c) NA
(d) Na
Soln:
Answer is (d) Na
Explanation:
Sodium word is derived from Latin word Natrium hence the chemical name of sodium is Na.
5. Which of the following would weigh the highest?
(a) 0.2 mole of sucrose (C12 H22 O11)
(b) 2 moles of CO2
(c) 2 moles of CaCO3
(d) 10 moles of H2O
Soln:
Answer is (c) 2 moles of CaCO3
Explanation:
Weight of a sample in grant = Number of moles x Molar mass
(a) 0.2 moles of Ci2H220„ = 0.2 x 342 = 68.4 g
(b) 2 moles of CO is 2 x 44 is. 88 g
(c) 2 moles of CaCO? 2 x IIMI - 200 g (4) 10 moles of I-120 = 10 x 18 - 1St g
Hence, option (c) is correct.
6. Which of the following has maximum number of atoms?
(a) 18g of H2O
(b) 18g of O2
(c) 18g of CO2
(d) 18g of CH4
Soln:
Answer is (d) 18g of CH4

Explanation:
Number of atoms = Mass of
substance × Number of atoms in the molecule/ Molar mass × NA
(a) 18 g of water =18 x3/18 ×NA = 3 NA
(b) 18 g of oxygen = 18 x2 /32 × NA = 1.12 NA
(c) 18 g of CO2 = 18 x3/44 × NA = 1.23 NA
(d) 18 g of CH4 =18 x5 /16 × NA = 5.63 NA
Note: NA = 6.023×1023
7. Which of the following contains maximum number of molecules?
(a) 1g CO2
(b) 1g N2
(c) 1g H2
(d) 1g CH4
Soln:
Answer is (c) 1g H2
Note: NA = 6.023×1023
Explanation:
1 g of H2 = ½ x NA = 0.5 NA = 0.5 × 6.022 × 1023 = 3.011 × 1023
8. Mass of one atom of oxygen is
(a) 23 16 g 6.023 10 ×
(b) 23 32 g 6.023 10 ×
(c) 23 1 g 6.023 10 ×
(d) 8u
Soln:
Answer is (a) 23 16 g 6.023 10 ×
Explanation:
Mass of one atom of oxygen = Atomic mass/NA = 16/6.023 × 1023 g
Note: NA = 6.023×1023

9. 3.42 g of sucrose are dissolved in 18g of water in a beaker. The number of oxygen atoms in the solution
are
(a) 6.68 × 1023
(b) 6.09 × 1022
(c) 6.022 × 1023
(d) 6.022 × 1021
Soln:
Answer is (a) 6.68 × 1023
Explanation:
1 mol of sucrose ( C12H22O11) contains = 11× NA atoms of oxygen, where NA = 6.023×1023
0.01 mol of sucrose (C12 H22 O11) contains = 0.01 × 11 × NA atoms of oxygen
= 0.11× NA atoms of oxygen
= 18 g/(1x2+ 16)gmol-1
=18 g /18 gmol-1
= 1mol
1mol of water (H2O) contains 1×NA atom of oxygen
Total number of oxygen atoms =
Number of oxygen atoms from sucrose + Number of oxygen atoms from water
= 0.11 NA + 1.0 NA = 1.11NA
Number of oxygen atoms in solution = 1.11 × Avogadro’s number
= 1.11 × 6.022 ×10”23 = 6.68 × 1023
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
Soln:
Answer is (c) when energy is either given to, or taken out from the system
Short Answer Questions
11. Which of the following represents a correct chemical formula? Name it.
(a) CaCl
(b) BiPO4
(c) NaSO4
(d) NaS

Soln:
Answer is (b) BiPO4, Its name is Bismuth Phosphate
Explanation:
Bismuth phosphate is right because Both ions are trivalent Bismuth phosphate(Bi3+- Trivalent anion. anion is an
ion that is negatively charged).
12. Write the molecular formulae for the following compounds
(a) Copper (II) bromide
(b) Aluminium (III) nitrate
(c) Calcium (II) phosphate
(d) Iron (III) sulphide
(e) Mercury (II) chloride
(f) Magnesium (II) acetate
Soln:
Answers are
(a) Copper (II) bromide- CuBr2
(b) Aluminium (III) nitrate = Al(NO3)3
(c) Calcium (II) phosphate - Ca3(PO4)2
(d) Iron (III) sulphide - Fe2S3
(e) Mercury (II) chloride - HgCl2
(f) Magnesium (II) acetate- Mg(CH3COO)2
13. Write the molecular formulae of all the compounds that can be formed by the combination of
following ions Cu2+
, Na+
, Fe3+
, C1–
SO4
-2
, PO4
-3
Soln:
Answers are
CuCl2/ CuSO4/ Cu3 (PO4) 2
NaCl/ Na2SO4/ Na3 PO4
FeCl3/ Fe2(SO4) 3 / FePO4
14. Write the cations and anions present (if any) in the following compounds
(a) CH3COONa
b) NaCl
(c) H2
(d) NH4NO3

Soln:
a) In CH3COONa-CH3COO is anion and Na is cation.
b) In NaCl-Cl anion Na is cation
c) In H2 both the ions are cations as they share electrovalent bond between them
d) In NH4NO3- NO3 is anion NH4 is cation
15. Give the formulae of the compounds formed from the following sets of elements
(a) Calcium and fluorine
(b) Hydrogen and sulphur
(c) Nitrogen and hydrogen
(d) Carbon and chlorine
(e) Sodium and oxygen
(f) Carbon and oxygen
Soln:
(b) Hydrogen and sulphur- H2S- Hydrogen Sulphide
(c) Nitrogen and hydrogen- NH3- Ammonia
(d) Carbon and chlorine – CCl4- Carbon Tetra chloride
(e) Sodium and oxygen – Na2O-Sodium Oxide
(f) Carbon and oxygen- CO2 ; CO- Carbon-di-oxide; Carbon Monoxide
16. Which of the following symbols of elements are incorrect? Give their correct symbols
(a) Cobalt CO
(b) Carbon c
(c) Aluminium AL
(d) Helium He
(e) Sodium So
Soln:
Cobalt CO is wrong, correct symbol is Co
Carbon c is wrong, correct symbol is C
Alluminium AL is wrong, correct symbol is Al
Helium He is the right symbol
Sodium So is wrong, correct symbol is Na

17. Give the chemical formulae for the following compounds and compute the ratio by mass of the
combining elements in each one of them. (You may use appendix-III).
(a) Ammonia
(b) Carbon monoxide
(c) Hydrogen chloride
(d) Aluminium fluoride
(e) Magnesium sulphide
Soln:
Sl. No. Compounds Chemical formula Ratio by mass of the
combining elements
(a) Ammonia NH3 N:H=14:3
(b) Carbon monoxide CO C:O= 12:16=3:4
(c) Aluminium fluoride HCl H:Cl= 1:35.5
(d) Aluminium fluoride AlF3 Al:F=27:57=9:19
(e) Magnesium sulphide MgS Mg:S= 24:32=3:4
18. State the number of atoms present in each of the following chemical species
(a) CO3-2
(b) PO4-3
(c) P2 O5
(d) CO
Soln:
(a) CO3-2
- 1+3=4
(b) PO4-3
-1+4=5
(c) P2 O5
– 2+5=7
(d) CO – 1+1=2
19. What is the fraction of the mass of water due to neutrons?
Soln:
Mass of 1 mole of a substance is equal to its relative atomic or molecular mass in grams.
Mass of one mole (Avogadro Number) of neutrons =1g
Mass of one neutron = 1/ Avogadro number(NA) g
Mass of one molecule of water = Molar mass / NA = 18/ NA g

The molar mass of water is
18.015 g/mol. This was
calculated by multiplying the atomic weight of hydrogen (1.008) by two and adding the result to the weight
for one oxygen (15.999)
Mass of one molecule of water = Molar mass / NA = 18/ NA g
Avogadro number(NA) =6.022 x 1023mol¯1
There are 8 neutrons in one atom of oxygen
Number of neutrons in oxygen= number of oxygen - Atomic number of oxygen
Oxygen's atomic weight= 15.9994
increases with an increase in temperature.
Therefore the mass is 16
Therefore number of neutrons= 16 - 8 = 8
Mass of one neutron = 1/ Avogadro number(NA) g
Mass of 8 neutrons = 8/ Avogadro number(NA) g
Fraction of mass of water due to neutrons = 8/18 g
20. Does the solubility of a substance change with temperature? Explain with the help of an
example.
Soln:
Solubility is the ability of a solute to get dissolved in 100g solvent. Solubility of a given solute to
dissolve in specific solvent depends on the temperature. With Increase in temperature solubility of
liquids and solids increase. In the same way solubility of gases decreases with increase in temperature.
Ex: Sugar dissolves faster in hot water than in cold water.
21. Classify each of the following on the basis of their atomicity.
(a) F2
(b) NO2
(c) N2O
(d) C2H6
(e) P4
(f) H2O2
(g) P4O10
(H) O3
(i) HCl
(j) CH4
(k) He
(l) Ag

Soln:
a) Monoatomic are inert gases that do not combine and exist as monoatomic gases
b) Diatomic- (a) 2- diatomic- NO2 = 1+ 2 = 3., HCl = 1+ 1 = 2
c) Triatomic-N2O = 2 + 1 = 3., NO2 = 1+ 2 = 3, O3 = 3
d) Tetraatomic- H2O2 = 2 + 2 = 4, P4O10 = 4 + 10 = 14, P4 = 4, CH4 = 1+ 4= 5
e) Octa atomic- C2H6 = 2 + 6 = 8
f) Polyatomic.
22. You are provided with a fine white coloured powder which is either sugar or salt. How would
you identify it without tasting?
To examine if the fine white coloured powder is sugar pr salt we can conduct two experiments.
Soln:
1. Heating: Upon heating sugar melts to liquid form because sucrose has a decomposition point and
melting point at temperatures between 190 to 192 degrees Celsius. This will turn sugar to light brown colour.
Upon heat further sugar gets charred to black colour.
Salt has a melting point of 841 degrees Celsius and 1545.8 degrees Fahrenheit. If we don’t heat it to that
point nothing change is observed.
2. Electric conductivity:
If we dissolve the given substance in water we can check for electric conductivity to check whether the
substance is sugar or salt. If it is salt it conducts electricity. Because salt (NaCl) has positive sodium ions
and negative chloride ions hence salt conducts electricity. But sugar don’t conduct electricity as sugar has
only positive ions.
23. Calculate the number of moles of magnesium present in a magnesium ribbon weighing 12 g.
Molar atomic mass of magnesium is 24g mol–1.
Soln:
Number of moles = weight
atomic weight
= 12 = 0.5 moles
24
Long Answer Questions
24. Verify by calculating that (a) 5 moles of CO2 and 5 moles of H2O do not have the same mass. (b) 240 g
of calcium and 240 g magnesium elements have a mole ratio of 3:5.
Soln:
(a) Molar mass of CO2 =12 + 2 x 16= 12 + 32 = 44 g mol-1
5 moles of CO2 have mass = 44 x 5 = 220 g
Similarly, molar mass of H2O = 2x 1 + 16 = 18 g mol -1

5 moles of H2O have mass =18
x 5 = 90 g
It is verified that 5 moles of CO2 and 5 moles of H2O are not same.
(b) Number of moles = w/ atomic weight
Atomic weight of Ca= 40 amu
Number of moles in 240g Ca metal 240/ 40 = 6
Number of moles in 240g of Mg metal 240/ 24 = 10
Atomic weight of Mg = 24amu
Ratio 6:10
25. Find the ratio by mass of the combining elements in the following compounds. (You may use
Appendix-III) (a) CaCO3 (d) C2H5OH (b) MgCl2 (e) NH3 (c) H2SO4 (f) Ca(OH)2
Soln:
a) CaCO3
Ca: C : O × 3
40 : 12 : 16 × 3
40: 12 : 48
10 : 3 : 12
(b) MgCl2
Mg : Cl × 2
24: 35.5 × 2
24: 71
(c) H2SO4
H x 2 : S : O × 4
2: 32 : 16 × 4
2 : 32 : 64
1: 16: 32
(d) C2H5OH
C × 2 : H × 6 : O
12 × 2 : 1 × 6 : 16
24 : 6 : 16
12 : 3 : 8

(e) NH3
N : H × 3
14 : 1 × 3
14: 3
(f) Ca(OH)2
Ca : O × 2 : H × 2
40 : 16 × 2 : 1 × 2
40 : 32 : 2
20 : 16 : 1
26. Calcium chloride when dissolved in water dissociates into its ions according to the following
equation. CaCl2 (aq) → Ca2+ (aq) + 2Cl– (aq) Calculate the number of ions obtained from CaCl2
when 222 g of it is dissolved in water.
Soln:
1 mole of calcium chloride = 111g
Therefore 222g of CaCl2 is equivalent to 2 moles of CaCl2
Since 1 formula unit CaCl2 gives 3 ions,
therefore, 1 mol of CaCl2 will give 3 moles of ions 2 moles of CaCl2 would give 3×2=6 moles of ions.
No. of ions = No. of moles of ions × Avogadro number
= 6 × 6.022 ×1023
= 36.132×1023
=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.
Soln:
Sodium atom and ion differ by one electron. For 100 moles each of sodium atoms and ions there would be a
difference of 100 moles of electrons.
Mass of 100 moles of electrons= 5.48002 g
Mass of 1 mole of electron = 5.48002 / 100 g
Mass of one electron = 5.48002 /100 × 6.022 ×1023
= 9.1 ×1028
g
= 9.1×10-31
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.
Soln:
Molar mass of HgS = The molar mass of Hg + the molar mass of S
= 200.6 + 32 = 232.6 g mol–1
1molecule of HgS contains 1 atom of Hg
232.6 g of HgS contains 200.6 g of Hg
Therefore, Mass of Hg in 225 g of HgS = 200.6 X 225 = 194.04g
232.6
29. The mass of one steel screw is 4.11g. Find the mass of one mole of these steel screws. Compare this value
with the mass of the Earth (5.98 × 1024
kg). Which one of the two is heavier and by how many times?
Soln:
One mole of screws weigh = 2.475 ×1024
g
= 2.475×1021
kg
Mass of the Earth / Mass of 1 mole of screws = 5.98 ×1024
kg
2.475×1021
= 2.4 ×10
Mass of earth is 2.4×103
times the mass of screws
The earth is 2400 times heavier than one mole of screws
30. A sample of Vitamin C is known to contain 2.58 ×1024 oxygen atoms. How many moles of oxygen atoms
are present in the sample?
Soln:
We know,
1 mole = 6.022 x 1023
The number of moles= Given number of particles
Avogadro Number
n= 2.58x1024
6.022x1023
n= 4.28 mol
31. Raunak took 5 moles of carbon atoms in a container and Krish also took 5 moles of sodium atoms in
another container of same weight.
(a) Whose container is heavier?
(b) Whose container has more number of atoms?

Soln:
a) Mass of sodium atoms carried by Krish = (5 ×23) g = 115 g
Atomic weight of Na = 23
While mass of carbon atom carried by Raunak = (5 ×12) g = 60g
b) Thus, Krish’s container has more number of atoms
32. Fill in the missing data in the Table 3.1
Species
property
H2O CO2 Na atom MgCl2
No of Moles 2 - - 0.5
No of particles - 3.011x1023 - 0
Mass 36g - 115g 0
Soln:
Species property H2O CO2 Na atom MgCl2
No of Moles 2 0.5 5 0.5
No of particles 12.044x1024 3.011x1023 3.011x1023 3.011x1023
Mass 36g 22g 115g 47.5g
33. The visible universe is estimated to contain 1022 stars. How many moles of stars are present in the
visible universe?
Soln:
Number of moles of stars = 1022
6.023 ×1023
= 0.0166 moles
34. What is the SI prefix for each of the following multiples and submultiples of a unit?
(a) 103
(b) 10–1
(c) 10–2
(d) 10–6
(e) 10–9
(f) 10–12

Soln:
a) 103 = 1000= kilo
(b) 10–1 =1/10= 0.1= deci
(c) 10–2 =1/100 = 0.01= centi
(d) 10–6 = 0.000 001= micro
(e) 10–9 =0.000 000 001 = nano
(f) 10–12=0.000 000 000 001 = pico
35. Express each of the following in kilograms
(a) 5.84×10-3 mg
(b) 58.34 g
(c) 0.584g
(d) 5.873×10-21g
Soln:
(a) 5.84 × 10–3 mg = 5.84 ×10–9 kg
(b) 58.34 g =5.834 ×10–2 kg
(c) 0.584g =5.84 ×10–4 kg
(d) 5.873×10-21g=5.873 ×10–24 kg
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)
Soln:
Mg2+ ion and Mg atom differ by two electrons.
103 moles of Mg2+ and Mg atoms would differ by
103
× 2 moles of electrons
Mass of 2 ×103
moles of electrons = 2×103 × 6.023 ×1023 × 9.1 ×10–31
kg
2×6.022 × 9.1×10–5
kg
109.6004 ×10–5
kg
1.096 × 10–3
kg
37. Which has more number of atoms? 100g of N2 or 100 g of NH3
Soln:
No. of moles of atoms = weight / atomic weight.
For N2
100 gms of N₂ = 100/2 x 14 moles = 100/28 moles
Number of molecules = 100 / 28 x 6.022 x 10²³
Molar mass of N₂ = 2 x molar mass of monoatomic N
Molar mass of N₂ = 2 x 14.0067 = 28 moles.
Number of molecules = 100/28 x 6.022 x 10²³
No. of atoms = 2 x 100/28 x 6.022 x 10²³ = 43.01 x 10²³
For NH₃

100 gm of NH₃ = 100/17 moles
Number of molecules = 100/17 x
6.022 x 10²³ molecules
No. of atoms in NH₃ = (1 + 3) = 4 x 100/17 x 6.022 x 10²³ =
141.69 x 10²³ atoms.
Therefore, NH₃ has more atoms than N₂.
38. Compute the number of ions present in 5.85 g of sodium chloride.
Soln:
58.5 g NaCl contains 6.023 x 1023
molecules
therefore 58.5 g NaCl contains 12.046 x 1023
ions.
Hence, 5.85 g NaCl contains 5.85 x 12.046 x 1023
58.5
= 1.2046 x 1023
ions
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?
Soln:
One gram of gold sample will contain 90 = =0.9g of gold
100
Number of moles of gold=mass of gold
atomic mass of gold
= 0.9
197
= 0.0046
One mole of gold contains NA atoms = 6.022×103
Therefore, 0.0046 mole of gold will contain= 0.0046 × 6.022
=2.77×1021
40. What are ionic and molecular compounds? Give examples.
Soln:
While forming some compounds, atoms gain or lose electrons, and form electrically charged particles called ions.
Compounds that are formed by the attraction of cations and anions are called as ionic compounds.
Ex : 2Na + Cl2 → 2Na+ Cl- → 2NaCl (sodium chloride- common salt.)
Sodium is a group 1 metal, thus forms a +1 charged cation. Chlorine is a non-metal, and has the ability to form a -
1 charged anion.
Compounds formed due to bonding of uncharged ions are called as molecular compounds and the bonding
between them is called covalent bonding. Molecular compounds are formed by sharing of electrons between the
two atoms and the elements are held together by covalent bonds.

Ex: 2C + O2 → 2CO ( Carbon monoxide)
41. Compute the difference in masses of one mole each of aluminium atoms and one mole of its ions. (Mass
of an electron is 9.1×10–28 g). Which one is heavier?
Soln:
Mass of one mole of Aluminium atom = {13 × mass of each electron + 13 × mass of each proton + 14 × mass of
each neutron} × Avogadro's constant.
We know, if atoms convert into ions, only transfer of electrons takes place, in Al+3 ion, Aluminium atom loss
three electrons,
So,
Mass of Al+3 ={10× mass of each electron +13 × mass of each proton + 14 × mass of each neutron} Avogadro's
constant
Now,
You see mass of aluminium atom is greater than aluminium ion by 3 electrons
Difference in mass =Mass of Aluminium atom -mass of aluminium ion
= 3 × mass of each electron x Avogadro's constant .
= 3 × 9.1 × 10-28
x 6.023 x 1023
=27.3 x 10-28
x 6.023 x 1023
g
=164.4 x 10-5
g
= 1.644 x 10-3
g
= 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.
Soln:
Mass of silver = m g
Mass of gold = m /100g
Number of atoms of silver = Mass/ Atomic mass X NA
= m /108NA Number of atoms of gold
= m/ 100 X197
Ratio of number of atoms of gold to silver = Au : Ag
= m/ 100 X 197 X NA : m/ 108 NA
= 108 : 100×197
= 108 : 19700 =
1 : 182.41
43. A sample of ethane (C2H6) gas has the same mass as 1.5 ×1020 molecules of methane (CH4). How many
C2H6 molecules does the sample of gas contain?
Soln:
6.02 x 1023
molecules of methane = 1 mole
Hence 1.5 x 1020
molecules of methane = (1.5 x 1020
x 1) ÷ (6.02 x 1023
) moles
= 2.49 x 10 -4
moles
Molar mass of Methane (CH4) = 12 + 1x4 = 16 g

Mass of methane = molar mass x no. of moles = 16 x 2.49 x 10 -4
= 3.984 x 10-3
g (This is the same mass as
Ethane)
Ethane (C2H6) = 12x2 + 1x6 = 30
If 30 g of Ethane has 6.02 x 1023
molecules
So 3.984 x 10 -3
g = (3.984 x 10-3
x 6.02 x 1023
) ÷ 30
= 8 x 1019
molecules of Ethane
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 ————.
(b) A group of atoms carrying a fixed charge on them is called ————.
(c) The formula unit mass of Ca3 (PO4) 2 is ————.
(d) Formula of sodium carbonate is ———— and that of ammonium sulphate is ————.
Soln:
Answers
a) Law of conservation of mass
b) Ions
c) 310
Explanation
3 × atomic mass of Ca+ 2 × atomic mass of phosphorus + 8 × atomic mass of oxygen) = 310
3 × 40 + 2 × 31 + 8 × 16 = 120 + 62 + 128 = 310
d) Na2 CO3 and (NH4) 2 SO4
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.
Across Down
The element used by Rutherford during his
α–scattering experiment
A white lustrous metal used for making
ornaments and which tends to get tarnished
black in the presence of moist air
An element which forms rust on exposure to
moist air
Both brass and bronze are alloys of the
element
A very reactive non–metal stored under
water
The metal which exists in the liquid state at
room temperature
Zinc metal when treated with dilute
hydrochloric acid produces a gas of this
element which when tested with burning
splinter produces a pop sound.
An element with symbol Pb

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 7. He 2. H 8. F 3. Ar 9. Kr 4. O 10. Rn 5. Xe 11. Ne 6. N
Soln:
b) Six : Helium (He); Neon ( Ne); Argon (Ar); Krypton (Kr); Xenon (Xe); Radon (Rn).

47. Write the formulae for the following and calculate the molecular mass for each one of them.
(a) Caustic potash
(b) Baking powder
(c) Lime stone
(d) Caustic soda
(e) Ethanol
(f) Common salt
Soln:
The formulae for the following and calculate the molecular mass for each one of them.
Sl No Compound Formula Molecular mass
A Caustic Potash KOH 39+16+1=56u
B Baking powder NaHCO3 23+1+12+3x16+84u
C Lime stone CaCO3 40+12+3x16+100u
D Caustic soda NaOH 23+16+1+40u
E Ethanol C2H5OH 2x2+5x1+16+1+46u
F Common Salt NaCl 23+35.5=58.5
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 C6 H12 O6 . 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
.
Soln:
6CO2 + 6 H2 O Chlorophyll /Sunlight → C6 H12 O6 + 6O2
1 mole of glucose needs 6 moles of water 180 g of glucose needs (6×18) g of water 1 g of glucose will need 108/
180 g of water.
18 g of glucose would need (108 /180) × 18 g of water = 10.8 g
Volume of water used = Mass
Density
= 10.8 g/ 1g cm-3
=10.8 cm3

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