# Atoms and Molecules : NCERT Exemplar

Multiple Choice Questions

Q.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)

Sol. (d)

(ii) 20 moles of water = 20 ×18 g = 360 g of water, because mass of 1 mole of water is the same as its molar mass, i.e., 18 g. Molarmassis the mass of one mole of a substance.

The molecular formula of water is H2O, which means that water is made up of two hydrogen and one oxygen atoms.

The standard atomic weight of hydrogen is 1.00794 and that of oxygen is 15.9994. So the molecular weight will be:

Molecular Weight of H2O = (2 × 1.00794) + (1×15.994) = 18.00988

So the molar mass of water is 18.00988 gm or 0.01800988 Kg.

(iv) 1.2044 × 1025 molecules of water contains

1.2044 × 1025/NA number of moles,

Where NA = 6.023 × 1023

Therefore,

1.2044 × 1025 /6.023×1023

= 0.199 x 102 moles  = 20 moles

Q.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

Sol. (a)

Q.3 The chemical symbol for nitrogen gas is

(a) Ni

(b) N2

(c) N+

(d) N

Sol. (b)

Q.4 The chemical symbol for sodium is

(a) So

(b) Sd

(c) NA

(d) Na

Sol. (d)

Q.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

Sol. (c)

Weight of a sample in gram = number of moles × molar mass

(a) 0.2 moles of C12H22O11 = 0.2 × 342 = 68.4 g

(b) 2 moles of CO2 = 2 × 44 = 88 g

12.011 + 2(16.00) = 12.011 + 32.00 = 44.011g/mol.

(c) 2 moles of CaCO3 = 2 × 100 = 200 g

2(40 + 12 + 3(16) = 2(40 +  12 + 48)  = 2 × 100 = 200g

(d) 10 moles of H2O = 10  × 18 = 180 g

10(1 + 1 + 16) = 10(18) = 10(18)  = 180 g

Q.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

Sol. (d)

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

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

(a) 1g CO2

(b) 1g N2

(c) 1g H2

(d) 1g CH4

Sol. (c)

1 g of H2 = ½ x NA = 0.5 NA = 0.5 × 6.022 × 1023  = 3.011  × 1023

Q.8 Mass of one atom of oxygen is Sol. (a)

Mass of one atom of oxygen = Atomic mass/N= 16/6.023  × 1023 g

Q.9 3.42 g of sucrose are dissolved in 18g of water in a beaker. The numbers of oxygen atoms in the solution are

(a) 6.68 × 1023

(b) 6.09 × 1022

(c) 6.022 × 1023

(d) 6.022 × 1021

Sol. (a) = 3.42 g/342 g mol-1

= 0.01mol

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 ×1023 = 6.68 × 1023

Q.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

Sol. (c)

Q.11 Which of the following represents a correct chemical formula? Name it.

(a) CaCl

(b) BiPO4

(c) NaSO4

(d) NaS

Sol. (b) BiPO4— Both ions are trivalent Bismuth phosphate(Bi3+- Trivalent anion. anion is an ion that is negatively charged)

(PO43- -Trivalent cation. A cation has a net positive charge)

Q.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

(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

Q.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-, PO43-

CuCl2/ CuSO4/ Cu3 (PO4) 2

NaCl/ Na2SO4/ Na3 PO4

FeCl3/ Fe2(SO4) 3 / FePO4

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

(a) CH3COONa

(b) NaCl

(c) H2

(d) NH4NO3

Anions                                    Cations

(a) CH3 COO-                                 Na

(b) Cl                                               Na+

(c) It is a covalent compound

(d) NO 3                                           NH4

Q.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

(a) Calcium and fluorine -   CaF2

(b) Hydrogen and sulphur-   H2S

(c) Nitrogen and hydrogen-   NH3

(d) Carbon and chlorine –  CCl4

(e) Sodium and oxygen – Na2O

(f) Carbon and oxygen- CO2 ; CO

Q.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

(a) Cobalt CO -Incorrect, the correct symbol of Cobalt is Co

(b) Carbon c - Incorrect, the correct symbol of Carbon is C

(c) Aluminium AL-Incorrect, the correct symbol of Aluminium is Al

(d) Helium He - Correct

(e) Sodium So- Incorrect, the correct symbol of Sodium is Na

Q.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

(a) Ammonia

NH3

N  :  H × 3

14  : 1 × 3

14 :  3

(b) Carbon monoxide

CO

C : O

12: 16

3 : 4

(c) Hydrogen chloride

HCl

H : Cl

1: 35.5

(e)Aluminium fluoride

AlF3

Al  : F × 3

27 : 19 × 3

27 : 57

9 : 19

(f) Magnesium sulphide

MgS

Mg : S

24 : 32

3: 4

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

(a) CO3 2–

(b) PO4 3–

(c) P2O5

(d) CO

(a) CO3 2–  = 1+3 =4

(b) PO4 3–  = 1 + 4= 5

(c) P2O5  = 2 + 5 = 7

(d) CO = 1 + 1 =2

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

The 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/ Ng

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/ Ng

Avogadro  number(NA)  =6.022 x 1023 mol¯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

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

Solubility refers to the ability of a given volume of solvent to dissolve a substance.

The solubility of a given solute in a given solvent typically depends on temperature. Solubility of solids and liquids usually increase as the temperature increases, but the solubility of gases decrease with increasing temperature.

For example,

(i)Take some salt  in a bowl.  Bring the temperature of the water to 20°C by adding some ice cubes to the water. Once the temperature of the water has stabilized, 100ml of the water is measured using the measurement cylinder and poured a beaker.

(ii)Sugar is added, 1 mg at a time, to the beaker and it is mixed using the spatula. The sugar is continuously added until it is not able to dissolve in the water anymore. The amount of sugar that has been added is calculated and recorded.

(iii)The beaker is now placed on the hot plate and the temperature of the solution is brought to 40°C. The salt is again added, 10mg a time, until it does not dissolve in the water anymore. The total salt added (including the amount of salt added previously) is calculated and recorded

(iv)Step iii is again repeated at 60°C, 80°C and 100°C and the results are recorded.

The results show that as the temperature of the water increases, the amount of salt dissolving in the solution is also higher.

Q.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

Atomicity is the number of atoms present in a molecule

(a) 2- diatomic

(b) NO= 1+ 2 = 3. Triatomic

(c)N2O = 2 + 1 = 3. Triatomic

(d) C2H6  = 2 + 6 =  8 Octa atomic

(e) P4  =  4 Tetra

(f) H2O2  = 2 + 2  = 4. Tetra

(g) P4O10 =  4 +  10 =  14

(h) O3 = 3

(i) HCl = 1+ 1 = 2

(j) CH4 = 1+ 4= 5

(k) 1 (monatomic (inert gases)do not combine and exist as monoatomic gases)

(l) Polyatomic.

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

(i)On heating the powder, if the white powder is sugar, it will melt down to a liquid form (sucrose has a decomposition point and melting point at temperatures between 190 to 192 degrees Celsius) and turned a light brown colour. Once the sugar had been heating for awhile, it should have turned black and started producing wisps of smoke.

Salt (sodium chloride) has a melting point of 841 degrees Celsius and 1545.8 degrees Fahrenheit. If we don’t heat it to that point nothing much happens.

(ii)Alternatively, the powder may be dissolved in water and checked for its conduction of electricity. If it conducts electricity, it is a salt. When table salt is dissolved in water, the solution conducts very well, because the solution contains ions. The ions come from the table salt, whose chemical name is sodium chloride. Sodium chloride contains sodium ions, which have a positive charge, and chloride ions, which have a negative charge. Because sodium chloride is made up of ions, it is called an ionic substance.
Sugar is made of uncharged particles called molecules. When sugar is dissolved in water, the solution does not conduct electricity, because there are no ions in the solution.

Q.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.

Number of moles = w/ atomic weight  =  12/24 =  0.5 mol

Q.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.

(a) CO2 has molar mass = 12.011 + 2(16.00) = 12.011 + 32.00 = 44.011g/mol.

5 moles of CO2 have molar mass = 44 × 5 = 220 g

Molar mass of H2O  = H × 2 = 1.01588 x 2 = 2. 01588 g
O × 1 = 1 x 15.9994 g = 15.9994 g

Molar mass of H2O = 2. 01588 + 15.9994 = 18.0153 g/mol.

5 moles of H2O have mass = 18 × 5 g = 90 g

(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

3: 5

Q.25 Find the ratio by mass of the combining elements in the following compounds. (You may use Appendix-III)

(a) CaCO3

(b) MgCl2

(c) H2SO4

(d) C2H5OH

(e) NH3

(f) Ca(OH)2

(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

Q.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

Molar mass of CaCl2 = 40.078 + 35.453 × 2  = 110.984 g/mol

1 mole of calcium chloride = 111g

∴  222g of CaCl2 is equivalent to 2 moles of CaCl2

Since 1 formula unit CaCl2 gives 3 ions(one Ca2 + cation and two Cl- anions),

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

Q.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.

A sodium atom and ion, differ by one electron. Sodium loses one electron to form sodium ion.

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

As 1 mole of electron have 6.022 × 1023 electrons(1 mole = 6.022 ×1023 electrons)

Therefore mass of 6.022 × 1023 electrons = 5.48002/ 100 g

Mass of one electron = 5.48002/ 100 × 6.022 ×10 = 9.1 ×10-28  g = 9.1×10-31  kg

Q.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.

Molar mass of HgS = 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 / 232.6)  × 225 = 194.04g

Q.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 × 1024kg). Which one of the two is heavier and by how many times?

One mole of screws weigh = 2.475 ×1024g = 2.475×1021 k

Mass of the Earth/ Mass of 1 mole of screws  = 5.98 ×1024 kg / 2.475 ×1021  kg = 2.4 x 103

Therefore, we can say that the mass of earth is 2.4×103 times the mass of screws.

The earth is 2400 times heavier than one mole of screws.

Q.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?

1 mole of oxygen atoms = 6.023×1023 atoms

Therefore,

Number of moles of oxygen atoms = 2.58 × 1024/6.023 ×1023  = 4.28 mol

4.28 moles of oxygen atoms

Q.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?

(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

Thus, Krish’s container is heavy

(b) Both the bags have same number of atoms as they have same number of moles of atoms

Q.32 Fill in the missing data in the Table .  Q.33 The visible universe is estimated to contain 1022 stars. How many moles of stars are present in the visible universe?

Number of moles of stars = 1022 / 6.023 ×1023  = 0.0166 moles

Q.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

(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

Q.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

(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

Q.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)

A 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 × 10–5kg  = 109.6004 ×10–5 kg  = 1.096 × 10–3kg

Q.37 Which has more number of atoms? 100g of N2 or 100 g of NH3

No of moles of atoms = w/ atomic weight

100 g of N2 = 100/ 2 x 14 moles

100 g of N2 =100/28 moles

Number of molecules = 100/ 28 × 6.022 ×1023

Molar mass of N2= 2 × molar mass of monatomic N = 2 × 14.0067

Molar mass of N2 = 28.01340

Molar mass of N2 = 28

To find number of molecules, number of moles X multiply by Avogadro's number

Number of molecules = 100/ 28 × 6.022 ×1023

Number of atoms in N2 =  1 + 1 = 2 = 2 ×100/ 28 × 6.022 ×1023 = 43.01×1023

(ii) 100 g of NH3 = 100/ 17 moles

To find number of molecules, number of moles × Avogadro's number

Number of molecules = 100/ 17 × 6.022 × 1023 molecules

Number of atoms in NH3  =  1 +  3 =  4 =  100/ 17× 6.022 ×1023 × 4 atoms =  141.69 ×1023

Therefore,NH3 would have more atoms

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

# Atomic weight; Na = 22.99 g, Cl = 35.45 g

Atomic weight of NaCl= 23 + 35.5 = 58.5g.

5.85 g of NaCl = 5.85/ 58.5 = 0.1moles

Each NaCl particle is equivalent to one Na+ and one Cl = 2 ions

Total moles of ions = 0.1 × 2= 0.2 moles

Number of ions= total number of moles of ions X Avogadro's number

Therefore,

No. of ions= 0.2 × 6.022 ×1023

1.2042 ×1023 ions

Q.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?

Given, the sample contains  90% of  Gold (Au).

One gram of gold sample will contain 90% of Gold i.e. 90/100=0.9g of gold

Number of moles of gold = Mass of gold/ Atomic weight of gold

Atomic weight of gold = 196.967 grams/mol= 197 grams/mol

Therefore,

Number of moles of gold = 0.9/197 = 0.0046

One mole of gold contains = 6.022 ×1023

∴ 0.0046 mole of gold will contain =0.0046  × 6.022 ×1023  = 2.77 ×1021

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

Chemical elements can join with each other to form chemical compounds.

Ionic compounds: Ionic compounds are formed by the attraction between positive and negative ions. Positively charged ions are known as cations and negatively charged ions are known as anions. Ionic compounds are formed. The compounds formed by the transfer of electrons are called as ionic compound. The bond found in these compounds are ionic in nature.

These compounds are pure substances which are formed be a metal and a non-metal. Cations are usually formed by metal atoms and the anions are formed by nonmetal atoms.  They are good conductors of electricity when in a molten or aqueous state, have high melting point,  soluble in water. Ionic compounds exist as crystals. Examples of ionic compounds: sodium chloride,  sodium bromide, potassium chloride etc.

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 nonmetal, and has the ability to form a -1 charged anion.

Molecular Compounds: Molecular compounds are formed by uncharged atoms. They are also called as covalent compounds, as they are formed by sharing of electrons between the two atoms and the elements are held together by covalent bonds.

These, are the pure substances, formed by non-metals. It has a low melting point and cannot conduct electricity regardless of state. Molecular compounds can exist as either solid, liquid or gaseous state. Eg: carbon dioxide, carbon monoxide, hydrogen chloride, water, methane etc.

Ex: 2C + O2 → 2CO ( Carbon monoxide)

Q.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?

Mass of 1 mole of aluminium atom = the molar mass of aluminium = 27 g mol–1

An aluminium atom loses three electrons to become an ion, Al3+

For one mole of A13+ ion, three moles of electrons are to be lost

Therefore, total number of electrons lost =3

The mass of three moles of electrons = 3 × (9.1×10–28) × 6.022×1023 g

= 27.3 × 6.022 ×10–5 g

= 164.400 ×10–5 g

= 0.00164 g

Molar mass of Al3+ = molar mass of aluminium – mass of electrons

(As it loses electrons to become Al3+) =(27–0.00164) g mol–1 = 26.9984 g mol–1

Therefore, difference in masses of one mole each of aluminium atoms and one mole of its ions

= 27 – 26.9984 = 0.0016 g

Q.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.

Mass of silver = ‘m’ g                         … (i)

Mass of gold= m/100 g                     … (ii)

Number of atoms of silver(Ag) = Mass / Atomic mass × Avogadro's number(NA)

= m/108 × NA                                (from (i)

Atomic weight of silver = 108 g/mol

Number of atoms of gold(Au) = Mass / Atomic mass × Avogadro's number(NA)

= m/100 ×197 × NA                  (from (ii)

Ratio of number of atoms of gold to silver = Au : Ag

= m/100 × 197 × N :  m/108 × NA

= 108 : 100 × 197

= 108 : 19700

= 1 : 182.41

Q.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?

Mass of 1 molecule of CH4 = 16 g/ NA

The atomic weight of carbon is 12, the atomic weight of hydrogen is 1

weight of CH4 = 12 +  4 × 1 = 16

Mass of 1.5 ×1020 molecules of methane  = 1.5 ×1020  × 16/ NA

Mass of 1 molecule of C2H6 =30 g/ NAC

C- 12 ;  H-1 ;  Therefore C2H6= 2 × 12 + 6 × 1 =  30

Mass of molecules of C2H6 is  = Mass of CH4(given)

Therefore,

Mass of molecules of C2H6 =1.5 ×1020 x 16/ NA

Number of molecules of C2H6 = {1.5 ×1020 x 16/ NA}/ 30 g/ NA

= {1.5 ×1020  ×16/ NA} × NA/30

= 0.8 × 1020

Q.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 ________.

(a) Law of conservation of mass.

(b) ions.

(c) 310

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      (NH4) 2 SO4

Q.45 Complete the following crossword puzzle Figure by using the name of the chemical elements. Use the data given in Table.  Down

Silver

Copper

Mercury

Lead

Across

Gold

Iron

Phosphorous

Hydrogen Q.46 (a) In this crossword puzzle Figure, names of 11 elements are hidden. Symbols of these are given below. Complete the puzzle.

Cl 7. He

H 8. F

Ar 9. Kr

O 10. Rn

Xe 11. Ne

N (a) (b) Six : Helium (He); Neon ( Ne); Argon (Ar); Krypton (Kr); Xenon (Xe); Radon (Rn).

Q.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

(a) KOH = (39 +16+1) = 56 g mol–1

(b) NaHCO3 = 23 + 1 + 12 + (3 × 16) = 84 g mol–1

(c) CaCO3 = 40 + 12 + (3 × 16) = 100 g mol–1

(d) NaOH = 23+16+1 = 40 g mol–1

(e) C2H5OH = C2H6O = 2 × 12 + (6 × 1) + 16 = 46 g mol–1

(f) NaCl = 23 + 35.5 = 58.5 g mol–1

Q.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

6CO2(g) + 6 H2O(l)  → C6 H12 O6 + 6O2(g)

1 mole of glucose needs 6 moles of water

Therefore, 180 g of glucose needs (6×18) g of water

Molar mass of water= 2+16= 18

Number of moles= given mass /molar mass

1 g of glucose will need 108/ 180 g of water.

C6 H12 O6 = C – 6 × 12 + 12 × 1 + 6 × 16 = 180g

18 g of glucose would need=   (108/180) × 18g of water = 10.8g

Volume = Mass /density

Therefore, Volume = 10.8 g /1 g cm–3  = 10.8 cm3

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