Solution

CHAPTER 1 BASIC CONCEPT OF CHEMISTRY

Class 11 - Chemistry
Section A
1. (a) 18.24
Explanation: 18.24
2. (c) 5, 4
Explanation: 5, 4
3. (c) Good precision but poor accuracy
Explanation: Good precision but poor accuracy
4. (c) Student D
Explanation: Student A and C have reported values that are neither precise nor accurate. Student B has reported values that are
precise but not accurate. Student D has reported values that are both precise and accurate.
5. (b) Transmutation
Explanation: The process of interconversion of elements through changes in atomic nuclei is called transmutation.

6. (d) 135.5 × 1022

Explanation: 4.4 g of CO2 = 4.4

44
= 0.1 mol

= 0.1 × 6.022 × 1023 molecules

= 6.022 × 1022 molecules
= 6.022 × 1022 × 22
= 132.5 × 1022 electrons (∵ 1 molecule of CO2 contain 22 electrons)

7. (c) molecules present in 1 mL of gas at STP

Explanation: 22.4 litre contains 6.023 × 1023 molecules at STP and Loschmidt number represents molecules of gas in 1 mL at
STP.
8. (a) All of these
Explanation: All of these
9. (d) Number of molecules ⟶ Moles × Molar mass
Explanation: Number of molecules = Moles × Avogadro’s number
10. (a) 4M
Explanation: Molecular mass = Vapour density × 2
∴ Vapour density of Y =
M

∴ Vapour density of X = 4 × M

2
= 2M
∵ (Vapour density of X is four times that of Y)
∴ Molecular mass of X

= Vapour density of X × 2 = 2M × 2
= 4M
11. (c) 24 g of C (12)
Explanation: 24 g carbon has 2 mole atoms. Rest all have 1 mole atoms.
12. (c) 1 m
W × 1000
Explanation: M olality =
B

MB WA

Mass of HCl WB = 18.25

Molar mass of HCl MB = 36.5,
MA = 18
18.25 × 1000
M olality = = 1m
36.5 × 500

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13. (c) C H O2 4 2

Explanation: Molecular mass = 2 × vapour density = 2 × 30 = 60

Empirical formula mass =12+2+16=30
⇒ n = 60/30 = 2

Molecular formula is (CH2 O)

2
= C2 H4 O2

14. (b) 0.6, 0.9, 0.3

Explanation: (H3PO3) diabasic 0.6
N(H3PO4) tribasic 0.9
N(H3BO3) monobasic 0.3 N

15. (d) the exact number of different types of atoms present in a molecule of a compound
Explanation: Molecular formula is the exact no. of atoms present in a molecule of a compound.
Molecular formula of a compound is related with its empirical formula as,
Mollecular formula = (Empirical formula)n
where n represents a positive integer.
16. (c) 1,6,3,2
Explanation: The given equation gets stoichiometrically balanced when the coefficients ( 1, 6, 3, 2 ) starting from LHS to
RHS are inserted in given blank spaces.
Thus, the balanced equation for the reaction is,
Ba3N2 + 6H2O ⟶ 3Ba(OH)2 + 2NH3

17. (c) Base = Weak, Acid = Strong, Endpoint = Yellow to pinkish-red

Explanation:
Methyl orange show Pinkish colour towards more acidic medium and yellow orange colour towards basic or less acidic media.
Its working pH range is

Weak base have the pH range greater than 7. When methyl orange is added to this weak base solution it shows yellow orange
colour.
Now when this solution is titrated against strong acid the pH move towards more acidic range and reaches to end point near 3.9
where yellow orange colour of methyl orange changes to Pinkish red resulting to similar change in colour of solution as well.

18. (c) 40 mL
Explanation: Meq. Of H3PO3 = Meq. of KOH
20× 0.1× 2 = V × 0.1
∴ V= 40 mL
19. (d) 0.075
Explanation: Cone, of Na+ = 100×0.1

200
+
100×0.1×2

200
= 0.15 M
∴ Ionic strength = 1

2
cz
2
=
1

2
× [0.15 × 1 ]
2
= 0.075
20. (a) 1.05
Explanation: I.S. =
1

2
Σcz
2

0.1×100
[Al2(SO4)3] after mixing = 200
= 0.05 M
0.2×100
[Na2SO4] after mixing = 200
= 0.1 M
1 2
∴ I.S. = Σcz
2

= 0.05 × 2 × 32 + 0.05 × 3 × 22 + 0.1 × 2 × 12 + 0.1 × 1 × 22

=
2.1

2
= 1.05
21. S.I. Unit of mass is kilogram. it's symbol is kg.

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Chemistry Assignment/Work Sheet
22. The SI unit of volume (Length × breadth× height) = m3 . On the other hand, Litre (L) is the common unit which is not an SI unit.
23. Precision means the closeness of various measurements for the same quantity. Accuracy is the agreement of a particular value to
the true value of the result.
24. According to Dalton's atomic theory,
' an atom is a tiny or ultimate particle of a matter , which can not be divided further .'
Section B
25. Since,
the number of atoms in 1 mol of 12C atoms = 6.022 × 1023 atoms = Atomic mass of carbon - 12 in gms. = 12 g
Thus, 6.0022 × 1023 atoms of 12C have mass = 12 g
∴ 1 atom of 12C will have mass = 12

23
g
6.022×10

= 1.9927 × 10-23 g
26. i. Atomic mass of Fe = 56 g
56 g of Fe contains 6.022 × 10 atoms = 1 mole
23

56 g of Fe contains = 1 mole
7.85 g of Fe contains = × 7.85 = 0.14 moles
1

56

ii. Atomic mass of Ca = 40 g

40 g of Ca = 40 × 103 mg of Ca
40 g of Ca = 1 mole
4 × 104 mg of Ca = 1 mole
7.9 mg of Ca = 7.9

4
= 1.97 × 10-4 moles
4×10

27. Calculations
Step 1.
Molar mass of methanol
= [( 1 × 12 ) + ( 4 × 1 ) + (1 × 16 )]
= 32 g mol-1
or, = 0.032kg mol-1
Step 2.
Molarity of the given methanol solution
Applying the relation,
M1 × V1 = M2V2
(Given solution) (Solution to be prepared)
24.78 × V1 = 0.25 × 2.5 L
∴ V1 = 0.2522 L
or, = 25.22 mL
So, the volume needed for making 2.5 L of its (ie. methanol) solution is 25.22 ml
28. We know that; Molecular formula = n × (Empirical formula) ; Where, n= Whole number value.
i. Example of molecules having same molecular formula and empirical formula: Carbon dioxide (CO2) and Water (H2O)
ii. Example of molecules having different molecular formula and empirical formula:
a) Hydrogen peroxide: molecular formula is H2O2 and empirical formula is HO.
b) Glucose : Molecular formula is (C6H12O6) and empirical formula is CH2O.
29. The mass per cent of iron (Fe) = 69.9 % ( given )
The mass per cent of oxygen (O) = 30.1%, ( given)
Number of moles of iron present in the oxide = [69.9 / 55.85]
Number of moles of oxygen present in the oxide = [30.1 / 16]
The ratio of iron to oxygen in the oxide, = 1.25 : 1.88
or, = (1.25 / 1.25) : (1.88 / 1.25)
= 1 : 1.5
So, a whole number ratio
=2:3

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Chemistry Assignment/Work Sheet
 hence, the empirical formula of the oxide is Fe2O3
The empirical formula mass of Fe2O3 = 2 × 55.85 + 3 × 16.00

= 159.7 g mol-1
Molar mass
n= Empirical formula mass

n = 159.69 g / 159.7g = 0.9999

= 1 (approx.)
The molecular formula of a compound is obtained by multiplying the empirical formula with this positive integer (n)
Thus, as per the empirical formula (Fe2O3 ) of the given oxide,
n = 1.
Hence, the molecular formula is same as empirical formula, Fe2O3
The molecular formula of the oxide is Fe2 O3
30. Given, Mass of chlorophyll = 2.0 g
Percentage of Mg = 2.68 %.
So, Mass of Mg = 2.68 g
Therefore, Mass of Mg in 2.0 g of chlorophyll = 2.68×2.0

100
= 0.054g

By applying mole concept, we have:

6.022 x 1023 atoms of magnesium has mass = 24 g
∴ 24 g of Mg contains = 6.022 × 1023 atoms
= 1.3× 1021 Mg atoms
23

∴ 0.054 g of Mg contains = 6.022×10

24
× 0.054

Section C
31. Given, Molarity of solution, M = 3 mol L-1
Mass of NaCl in 1 L solution = 3 × 58.5 = 175.5 g
Mass of 1 L solution = Volume × density of solution = 1000 mL × 1.25 g/mL = 1250 g (since density = 1.25 g mL-1 )
Mass of water solution = 1250 -175.5 = 1074.5 g = 1.0745 kg.
3mol
Now, Molality of solution = number of moles of solute

mass of solvent in kg
= 1.0745kg
= 2.79m .
32. Molecular mass of Ca3(PO4)2 = 3× Ca + 2(1× P +4 × O ) = 3 × 40 + 2(1 × 31 + 4 × 16)= 310u
3×( atomic mass of calcium )
Mass percent of calcium= × 100
molecular mass of Ca3 (P O4 ) 2

= 120u

310u
× 100 = 38.71%
2×( atomic mass of phosphorus)
Mass percent of phosphorus= × 100
molecular mass of Ca3 (P O4 )
2
2×31u
= 310u
× 100 = 20%
8×(Atomic mass of oxygen)
Mass percent of oxygen = × 100
molecular mass of Ca3 (P O4 )
2
8×16u
= 310u
× 100 = 41.29%
33. Ratio of metal and oxygen in first oxide, M3O4 = 72.4 : 27.6
Ratio of metal and oxygen in second oxide = 70:30
Let molecular mass of metal =M
Therefore, the percentage by weight of the metal in the oxide = 3 × M ×100

3×M +4×O
= 72.4

3 × M ×100 72.4
=
3×M +4×16 1

300M = 217.2 M + 4633.6

⇒ 300M − 217.2M = 82.8M = 4633.6

4633.6
⇒ M = = 55.96 ≈ 56
82.8

Moles of of metal in second oxide = 70/56=1.25

Moles of oxygen in second oxide =30/16= 1.875
Ratio of moles of metal and oxygen in second oxide =1.25 : 1.875 = 1 :1.5 =2:3
Hence, Formula of second oxide =M2O3.
34. The oxidation of oxalic acid, (COOH)2, by potassium permanganate, KMnO4 takes place in the following steps
I. KMnO4 is a strong oxidising agent reacts with dil. H2SO4 to produce nascent oxygen as given by chemical equation:
KMnO4 + H2 SO4 ⟶ K2 SO4 + MnSO4 + 3H2 O + [O]

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Chemistry Assignment/Work Sheet
 By balancing this skeleton equation by hit and trial method, we get
2KMnO4 + 3H2 SO4 ⟶ K2 SO4 + 2MnSO4 + 3H2 O + 5[O] .....................(i)
II. Oxalic acid is oxidised to CO2 and H2O by the nascent oxygen produced in equation (i). The balanced partial equation for this
reaction is
(COOH)2 + [O] ⟶ 2CO2 + H2 O ..........................(ii)
III. Now, (i) + 5 [(ii)] , We get
2KMnO4 + 3H2 SO4 + 5(COOH)2 ⟶ K2 SO4 + 2MnSO4 + 10CO2 + 8H2 O

This represents the balanced chemical equation for the above reaction.

35. i. Number of moles of iron (55.8Fe) = mass of iron

atomic mass
= 7.85

55.8
= 0.141mol

ii. Number of moles of silicon(28.1Si) =

−3
mass of silicon 4.68×10 −4
= = 1.66 × 10 mol
atomic mass 28.1

iii. Number of moles of carbon(12C) =

−6
mass of carbon

atomic mass
= 65.6×10

12
= 5.47 × 10 −6
mol

36. The given chemical equation is:

4HC l(aq) + M nO2 → 2H2 O(l) + M nC l2 (aq) + C l2 (g)
4×36.5 87g

It is clear from balanced chemical equation,

87 g of MnO2 reacts with 146 g HCI
Therefore, 5 g of MnO2 will react with = 146 ×5

87
= 8.4 g HCl.

37. a. Moles of oxygen = 1.6

32
= 0.05 mol
1 mol of O2 at STP has volume = 22.4 L
0.05 mol of O2 at STP has volume = 22.4 × 0.05 = 1.12 L
V1 = 1.12 L p1 = 1 atm
1
V2 = ? p 2
=
2
= 0.5atm

According to Boyle's law: when pressure reduced to half, the volume is double.
p1V1 = p2V2
p1 V1
or V 2 =
p2
=
latm ×1.12L

0.5atm
= 2.24L

b. No. of molecules in 1.6g or 0.05 mol

23 22
= 6.022 × 10 × 0.05 = 3.011 × 10

38. i. 1 mol of He
= 6.022 × 1023 atoms
∴ 52 mol of He
= 52 × 6.022 × 1023 atoms
= 3.131 × 1025 atoms
ii. 1 atom of He
= 4 u of He;
(since mass number of He atom (24 He) is 4 u)
or, 4 u of He
= 1 atom of He
∴ 52 u of He

= × 52 atoms
1

= 13 atoms of He
iii. 1 mole of He
=4g
No. of atoms in 4 g of He
= 6.022 × 1023 atoms
No. of atoms in 52 g of He
23
6.022×10
= 4
× 52 atoms
= 7.8286 × 1024 atoms of He
Section D

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Chemistry Assignment/Work Sheet
39. The balanced chemical equation is :
BaCl2 (aq) + Na 2 SO4 (aq) ⟶ BaSO4 (s) + 2NaCl(aq)

Let us first calculate moles of Na2SO4 and BaCI2 0.5 M solution of NaSO4 means that 0.5 mol of Na2SO4 are present in 1000 ml
of solution.
1000 ml of solution contain Na2SO4 = 0.5 mol
0.5
250 ml of solution contain NaSO4 = 1000
× 250
= 0.125 mol
Moles of BaCI2 in solution = 10

208

(Mol. mass of BaCI2 = 137 + 2 × 35.5 = 208)

= 0.048
According to the balanced equation, 1 mol of BaCl2 reacts with 1 mol of Na2SO4.. Therefore, BaCl of BaCl2 limiting reactant, so
only 0.048 mol of Na2SO4 reacts with 0.048 mol of Na2SO4.
Now, according to the equation,
1 mol of BaCI2 produces BaSO4 = 1 mol
0.048 mol of BaCI2 produces BaSO4 = 1 × 0.048
= 0.048 mol
Amount of BaSO4 obtained = 0.048 × 233
(Mol. mass of BaSO4 = 233)
= 11.18 g
40. i. 0.50 m Na2CO3 means that 0.50 moles of Na2CO3 are dissolved in 1000 g of water.
0.50 M Na2CO3 solution means that 0.50 moles of na2CO3 are dissolved in 1000 mL of solution.
ii. C + O2 ⟶ CO2
1 Mol of carbon reacts with 1 mole of oxygen to form 1 mole of CO2.
a. 1 mol of CO2 or 44 g of CO2
b. since 16 g of dioxygen i.e. 0.5 mol of O2 are present, it is a limiting reagent. 0.5 mol of O2 will form 0.5 mol of CO2 i.e.
22 g
c. 16 g or 0.5 mol O2 is limiting reagent 0.5 mol of O2 will form 0.5 mol of CO2 i.e. 22 g

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