CHAPTER TEST-2 CHEMISTRY CLASS: XI-NEET/JEE

TOPIC: Some basic Concepts of Chemistry Dt: 08.07.25

(a) 0.04 (b) 0.01

1. What is the molarity of H2SO4 solution that has (c) 0.03 (d) 0.02
density 1.84g / mL and 89% by weight ?
10. In a mixture of gases, the volume content of a gas is
(a) 17.14 M (b) 16.7 M
(c) 18.4 M (d) 8.9 M
0.06% at STP. Calculate the number of molecules of the
gas in 1 L of the mixture.
2. The mass of one molecule of carbon dioxide is (a) 1.613  1023 (b) 6.023 1023
(a) 26.49 1024 g (b) 44g (c) 1.611027 (d) 1.611019
(c) 7.30 10−23 g (d) 22 g
11. Chlorine gas is prepared by reaction of H 2SO 4 with
MnO2 and NaCl. What volume of Cl2 will be produced
3. Which of the following pairs illustrates the law of
multiple proportions? at STP if 50 g NaCl is taken in the reaction?
(a) PH3 ,HCl (b) pbO,PbO2 (a) 1.915L (b) 22.4L
(c) H 2S,SO2 (d) CuCl2 ,CuSO4 (c) 11.2L (d) 9.57L

4. 10 g of MgCO3 decomposes on heating to 0.1 mole CO2 12. For the following reaction, the mass of water produced
and 4 g MgO. The percent purity of MgCO3 is (Given that from 445 g of C57H110O6 is :
atomic weights of Mg, C and O are 24, 12 and 16 u)
(a) 44% (b)54% 2C57H110O6(s) + 163O2(g) ⎯⎯→ 114CO2(g) + 110H2O(l)
(c)74% (d)84% (a) 490 g (b) 445 g
(c) 495 g (d) 890 g
5. A Compound Containing beryllium has the following
composition, Be = 6.1%, N = 37.8% Cl=48%, 13. 4.28g of NaOH is dissolved in water and the solution is
H = 8.1 %. One mole of the compound has mass of 148 made to 250 cc. what will be the molarity of the solution?
g and average atomic mass of beryllium is 9. The
(a) 0.615molL−1 (b) 0.428molL−1
molecular formula of the compound is :
(a) BeN4H12Cl2 (b) BeN2H10Cl (c) 0.99molL−1 (d) 0.301molL−1
(c) BeN4H2Cl3 (d) Be2N4 H10 CI2
14. 30 ml of acid solution is neutralized by 15 ml of a 0.2 N
6. The number of water molecules present in 0.20 g sample base. The strength of acid solution is
of CuSO4.5H2O (Molar mass = 249.7) is (a) 0.1 N (b) 0.15 N
(a) 1.2 × 1021 (b) 2.14 × 1021 (c) 0.3 N (d) 0.4 N
(c) 2.14 × 1022 (d) 1.2 × 1023
15. A compound contains 4.07% hydrogen, 24.27% C
7. Equal masses of oxygen, hydrogen and methane are taken and 71.65% C1 . Its molar mass is 98.96 g. The
in a container in identical condition. Find the ratio of the molecular formula is
volumes of the gases. CH2Cl
(a) O2 : H2 : CH4 1 : 16 : 2 (a) (b) C2H4Cl2
(b) O2 : H2 : CH4 1 : 8 : 1 (c) C2H2Cl2 (d) CH3C1
(c) O2 : H2 : CH4 16 : 1 : 8
(d) O2 : H2 : CH4 8 : 1 : 8 16. Two elements A and B have atomic weights 12 and
16. They form four compounds W, X, Y and Z in
8. How many grams of CaO are required to react with 852 which A is present in same amount while element B is
g of P4 O10 ? present in the ratio 1:2:3:4. If compound W has 24
(a) 852 g (b) 1008 g paris by weight of A and 32 parts of B, then compound
(c) 85 g (d) 7095 g X has 24 parts by weight of A and
(a) 64 parts of B (b) 32 parts of B
(c) 14 parts of B (d) 56 parts of B
9. 40 mL of 𝑥MKMnO4 solution is required to react
completely with 200 mL of 0.02M oxalic acid solution
in acidic medium. The value of 𝑥 is
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CH-XI-D.T-2-20260708
17. The final molarity of a solution made by mixing 50 mL (a) P2 Q 2 (b) PQ (c) P2 Q (d) PQ2
of 0.5M HCl, 150 mL of 0.25M HCl and water to make
the volume 250 mL is 26. Formation of polyethene from calcium carbide takes
(a) 0.5 M (b) 1 M (c) 0.75M (d) 0.25M
place as follows :
CaC2 + H2O → Ca(OH)2 + C2H2 ; C2H2 + H → C2H4
18. How many moles of potassium chlorate need to be heated n(C2H4) → (−CH2−CH2−)n.
to produce 11.2 litre oxygen at N.T.P. The amount of polyethylene possibly obtainable from
1 1 64.0 kg CaC2 can be
(a) mol (b) mol
2 3 (a) 28Kg (b) 14kg
1 2 (c) 21kg (d) 42 kg
(c) mol (d) mol
4 3
27. One litre of a gas at STP weight 1.16 g it can possible be
19. A compound of magnesium contains 21.9% magnesium (a) 𝐶2 𝐻2 (b) 𝐶𝑂
27.8% phosphorus and 50.3% oxygen. What will be the (c) 𝑂2 (d) 𝐶𝐻4
simplest formula of the compound?
(a) Mg 2 P2O7 (b) MgPO3
28. Equivalent weight of a bivalent metal is 37.2. The
(c) Mg 2 P2O2 (d) MgP2O4 molecular weight of its chloride is
(a) 412.2 (b) 216
20. What will be the molality of the solution containing (c) 145.4 (d) 108.2
9.125 g of HCl in 500 g of water).
(a) 0.l m (b) 1 m
(c) 0.5 m (d) 1 m 29. The ratio of the masses of methane and ethane in a gas
mixture is 4 : 5. The ratio of number of their molecules
21. The strength of 10– 2 M Na2CO3 solution in terms of in the mixture is
(a) 4:5 (b)3:2 (c)2:3 (d)5:4
molality will be (density of solution = 1.10 g mL–1).
(Molecular weight of Na2CO3 = 106 g mol–1)
30. The decomposition of a certain mass of CaCO3 gave
(a) 9.00 × 10 – 3 (b) 1.5 × 10–2 11.2dm3 of CO2 gas at STP. The mass of KOH required
(c) 5.1 × 10–3 (d) 11.2 × 10– 3 to completely neutralise the gas is
(a) 56 g (b) 28 g (c) 42 g (d) 20 g
M
22. A solution of FeCl3 is its molarity for Cl– ion will be
30 31. 36.5 % HCl has density equal to 1.20 g mL–1. The
: molarity (M) and molality (m), respectively, are
M M (a) 15.7, 15.7 (b) 12, 12
90 30 M M (c) 15.7, 12 (d) 12, 15.7
(a) (b) (c) (d)
10 5
32. 2.82 g of glucose is dissolved in 30 g of water. The mole
23. 500 mL of a glucose solution contains 6.02 × 1022 fraction of glucose in the solution is.
molecules. The concentration of the solution is (a) 0.01 (b) 0.99 (c) 0.52 (d) 1.66

(a) 0.1 M (b) 1.0 M 33. The number of water molecules in 1 litre of water is
(c) 0.2 M (d) 2.0 M (a) 18 (b) 18 1000
(c) N A (d) 55 .55 N A
24. A mixture having 2 g of hydrogen and 32 g of oxygen
occupies how much volume at NTP?
34. The ratio of the masses of methane and ethane In a gas
(a) 44.8L (b) 22.4L
mixture is 4 : 5. The ratio of number of their molecules
(c) 11.2L (d) 67.2L
in the mixture is:
(a) 4 : 5 (B) 3 : 2 (c) 2 : 3 (d) 5 : 4
25. Two elements ‘P’ and ‘Q’ combine to form a compound.
35. How much oxygen is required for complete combustion
Atomic mass of ‘P’ is 12 and ‘Q’ is 16. Percentage of ‘P’
of 560 g of ethane?
in the compound is 27.3 what will be the empirical
(a) 6.4 kg (b) 1.92kg (c) 2.8kg (d) 9.6kg
formula of the compound?
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CH-XI-D.T-2-20260708
 CHAPTER TEST-2 KEY CHEMISTRY CLASS: XI-NEET/JEE
TOPIC: Some basic Concepts of Chemistry Dt: 08.07.25

= 148 = n = 1
1. (b) Mass of solute = 89 g Molecular formula = BeN4Cl2H12
Mass of solution = 100 g 6. (b)

Volume of solution 1.Calculate the molar mass of CuSO4 ⋅ 5H2 O :

Mass 1.00 Molar mass = 63.5 + 32 + (4 × 16) + 5 × (2 × 1 + 16)
= = = 54.34mL
Density 184 Molar mass = 63.5 + 32 + 64 + 50 = 209.5 g/mol
2. Calculate the number of moles of CuSO4 ⋅ 5H2 O in 0.20 g
89 1000
Molarity =  :
98 54.34 mass 0.20 g
= 16.7M Number of moles = molar mass
= 249.7 g/mol ≈ 8.01 ×
10−4 mol
2. (c) 3. Determine the number of moles of water molecules
As we know, (5H2 O) :
No. of molecules = No. of moles × Avogadro Number Since each formula unit of CuSO4 ⋅ 5H2 O contains 5 water
No. of moles =
No.of molecules molecules:
Avogadro number
Moles of water molecules = 8.01 × 10−4 mol × 5 =
Atomic Mass of C = 12u
4.005 × 10−3 mol
Atomic Mass of O = 16u
4. Calculate the number of water molecules:
Molecular Mass of CO2 = 12 + (2 × 16) = 44u
44
Number of molecules = 4.005 × 10−3 mol × 6.022 × 1023
Mass of One molecule of CO2 = molecules /mol
6.023×1023
= 7.30 × 10−23 g Number of molecules = 2.41 × 1021 molecules
3. (b) Therefore, the number of water molecules present in the
the law of multiple proportions.: two elements combine to 0.20 g sample of CuSO4 ⋅ 5H2 O is approximately 2.41 ×
form more than one compound and the masses of one element 1021.
that combine with a fixed mass of the other element is in the Answer: (b) 2.14 × 1021
ratio of small whole numbers. 7. (a) Suppose each gas has a mass of X g.
So all elements are same only the mass ratio of reactants Therefore, O2 : H2 : CH4
varies the product formation. Weight – X X X
thus in PbO and PbO2 the reactants are Pb and O2 that X X X
combine in different ratio. other options do not have all the No. of moles – 32 2 16
elements common. X X X
4. (d)84% Volume of ratio – 32 : 2 : 16
MgCO3 decomposes as below: Hence, O2 : H2 : CH4 1 : 16 : 2
MgCO3 → MgO + CO2
8. (b) : 6CaO + P4 O10 → 2Ca 3 (PO 4 ) 2
mole ratio =1:1:1
Thus 0.1 mole of CO2 will be produced by 0.1 mole of 1 mole of P4 O10 = molar mass of P4 O14 = 284g
MgCO3. 852
P4 O10 = = 3mol
Mass of MgCO3 in 10 g sample = 0.1 x 84 = 8.4 g 852 g of 284
8.4
1 mole of P4 O10 reacts with 6 moles of CaO
Percent purity = 10 x100 = 84%
5. (a)Element % %/A Simplest ratio 3 moles of P4 O10 reacts with 18 moles of CaO
Be 6.1 6.1/9 = 0.677 1 Mass of 18 moles of CaO = 18  56 = 1008g
N 37.8 37.8/14 = 2.7 4 9. (a)
Cl 48 48/35.5 = 1.35 2
H 8.1 8.1/1 = 8.1 12
Empirical formula = BeN4Cl2H12
= 9 + 56 + 71 +12

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CH-XI-D.T-2-20260708
 2KMnO4 + 3H2 SO4 + 5(COOH)2 → = ( CH2Cl )2 = C2H4Cl2
K 2 SO4 + 2MnSO4 + 8H2 O + 10CO2 Thus, Molecular formula
𝑀1 𝑉1 𝑀2 𝑉2 𝑥 × 40 mL 0.02 × 200 mL 16. (a)
∴ = ⇒ =
𝑛1 𝑛2 2 5 In W A:B 1:1
⇒ 𝑥 = 0.04M In X A:B 1: 2
10. (d) : Volume of gas in 1 L In Y A:B 1: 3
0.06 In Z A:B 1: 4
= = 6 10 −4 L
100 W 24g of A combines with 32 g of B. Then
In compound
Number of molecules of CO 2 = n  N A X24g of A will combine with 64 g of B.
in
6 10−4
 6.023 1023 = 1.611019
22.4
11. (d) 17. (d) : M1V1 + M 2 V2 = MV
2NaCl + 3H2 SO4 + MnO2 0.5  50 + 0.25 150 = M  250
→ 2NaHSO4 + MnSO4 + Cl2 + 2H2 O 25 + 37.5
= 0.25M
Molar mass of NaCl = 23 + 35.5 = 58.5 g M= 250
According to the balanced reaction 58.5 × 2 = 117 g of 18. (b)
NaCl produces 1 mol of Cl2 i.e. 22.4 L at STP. 11.2
At NTP, 11.2 L of oxygen corresponds to 22.4 = 0.5 mol
22.4
∴ 50 gNaCl will produce 117
× 50 = 9.57 LCl2 at STP heat
2KClO3 → 2KCl + 3O2
option d is correct
Thus, 2 moles of potassium chlorate on heating gives 3 moles
12. (c)2C57H110O6(s) + 163O2(g) ¾® 114CO2(g) + 110H2O
of oxygen.
(l) 2 1
Thus, 3 × 0.5 = 3 moles of potassium chlorate on heating
1
445 g = 2 mole gives 11.2 L of oxygen at NTP.
110 1 110  18 Hence, the correct option is b
g 19. (a) :
2 × 2 mole = 4 = 495 g.
Elemen Percentag Molar Relativ Simple
4.28
= = 0.107 t e ratio e whole
13. (b) : No. of moles of NaOH 40
Ratio numbe
3
Volume of solution = 250 cm r
n 0.107 ratio
M= = 1000 = 0.428mol L−1
V in L 250
14. (a) Acid base Mg 21.9 21.9 / 24 1 2
= 0.91
N1 V1
= N 2 V2 ; N1  30 = 0.2  15 ; N1 = 0.1 N
P 27.8 27.8 / 31 1 2
15. (b)
Element % Composition Number of moles Simplest molar ratio Simplest whole number
= 0.90
C 24.27 24.27 / 12 = 2.02 2.02
=1
1 O 50.3 50.3 / 16 3.48 7
2.02
H 4.07 4.07 / 1 = 4.07 4.07 2 = 3.14
=2
2.02
Cl 71.65 71.65 / 35.5 = 2.02 2..02
=1
1 Formula of the compound= Mg2 P2 O 7
202
9.125
= CH2Cl HCl = = 0.25
Thus, empirical formula 20. (c) Moles of 36.5 mole
Empirical formula mass = 12 + 2 + 35.5 = 49.5 Molesof solute
Molality = 1000
Molar mass of compound = 98.96 Massof solvent
Molar mass 98.96 0.25
n= = =2 = 1000 = 0.5m
Empiricalformula mass 49.5 500
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 M  1000 = 25.984 ≈ 26
21. (a)Explanation : m = (1000  d − M  M.Wt.) where ‘m’ is This molecular weight indicates that given
molality, M is molarity. compound is 𝐶2 𝐻2.
10−2  1000 28. (c)Equivalent weight of bivalent metal = 37.2
−2
= (1000  1.1− 10  106)  Atomic weight of metal = 37.2  2 = 74.4
10 10  Formula of chloride = MCl 2
= 1100 − 1.6 = 1099.4 = 9.00 × 10–3 [Take Hence, molecular weight of chloride
1099.4 = 1100] (MCl 2 ) = 74 .4 + 2  35 .5 = 145 .4
29. (b) 3:2

( )
 M M
 30 
22. (c)Molarity of Cl– = 3 (molarity of FeCl3) = 3 = 10 nCH4 mCH4 MC2H6
= = 3:2
(m )( M )
.
6.02  1022 1 nC2 H6 CH4 CH4
23
23. (c)Molarity = 6.02  10 ´ 1/ 2 = 0.2
24. (a) : 2g of H 2 = 1 mole, 32 g of O 2 =1 mole 30. (b) : Weight of 11.2dm3 of CO2 gas at STP is 44/2 =
Total volume of 2 moles of gases at NTP= 2 22.4L = 22 g
44.8L KOH + CO2 → KHCO3
25. (d) : 56 g 44 g
Element % No. of Mole Whole KOH required for complete neutralisation of 22 g
56
moles Ratio No. CO2 is = × 22 = 28 g
44
ratio % by weight  10  d 36.5  10  1.2
Mw 2 36.5
31. (d)M = = = 12 M
P 27.3 27.3 / 12 1 1
36.5  1000 1000
= 2.27 36.5  (100 − 36.5)
m= = 63.5 = 15.7 m
Q 72.7 72.7 /16 2 2 32. (a)
= 4.54 Solution contains glucose C6 H12 O6 and water
Molar mass of glucose = 18og
Empirical formula = PQ2 Molar mass of water = 18 g
26. (a)CaC2 + H2O ⎯⎯→ Ca(OH)2 + C2H2 ⎯⎯→ C2H4 mole fraction of glucose = moles of glucose /( moles of
... (a) gluocose + moles of water)
moles of glucose, n = mass / molar mass = 2.82/180 =
nC2H4 ⎯⎯→ ... (b)
0.016
From equation (a)
moles of water, n = mass / molar mass = 30/18 = 1.67
mole of CaC2 = mole of C2H4
mole fraction of glucose = 0.016/(0.016 + 1.67) =
64  103 0.0095 ≈ 0.01
64 = mole of C2H4 M
d=
From equation (b) 33. (d) V (d = density, M= mass, V =volume)
mole of C2H4 mole of polymer Since d = 1
n = 1 So, M = V
103 wt. of polymer 18gm = 18ml
n = n(28) 18ml = NA molecules (NA = avogadro's no.)
wt of polymer = 28 ´ 103 g = 28 Kg NA
=  1000
1000ml 18 = 55.555 NA.
27. (a)∵ 1L of gas at S.T.P. weight 1.16g
∴ 22.4 L of gas at S.T.P. weight = 22.4 × 1.16
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 nCH4 mCH4 (MC2H6 )
nC2H6 (m )(M )
34. (b) = CH CH
4
=3:2 4

C2 H4 + 3O2 → 2CO2 + 2H 2 O
35. (b) : 28g 332=96 g
28 g of C 2 H 2 requires 96g of O 2
96
 560
C H
560 g of 2 4 requires 28
= 1920g or 1.92kg of O 2

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