NEET : Chemistry Some Basic Concepts of Chemistry
20. If V mL of the vapours of substance at NTP weight 29. An oxide of sulphur contains 50% of sulphur
Exercise - I
W g. Then molecular weight of substance is:- in it. Its empirical formula is -
QUESTIONS BASED ON MOLES 10. The weight of one atom of Uranium is 238 V (1) SO2 (2) SO3 (3) SO (4) S2O
(1) (W/V) × 22400 (2) × 22.4 30. A hydrocarbon contains 80% of carbon, then
1. The number of atoms present in 16 g of amu. Its actual weight is .... g. W
oxygen is (1) 1.43 × 1026 (2) 3.94 × 10–22 W 1 the hydrocarbon is -
(1) 6.02 × 1011.5 (2) 3.01 × 1023 (3) 6.99 × 10–23 (4) 1.53 × 10–22 (3) (W - V) × 22400 (4) (1) CH4 (2) C2H4 (3) C2H6 (4) C2H2
V 22400
(3) 3.01 × 1011.5 (4) 6.02 × 1023 11. The actual weight of a molecule of water is - 31. Empirical formula of glucose is -
21. If 3.01 1020 molecules are removed from
2. The number of atoms in 4.25 g of NH3 is (1) 18 g (1) C6H12O6 (2) C3H6O3 (3) C2H4O2 (4) CH2O
98 mg of H2SO4, then the number of moles of 32. An oxide of metal M has 40% by mass of
approx:- (2) 2.99 × 10–23 g H2SO4 left are :– oxygen. Metal M has atomic mass of 24. The
(1) 1 1023 (2) 1.5 1023 (3) both (1) & (2) are correct (1) 0.1 10–3 (2) 0.5 10–3 empirical formula of the oxide is :-
(3) 2 1023 (4) 6 1023 (4) 1.66 × 10–24 g
(3) 1.66 10–3 (4) 9.95 10–2 (1) M2O (2) M2O3 (3) MO (4) M3O4
3. Which of the following contains maximum 12. What is the mass of a molecule of CH4 :–
22. A gas is found to have the formula (CO)x. It's 33. A compound contains 38.8% C, 16.0% H and
number of oxygen atoms ? (1) 16 g (2) 26.6 1022 g VD is 70. The value of x must be:- 45.2% N. The formula of the compound
(1) 1 g of O (3) 2.66 10–23 g (4) 16 NA g (1) 7 (2) 4 (3) 5 (4) 6 would be
(2) 1 g of O2 (1) CH3NH2 (2) CH3CN
13. Which of the following has the highest mass ? 23. Vapour density of gas is 11.2. Volume
(3) 1 g of O3 (3) C2H5CN (4) CH2(NH)2
(1) 1 g atom of C occupied by 2.4 g of this at STP will be -
(4) all have the same number of atoms 34. The simplest formula of a compound
(2) 1/2 mole of CH4 (1) 11.2 L (2) 2.24 L (3) 22.4 L (4) 2.4 L
4. The number of atoms present in 0.5 g atom of containing 50%(w/w) of element X(at wt. =
(3) 10 mL of water 24. The volume of a gas in discharge tube is
nitrogen is same as the atoms in – 10) and 50% of element Y(at wt. = 20) is:-
(4) 3.011 × 1023 atoms of oxygen 1.12 × 10–7 mL at STP. Then the number of
(1) 12 g of C (2) 32 g of S (1) XY (2) X2Y (3) XY2 (4) X3Y
14. Which of the following contains the least molecule of gas in the tube is -
(3) 8 g of oxygen (4) 24 g of Mg 35. Which of the following compound has same
number of molecules ? (1) 3.01 × 104 (2) 3.01 × 1015
5. Which of the following contains maximum empirical formula as that of glucose:-
(1) 4.4 g CO2 (2) 3.4 g NH3 (3) 3.01 × 1012 (4) 3.01 × 1016
number of atoms ? (1) CH3CHO (2) CH3COOH
(3) 1.6 g CH4 (4) 3.2 g SO2 25. The number of electron in 3.1 mg NO3¯ is
(1) 4 g of H2 (2) 16 g of O2 (3) CH3OH (4) C2H6
15. The number of molecule in 4.25 g of NH3 is - (NA = 6 × 1023)
(3) 28 g of N2 (4) 18 g of H2O 36. 2.2 g of a compound of phosphorous and
(1) 1.505 × 1023 (2) 3.01 × 1023 (1) 32 (2) 1.6 × 10–3
6. Number of neutrons present in 1.7 g of sulphur has 1.24 g of 'P' in it. Its empirical
ammonia is - (3) 6.02 × 1023 (4) None of these (3) 9.6 × 1020 (4) 9.6 × 1023
formula is -
(1) NA (2) NA/10 × 4 16. Elements A and B form two compounds B2A3 26. Given that one mole of N2 at NTP occupies
(1) P2S3 (2) P3S2 (3) P3S4 (4) P4S3
(3) (NA/10) × 7 (4) NA × 10 × 7 and B2A. 0.05 moles of B2A3 weight 9.0 g and 22.4 L the density of N2 is -
37. On analysis, a certain compound was found to
7. 5.6 L of oxygen at STP contains - 0.10 mole of B2A weight 10 g. Calculate the (1) 1.25 g L–1 (2) 0.80 g L–1
contain iodine and oxygen in the mass ratio of
(1) 6.02 × 1023 atoms (3) 2.5 g L–1 (4) 1.60 g L–1
atomic weight of A and B :- 254:80. The formula of the compound is :
(2) 3.01 × 1023 atoms 27. The number of carbon atoms present in a
(1) 20 and 30 (2) 30 and 40 (At mass I = 127, O = 16)
(3) 1.505 × 1023 atoms signature, if a signature written by carbon (1) IO (2) I2O (3) I5O2 (4) I2O5
(3) 40 and 30 (4) 30 and 20
(4) 0.7525×1023 atoms pencil, weighing 1.2 × 10–3 g is 38. The number of atoms of Cr and O are 4.8 × 1010
17. 5.6 L of oxygen at NTP is equivalent to –
8. Number of oxygen atoms in 8 g of ozone is - (1) 12.04 × 1020 (2) 6.02 × 1019 and 9.6 × 1010 respectively. Its empirical
(1) 1 mol (2) 1/2 mol
(3) 3.01 × 1019 (4) 6.02 × 1020 formula is –
6.02 1023 (3) 1/4 mol (4) 1/8 mol
(1) 6.02 × 1023 (2) (1) Cr2O3 (2) CrO2 (3) Cr2O4 (4) CrO5
2 18. 4.4 g of an unknown gas occupies 2.24 L of QUESTIONS BASED ON PERCENTAGE, EMPIRICAL 39. Insulin contains 3.4% sulphur by mass. The
6.02 1023 6.02 1023 volume at STP. The gas may be :- FORMULA & MOLECULAR FORMULA
(3) (4) minimum molecular weight of insulin is :
3 6 (1) N2O (2) CO 28. A compound of X and Y has equal mass of (1) 941.176 (2) 944 (3) 945.27 (4) None
9. Sum of number of protons, electrons and (3) CO2 (4) 1 & 3 both them. If their atomic weights are 30 and 20 40. Caffine has a molecular weight of 194. It
neutrons in 12g of 126 C is :- 19. Which contains least number of molecules :– respectively. Molecular formula of the contains 28.9% by mass of nitrogen Number
(1) 1.8 (2) 12.044 × 1023 (1) 1 g CO2 (2) 1 g N2 compound is :- of atoms of nitrogen in one molecule of it is :-
(3) 1.084 × 1025 (4) 10.84 × 1023 (3) 1 g O2 (4) 1 g H2 (1) X2Y2 (2) X3Y3 (3) X2Y3 (4) X3Y2 (1) 2 (2) 3 (3) 4 (4) 5
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, NEET : Chemistry Some Basic Concepts of Chemistry
QUESTIONS BASED ON STOICHIOMETRY 47. 1 L of CO2 is passed over hot coke. When the 55. 12 L of H2 and 11.2 L of Cl2 are mixed and 65. 0.126 g of an acid requires 20 mL of 0.1 N
41. An organic compound contains carbon, volume of reaction mixture becomes 1.4 L, exploded. The composition by volume of NaOH for complete neutralisation. Equivalent
hydrogen and oxygen. Its elemental analysis the composition of reaction mixture is– mixture is– weight of the acid is –
gives 38.71% of C and 9.67% of H. The (1) 0.6 L CO (1) 24 L of HCl (g) (1) 45 (2) 53 (3) 40 (4) 63
empirical formula of the compound would be :- (2) 0.8 L CO2 (2) 0.8 L Cl2 and 20.8 L HCl (g) 66. 2g of a base whose equivalent weight is 40
(1) CHO (2) CH4O (3) CH3O (4) CH2O (3) 0.6 L CO2 and 0.8 L CO (3) 0.8 L H2 and 22.4 L HCl (g) reacts with 3 g of an acid. The equivalent
42. The amount of water (g) produced by (4) None (4) 22.4 L HCl (g) weight of the acid is :
combustion of 286 g of propane is 48. 26 cc of CO2 are passed over red hot coke. 56. 10 mL of gaseous hydrocarbon on (1) 40 (2) 60 (3) 10 (4) 80
(1) 168 g (2) 200 g (3) 468 g (4) 693 g combustion give 40 mL of CO2(g) and 50 mL
The volume of CO evolved is :– 67. Equivalent weight of a divalent metal is 24.
43. In a gaseous reaction of the type of H2O (vap.). The hydrocarbon is -
(1) 15 cc (2) 10 cc (3) 32 cc (4) 52 cc The volume of hydrogen liberated at STP by
aA + bB ⎯→ cC + dD, (1) C4H5 (2) C8H10 (3) C4H8 (4) C4H10
49. If 1/2 mol of oxygen combine with Aluminium to 12 g of the same metal when added to excess
which statement is wrong ?
form Al2O3 then weight of Aluminium metal used QUESTIONS BASED ON EQUIVALENT WEIGHTS of an acid solution is -
(1) A litre of A combines with b litre of B to
in the reaction is (Al=27) – 57. Molecular weight of tribasic acid is W. Its (1) 2.8 litres (2) 5.6 litres
give C and D
equivalent weight will be : (3) 11.2 litres (4) 22.4 litres
(2) A mole of A combines with b moles of B (1) 27 g (2) 18 g (3) 54 g (4) 40.5 g
to give C and D W W 68. 0.84 g of a metal carbonate reacts exactly
50. If 0.5 mol of BaCl2 is mixed with 0.2 mol of (1) (2) (3) W (4) 3W
(3) A g of A combines with b g of B to give 2 3 with 40 mL of N/2 H2SO4. The equivalent
Na3PO4, the maximum number of moles of
C and D 58. A, E, M and n are the atomic weight, weight of the metal carbonate is -
Ba3(PO4)2 that can be formed is – equivalent weight, molecular weight and
(4) A molecules of A combines with b (1) 84 (2) 64 (3) 42 (4) 32
molecules of B to give C and D 3BaCl2 + 2Na3PO4 → Ba3(PO4)2 + 6NaCl valency of an element. The correct relation is: 69. 1.0 g of a metal combines with 8.89 g of
44. Assuming that petrol is octane (C8H18) and (1) 0.7 (2) 0.5 (3) 0.3 (4) 0.1 M Bromine. Equivalent weight of the metal is
(1) A = E × n (2) A =
has density 0.8 g mL–1. 1.425 L of petrol on 51. If 8 mL of uncombined O2 remain after E nearly (at.wt. of Br = 80)
complete combustion will consume. exploding O2 with 4 mL of hydrogen, the M (1) 8 (2) 9 (3) 10 (4) 7
(1) 50 mole of O2 (2) 100 mole of O2 (3) A = (4) M = A × n
number of mL of O2 originally were - n 70. 0.84 g of metal hydride contains 0.04 g of
(3) 125 mole of O2 (4) 200 mole of O2 hydrogen. The equivalent wt. of the metal is....
(1) 12 (2) 2 (3) 10 (4) 4 59. Sulphur forms two chlorides S2Cl2 and SCl2.
45. In a given reaction, 9 g of Al will react with
52. 4 g of hydrogen are ignited with 4 g of oxygen. The equivalent mass of sulphur in SCl2 is 16. (1) 80 (2) 40 (3) 20 (4) 60
3 The equivalent weight of sulphur in S2Cl2 is -
2Al + O2 → Al2 O3 The weight of water formed is - 71. When an element forms an oxide in which
2 (1) 8 (2) 16 (3) 32 (4) 64 oxygen is 20% of the oxide by mass, the
(1) 0.5 g (2) 3.5 g (3) 4.5 g (4) 2.5 g
(1) 6 g O2 (2) 8 g O2 (3) 9 g O2 (4) 4 g O2 60. If equivalent weight of S in SO2 is 8 then equivalent mass of the element will be –
46. The equation : 53. For the reaction A + 2B ⎯→ C, equivalent weight of S in SO3 is - (1) 32 (2) 40 (3) 60 (4) 128
3 5 mol of A and 8 mol of B will produce 82 83 2 3
2Al(s) + O2(g) → Al2O3(s) shows that :- (1) (2) (3) 8×2×3 (4) 72. 2.8 g of iron displaces 3.2 g of copper from a
2 (1) 5 mole of C (2) 4 mole of C 3 2 8 solution of copper sulphate. If the equivalent
3 (3) 8 mole of C (4) 13 mole of C 61. Which property of an element is not variable:
(1) 2 mol of Al reacts with mol of O2 to mass of iron is 28, then equivalent mass of
2 54. If 1.6 g of SO2 and 1.5 × 1022 molecules of H2S (1) Valency (2) Atomic weight copper will be –
7 are mixed and allowed to remain in contact in (3) Equivalent weight (4) None
produce mol of Al2O3 (1) 16 (2) 32 (3) 48 (4) 64
2 62. One g equivalent of a substance is present in -
a closed vessel until the reaction 73. If m1 g of a metal A displaces m2 g of another
3 (1) 0.25 mol of O2 (2) 0.5 mol of O2
(2) 2 g of Al reacts with g of O2 to produce 2H2S + SO2 ⎯→ 3S + 2H2O, metal B from its salt solution and if their
2 (3) 1.00 mol of O2 (4) 8.00 mol of O2
proceeds to completion. Which of the equivalent weight are E2 and E1 respectively
one mol of Al2O3 63. 0.45 g of acid (molecular wt. = 90) was exactly
3 following statement is true ? then the equivalent weight of A can be
(3) 2 g of Al reacts with L of O2 to produce neutralised by 20 mL of 0.5 N NaOH. Basicity
2 expressed by:-
(1) Only 'S' and 'H2O' remain in the reaction of the acid is -
1 mol of Al2O3 vessel. (1) 1 (2) 2 (3) 3 (4) 4 m m
(1) 1 × E2 (2) 2 × E2
3 (2) 'H2S' will remain in excess 64. 0.5 g of a base was completely neutralised by 100 m2 m1
(4) 2 mol of Al reacts with mol of O2 to
2 (3) 'SO2' will remain in excess mL of 0.2 N acid. Equivalent weight of the base is m1 m2
produce 1 mol of Al2O3 (4) None (1) 50 (2) 100 (3) 25 (4) 125 (3) × E1 (4) × E1
m2 m1
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20. If V mL of the vapours of substance at NTP weight 29. An oxide of sulphur contains 50% of sulphur
Exercise - I
W g. Then molecular weight of substance is:- in it. Its empirical formula is -
QUESTIONS BASED ON MOLES 10. The weight of one atom of Uranium is 238 V (1) SO2 (2) SO3 (3) SO (4) S2O
(1) (W/V) × 22400 (2) × 22.4 30. A hydrocarbon contains 80% of carbon, then
1. The number of atoms present in 16 g of amu. Its actual weight is .... g. W
oxygen is (1) 1.43 × 1026 (2) 3.94 × 10–22 W 1 the hydrocarbon is -
(1) 6.02 × 1011.5 (2) 3.01 × 1023 (3) 6.99 × 10–23 (4) 1.53 × 10–22 (3) (W - V) × 22400 (4) (1) CH4 (2) C2H4 (3) C2H6 (4) C2H2
V 22400
(3) 3.01 × 1011.5 (4) 6.02 × 1023 11. The actual weight of a molecule of water is - 31. Empirical formula of glucose is -
21. If 3.01 1020 molecules are removed from
2. The number of atoms in 4.25 g of NH3 is (1) 18 g (1) C6H12O6 (2) C3H6O3 (3) C2H4O2 (4) CH2O
98 mg of H2SO4, then the number of moles of 32. An oxide of metal M has 40% by mass of
approx:- (2) 2.99 × 10–23 g H2SO4 left are :– oxygen. Metal M has atomic mass of 24. The
(1) 1 1023 (2) 1.5 1023 (3) both (1) & (2) are correct (1) 0.1 10–3 (2) 0.5 10–3 empirical formula of the oxide is :-
(3) 2 1023 (4) 6 1023 (4) 1.66 × 10–24 g
(3) 1.66 10–3 (4) 9.95 10–2 (1) M2O (2) M2O3 (3) MO (4) M3O4
3. Which of the following contains maximum 12. What is the mass of a molecule of CH4 :–
22. A gas is found to have the formula (CO)x. It's 33. A compound contains 38.8% C, 16.0% H and
number of oxygen atoms ? (1) 16 g (2) 26.6 1022 g VD is 70. The value of x must be:- 45.2% N. The formula of the compound
(1) 1 g of O (3) 2.66 10–23 g (4) 16 NA g (1) 7 (2) 4 (3) 5 (4) 6 would be
(2) 1 g of O2 (1) CH3NH2 (2) CH3CN
13. Which of the following has the highest mass ? 23. Vapour density of gas is 11.2. Volume
(3) 1 g of O3 (3) C2H5CN (4) CH2(NH)2
(1) 1 g atom of C occupied by 2.4 g of this at STP will be -
(4) all have the same number of atoms 34. The simplest formula of a compound
(2) 1/2 mole of CH4 (1) 11.2 L (2) 2.24 L (3) 22.4 L (4) 2.4 L
4. The number of atoms present in 0.5 g atom of containing 50%(w/w) of element X(at wt. =
(3) 10 mL of water 24. The volume of a gas in discharge tube is
nitrogen is same as the atoms in – 10) and 50% of element Y(at wt. = 20) is:-
(4) 3.011 × 1023 atoms of oxygen 1.12 × 10–7 mL at STP. Then the number of
(1) 12 g of C (2) 32 g of S (1) XY (2) X2Y (3) XY2 (4) X3Y
14. Which of the following contains the least molecule of gas in the tube is -
(3) 8 g of oxygen (4) 24 g of Mg 35. Which of the following compound has same
number of molecules ? (1) 3.01 × 104 (2) 3.01 × 1015
5. Which of the following contains maximum empirical formula as that of glucose:-
(1) 4.4 g CO2 (2) 3.4 g NH3 (3) 3.01 × 1012 (4) 3.01 × 1016
number of atoms ? (1) CH3CHO (2) CH3COOH
(3) 1.6 g CH4 (4) 3.2 g SO2 25. The number of electron in 3.1 mg NO3¯ is
(1) 4 g of H2 (2) 16 g of O2 (3) CH3OH (4) C2H6
15. The number of molecule in 4.25 g of NH3 is - (NA = 6 × 1023)
(3) 28 g of N2 (4) 18 g of H2O 36. 2.2 g of a compound of phosphorous and
(1) 1.505 × 1023 (2) 3.01 × 1023 (1) 32 (2) 1.6 × 10–3
6. Number of neutrons present in 1.7 g of sulphur has 1.24 g of 'P' in it. Its empirical
ammonia is - (3) 6.02 × 1023 (4) None of these (3) 9.6 × 1020 (4) 9.6 × 1023
formula is -
(1) NA (2) NA/10 × 4 16. Elements A and B form two compounds B2A3 26. Given that one mole of N2 at NTP occupies
(1) P2S3 (2) P3S2 (3) P3S4 (4) P4S3
(3) (NA/10) × 7 (4) NA × 10 × 7 and B2A. 0.05 moles of B2A3 weight 9.0 g and 22.4 L the density of N2 is -
37. On analysis, a certain compound was found to
7. 5.6 L of oxygen at STP contains - 0.10 mole of B2A weight 10 g. Calculate the (1) 1.25 g L–1 (2) 0.80 g L–1
contain iodine and oxygen in the mass ratio of
(1) 6.02 × 1023 atoms (3) 2.5 g L–1 (4) 1.60 g L–1
atomic weight of A and B :- 254:80. The formula of the compound is :
(2) 3.01 × 1023 atoms 27. The number of carbon atoms present in a
(1) 20 and 30 (2) 30 and 40 (At mass I = 127, O = 16)
(3) 1.505 × 1023 atoms signature, if a signature written by carbon (1) IO (2) I2O (3) I5O2 (4) I2O5
(3) 40 and 30 (4) 30 and 20
(4) 0.7525×1023 atoms pencil, weighing 1.2 × 10–3 g is 38. The number of atoms of Cr and O are 4.8 × 1010
17. 5.6 L of oxygen at NTP is equivalent to –
8. Number of oxygen atoms in 8 g of ozone is - (1) 12.04 × 1020 (2) 6.02 × 1019 and 9.6 × 1010 respectively. Its empirical
(1) 1 mol (2) 1/2 mol
(3) 3.01 × 1019 (4) 6.02 × 1020 formula is –
6.02 1023 (3) 1/4 mol (4) 1/8 mol
(1) 6.02 × 1023 (2) (1) Cr2O3 (2) CrO2 (3) Cr2O4 (4) CrO5
2 18. 4.4 g of an unknown gas occupies 2.24 L of QUESTIONS BASED ON PERCENTAGE, EMPIRICAL 39. Insulin contains 3.4% sulphur by mass. The
6.02 1023 6.02 1023 volume at STP. The gas may be :- FORMULA & MOLECULAR FORMULA
(3) (4) minimum molecular weight of insulin is :
3 6 (1) N2O (2) CO 28. A compound of X and Y has equal mass of (1) 941.176 (2) 944 (3) 945.27 (4) None
9. Sum of number of protons, electrons and (3) CO2 (4) 1 & 3 both them. If their atomic weights are 30 and 20 40. Caffine has a molecular weight of 194. It
neutrons in 12g of 126 C is :- 19. Which contains least number of molecules :– respectively. Molecular formula of the contains 28.9% by mass of nitrogen Number
(1) 1.8 (2) 12.044 × 1023 (1) 1 g CO2 (2) 1 g N2 compound is :- of atoms of nitrogen in one molecule of it is :-
(3) 1.084 × 1025 (4) 10.84 × 1023 (3) 1 g O2 (4) 1 g H2 (1) X2Y2 (2) X3Y3 (3) X2Y3 (4) X3Y2 (1) 2 (2) 3 (3) 4 (4) 5
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, NEET : Chemistry Some Basic Concepts of Chemistry
QUESTIONS BASED ON STOICHIOMETRY 47. 1 L of CO2 is passed over hot coke. When the 55. 12 L of H2 and 11.2 L of Cl2 are mixed and 65. 0.126 g of an acid requires 20 mL of 0.1 N
41. An organic compound contains carbon, volume of reaction mixture becomes 1.4 L, exploded. The composition by volume of NaOH for complete neutralisation. Equivalent
hydrogen and oxygen. Its elemental analysis the composition of reaction mixture is– mixture is– weight of the acid is –
gives 38.71% of C and 9.67% of H. The (1) 0.6 L CO (1) 24 L of HCl (g) (1) 45 (2) 53 (3) 40 (4) 63
empirical formula of the compound would be :- (2) 0.8 L CO2 (2) 0.8 L Cl2 and 20.8 L HCl (g) 66. 2g of a base whose equivalent weight is 40
(1) CHO (2) CH4O (3) CH3O (4) CH2O (3) 0.6 L CO2 and 0.8 L CO (3) 0.8 L H2 and 22.4 L HCl (g) reacts with 3 g of an acid. The equivalent
42. The amount of water (g) produced by (4) None (4) 22.4 L HCl (g) weight of the acid is :
combustion of 286 g of propane is 48. 26 cc of CO2 are passed over red hot coke. 56. 10 mL of gaseous hydrocarbon on (1) 40 (2) 60 (3) 10 (4) 80
(1) 168 g (2) 200 g (3) 468 g (4) 693 g combustion give 40 mL of CO2(g) and 50 mL
The volume of CO evolved is :– 67. Equivalent weight of a divalent metal is 24.
43. In a gaseous reaction of the type of H2O (vap.). The hydrocarbon is -
(1) 15 cc (2) 10 cc (3) 32 cc (4) 52 cc The volume of hydrogen liberated at STP by
aA + bB ⎯→ cC + dD, (1) C4H5 (2) C8H10 (3) C4H8 (4) C4H10
49. If 1/2 mol of oxygen combine with Aluminium to 12 g of the same metal when added to excess
which statement is wrong ?
form Al2O3 then weight of Aluminium metal used QUESTIONS BASED ON EQUIVALENT WEIGHTS of an acid solution is -
(1) A litre of A combines with b litre of B to
in the reaction is (Al=27) – 57. Molecular weight of tribasic acid is W. Its (1) 2.8 litres (2) 5.6 litres
give C and D
equivalent weight will be : (3) 11.2 litres (4) 22.4 litres
(2) A mole of A combines with b moles of B (1) 27 g (2) 18 g (3) 54 g (4) 40.5 g
to give C and D W W 68. 0.84 g of a metal carbonate reacts exactly
50. If 0.5 mol of BaCl2 is mixed with 0.2 mol of (1) (2) (3) W (4) 3W
(3) A g of A combines with b g of B to give 2 3 with 40 mL of N/2 H2SO4. The equivalent
Na3PO4, the maximum number of moles of
C and D 58. A, E, M and n are the atomic weight, weight of the metal carbonate is -
Ba3(PO4)2 that can be formed is – equivalent weight, molecular weight and
(4) A molecules of A combines with b (1) 84 (2) 64 (3) 42 (4) 32
molecules of B to give C and D 3BaCl2 + 2Na3PO4 → Ba3(PO4)2 + 6NaCl valency of an element. The correct relation is: 69. 1.0 g of a metal combines with 8.89 g of
44. Assuming that petrol is octane (C8H18) and (1) 0.7 (2) 0.5 (3) 0.3 (4) 0.1 M Bromine. Equivalent weight of the metal is
(1) A = E × n (2) A =
has density 0.8 g mL–1. 1.425 L of petrol on 51. If 8 mL of uncombined O2 remain after E nearly (at.wt. of Br = 80)
complete combustion will consume. exploding O2 with 4 mL of hydrogen, the M (1) 8 (2) 9 (3) 10 (4) 7
(1) 50 mole of O2 (2) 100 mole of O2 (3) A = (4) M = A × n
number of mL of O2 originally were - n 70. 0.84 g of metal hydride contains 0.04 g of
(3) 125 mole of O2 (4) 200 mole of O2 hydrogen. The equivalent wt. of the metal is....
(1) 12 (2) 2 (3) 10 (4) 4 59. Sulphur forms two chlorides S2Cl2 and SCl2.
45. In a given reaction, 9 g of Al will react with
52. 4 g of hydrogen are ignited with 4 g of oxygen. The equivalent mass of sulphur in SCl2 is 16. (1) 80 (2) 40 (3) 20 (4) 60
3 The equivalent weight of sulphur in S2Cl2 is -
2Al + O2 → Al2 O3 The weight of water formed is - 71. When an element forms an oxide in which
2 (1) 8 (2) 16 (3) 32 (4) 64 oxygen is 20% of the oxide by mass, the
(1) 0.5 g (2) 3.5 g (3) 4.5 g (4) 2.5 g
(1) 6 g O2 (2) 8 g O2 (3) 9 g O2 (4) 4 g O2 60. If equivalent weight of S in SO2 is 8 then equivalent mass of the element will be –
46. The equation : 53. For the reaction A + 2B ⎯→ C, equivalent weight of S in SO3 is - (1) 32 (2) 40 (3) 60 (4) 128
3 5 mol of A and 8 mol of B will produce 82 83 2 3
2Al(s) + O2(g) → Al2O3(s) shows that :- (1) (2) (3) 8×2×3 (4) 72. 2.8 g of iron displaces 3.2 g of copper from a
2 (1) 5 mole of C (2) 4 mole of C 3 2 8 solution of copper sulphate. If the equivalent
3 (3) 8 mole of C (4) 13 mole of C 61. Which property of an element is not variable:
(1) 2 mol of Al reacts with mol of O2 to mass of iron is 28, then equivalent mass of
2 54. If 1.6 g of SO2 and 1.5 × 1022 molecules of H2S (1) Valency (2) Atomic weight copper will be –
7 are mixed and allowed to remain in contact in (3) Equivalent weight (4) None
produce mol of Al2O3 (1) 16 (2) 32 (3) 48 (4) 64
2 62. One g equivalent of a substance is present in -
a closed vessel until the reaction 73. If m1 g of a metal A displaces m2 g of another
3 (1) 0.25 mol of O2 (2) 0.5 mol of O2
(2) 2 g of Al reacts with g of O2 to produce 2H2S + SO2 ⎯→ 3S + 2H2O, metal B from its salt solution and if their
2 (3) 1.00 mol of O2 (4) 8.00 mol of O2
proceeds to completion. Which of the equivalent weight are E2 and E1 respectively
one mol of Al2O3 63. 0.45 g of acid (molecular wt. = 90) was exactly
3 following statement is true ? then the equivalent weight of A can be
(3) 2 g of Al reacts with L of O2 to produce neutralised by 20 mL of 0.5 N NaOH. Basicity
2 expressed by:-
(1) Only 'S' and 'H2O' remain in the reaction of the acid is -
1 mol of Al2O3 vessel. (1) 1 (2) 2 (3) 3 (4) 4 m m
(1) 1 × E2 (2) 2 × E2
3 (2) 'H2S' will remain in excess 64. 0.5 g of a base was completely neutralised by 100 m2 m1
(4) 2 mol of Al reacts with mol of O2 to
2 (3) 'SO2' will remain in excess mL of 0.2 N acid. Equivalent weight of the base is m1 m2
produce 1 mol of Al2O3 (4) None (1) 50 (2) 100 (3) 25 (4) 125 (3) × E1 (4) × E1
m2 m1
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