Q.1. The number of significant figures in π are:
(a) One
(b) Two
(c) Three
(d) Infinite.
Ans. (d)
Solution.
Significant figures in π are infinite it is irrational number (22/7) there are infinite digit in π = 3.1428571432479 (more than trillion digits in π)
Q.2. Given the numbers 786, 0.786 and 0.0786. The number of significant figures for the three numbers is:
(a) 3, 4 and 5 respectively
(b) 3, 3 and 3 respectively
(c) 3, 3 and 4 respectively
(d) 3, 4 and 4 respectively
Ans. (b)
Solution.
If zero at the left hand side of non zero digit it is not a significant figures as
0.004321 = 4
0.00231 = 3
786 = 3 (s.f), 0.786 = 3 (s.f), 0.0786 = 3 (s.f)
Q.3. In which of the following numbers all zeros are significant?
(a) 0.0005
(b) 0.0500
(c) 50.000
(d) 0.0050
Ans. (c)
Solution.
(a) 0.0005 = no zero considered in significant number
(b) 0.0500 = only two zero considered in significant number
(c) 50.000 = all zero considered in significant number
(d) 0.0050 = only one zero consider in significant number
Q.4. 108 ÷ 7.2 = 14.583. The correct answer to this problem in proper number of significant digits is
(a) 15
(b) 14.58
(c) 14.5
(d) None of these
Ans. (a)
Solution.
In the multiplication & division significant figures are adjusted according to the least significant figure
Q.5. 14.90 + 0.0070 + 1.0 + 0.081 = 15.9880. The sum to proper number of significant digit is
(a) 15.9
(b) 16.0
(c) 15.99
(d) 16
Ans. (b)
Solution.
In the result of addition & subtraction least significant figures are adjusted after the decimal place.
3.21 ( 3. Significant figure – 2 decimal places)
14.90 + 0.0070 + 1.0 + 0.081 = 15.9880
Answer will be reported after decimal place one digit
15.9880 → Rounding off upto 3 digit
More than 5 ↑ so, previous digit increased by one
= 16.0
Q.6. Which is the larger quantity?
(a) Mega
(b) Femto
(c) Milli
(d) Giga
Ans. (d)
Solution.
Q.7. The scientific notation of 0.0000000540 is
(a) 5.40 × 10^{7}
(b) 5.40 × 10^{8}
(c) 54.0 × 10^{7}
(d) 54.0 × 10^{9}
Ans. (b)
Solution.
Scientific notation
= 0.0000000540 = 5.40 × 10^{8}
Rule: N = N is the number between 1.000 & 9.999
n = exponent
1472.92 → decimal shifter to L.H.S
Then n = 3 = (+ve)
1.47292 × 10^{3}
Decimal shifted R.H.S than 10^{y} = y (−ve)
0.00762 = 7.62 × 10^{4}= 4 (−ve)
Q.8. 3g of a hydrocarbon on combustion with 11.2 g of oxygen produces 8.8g CO_{2} and 5.4g H_{2}O. The data illustrates the law of
(a) Conservation of mass
(b) Multiple proportions
(c) Definite proportions
(d) Reciprocal proportions
Ans. (a)
Solution.
Law of conservation of mass (lavoise)
In a general chemical reaction total weight of reactant is equal to total weight if product
Total weight g reactant = 14.2 g
= Total weight of product = 14.2 g
Q.9. The gram mole of a gas at N.T. P. occupies 22.4 L. This fact was derived from
(a) Law of gaseous volumes
(b) Avogadro’s hypothesis
(c) Berzelius hypothesis
(d) Dalton’s atomic theory
Ans. (b)
Solution.
Avogadro’s hypothesis at similar condition of temperature & pressure equal volume of different gates contains equal number of males or molecules
N_{2} + 3H_{2} ⇌ 2NH_{3}
Mole  1  3  2 
Molecule  1  3  2 
Volume  1  3  2 
Q.10. Among the following pairs of compound the one that illustrates the law of multiple proportion is
(a) NH_{3} and NF_{3}
(b) CO_{2} and CS_{2}
(c) CS_{2} and FeSO_{4 }
(d) SnCl_{2}, SnCl_{4}
Ans. (d)
Solution.
Sn + Cl_{2 }→_{ }SnCl_{2 } 118 : 71
Sn + 2Cl_{2} _{ }→ SnCl_{4 } 118 : 142
Ratio = 1:2
Q.11. The percentage of carbon and oxygen in samples of CO_{2} obtained by different methods were found to be the same. This illustrates the law of:
(a) Conservation of mass
(b) Constant proportions
(c) Multiple proportions
(d) Reciprocal proportions
Ans. (b)
Solution.
C + O_{2 }_{ }→ CO_{2}_{ } 12: 32
Sn + 2Cl_{2} → CaO + CO_{2} 12: 32
2NaHCO_{3} → Na_{2}CO_{3} + CO_{2} +H_{2}O 12 ∶ 32
Formation of a compound from different method / different source always contain same type of element in a fixed proportion by weight.
Q.12. One of the following combinations which illustrates the law of reciprocal proportions?
(a) N_{2}O_{3}. N_{2}O_{4}, N_{2}O_{5}
(b) NaCl, NaBr, NaI
(c) CS_{2}, CO_{2}, SO_{2}
(d) MgO, Mg(OH)_{2}
Ans. (c)
Solution.
Weight ratio = s/o = 32/32 + 1 : 1
Eq Ratio 8/8 = 1:1
The elements (S) & (O) combine separately with the third element (C) to form CS_{2} & CO_{2 }they combine directly with each other to form SO_{2}_{. }
CS_{2}/CO_{2} = SO_{2} weight ratio
12/72 x 32/12 = 32/32
= 1: 1 = 1:1
Equivalent ratio:
CS_{2}/CO_{2} = SO_{2}
8 : 8 = 8 : 8
1 : 1 = 1 : 1
⇒ 12/4 × 32/4
⇒ 3:8/3:8 = 1 : 1
S : O_{2}
⇒ 32/4 = 8
8 : 8 ⇒ 1 : 1
Equivalent ratio & weight ratio both are same it follow equivalent proportion & reciprocal proportion
Q.13. Two elements X and Y combine in gaseous state to form XY in the ratio 1: 35 .5 by mass.
The mass of Y that will be required to react with 2g of X is
(a) 7.1 g
(b) 3.55 g
(c) 71 g
(d) 35.5 g
Ans. (c)
Solution.
To determine the mass of Y required to react with 2g of X, we need to consider the given ratio of X and Y in the formation of XY.
According to the given ratio, the mass of X is 1 unit, and the mass of Y is 35.5 units (1 + 35.5 = 36.5 units total).
Now, if 1 unit of X reacts with 35.5 units of Y to form XY, we can set up a proportion to find the mass of Y required to react with 2g of X.
1 unit of X / 35.5 units of Y = 2g of X / mass of Y
Crossmultiplying, we get:
1 * mass of Y = 35.5 * 2
mass of Y = (35.5 * 2) / 1mass of Y = 71 g
Therefore, the mass of Y required to react with 2g of X is 71 g.
Q.14. Calculate the number of gram atoms in 2.3 g of sodium.
(a) 23
(b) 10
(c) 2.24
(d) 0.1
Ans. (d)
Solution.
Gram atom or mole = 2.3/2.3 = 0.1
Q.15. Calculate the mass of 2.5 gram atoms of oxygen
(a) 40 g
(b) 80.0 g
(c) 8 g
(d) 11.2 g
Ans. (a)
Solution.
2.5 gram atoms = 2.5 mole of oxygen
1 mole (o)  16 gm
2.5 (o)  16 × 2.5
= 40 gm
Q.16. Calculate the mass of 1.5 gram molecule of sulphuric acid
(a) 151 g
(b) 129.0
(c) 147.0 g
(d) 200 gm
Ans. (c)
Solution.
1.5 gram molecule H_{2}SO_{4}= 1.5 mole of H_{2}So_{4}
1 mole of H_{2}SO_{4}  98 g
1.5 mole  98 × 1.5
= 147 g
Q.17. 19.7 kg gold was recovered from a smuggler. The atoms of gold recovered are: (Au = 197)
(a) 100
(b) 6.02 × 10^{23}
(c) 6.02 × 10^{24 }
(d) 6.02 × 10^{25}
Ans. (d)
Solution.
mole = 19.7 Kg/197 = 197 × 100/197 = 100 mole
mole×N_{A} = total no. of atom
= 100 × 6.023 × 10^{23}
=^{ }6.023 × 10^{25}
Q.18. The molecular mass of CO_{2} is 44 amu and Avogadro’s number is 6.02 × 10^{23}. Therefore, the mass of one molecule of CO_{2} is:
(a) 7.31 × 10^{}^{23}
(b) 3.65 × 10^{}^{23}
(c) 1.01 × 10^{}^{23 }
(d) 2.01 × 10^{}^{23}
Ans. (a)
Solution.
N_{A} molecules of Co_{2} = 44 gm
1 molecules = 44/6.023×10^{23}
= 7.31 × 10^{23 }gm
Q.19. What is the mass of 3.01 x 10^{22} molecules of ammonia?
(a) 1 kg.
(b) 0.85 g
(c) 2 gm
(d) 5 mg
Ans. (b)
Solution.
N_{A} molecules of NH_{3} = 17 gm
3 × 10^{22 }moles = 17 × 3 × 10^{22}/6.023 × 10^{23}
= 0.85 gm
Q.20. The largest number of molecules is in:
(a) 28 g of CO
(b) 46 g of C_{2}H_{5}OH
(c) 36 g of H_{2}O
(d) 54 g of N_{2}O_{5}
Ans. (c)
Solution.
(1) 28/28 = 1 mole
(2) 46/46 = 1 mole
(3) 36/18 = 2 mole
(4) 54/108 = 0.5 mole
Q.21. The number of molecules in 89.6 litre of a gas at NTP are:
(a) 6.02 × 10^{23}
(b) 2 × 6.02 × 10^{23}
(c) 3 × 6.02 × 10^{23}
(d) 4 × 6.02 × 10^{23}
Ans. (d)
Solution.
89.6 litre of a gas ?
⇒ 22.4 litre  N_{A} molecules
⇒ 89.6 litre  89.6 / 22.4
= 4 × 6.023 × 10^{23}
Q.22. The density of a gaseous element is 5 times that of oxygen under similar conditions. If the molecule of the element is diatomic, what will be its atomic mass?
(a) 49.0
(b) 53.33
(c) 80
(d) 41.7
Ans. (b)
Solution.
CH_{4} = 6 + 4 = 10^{e }present
⇒ 1.6 / 16 = 0.1 mole
Total e^{} =mole × N_{A} × Total no. of e^{}
= 0.1 × 6.023 × 10^{23} × 10
= 6 × 10^{23} electrons
Q.23. The gram atoms present in 5 gram of Calcium are
(a) 0.125
(b) 0.21
(c) 0.117
(d) 0.512
Ans. (b)
Solution.
PV = nRT
P.T constant
P = d/Mw × RT WA/V = d
⇒ d_{1}/d_{2} = Mw_{1}/Mw_{2} ⇒ d/5d = 16/Mw_{2}
Mw_{2} = 16 × 580
Q.24. The gram atoms present in 5 gram of Calcium are
(a) 0.125
(b) 0.21
(c) 0.117
(d) 0.512
Ans. (a)
Solution.
Gram atoms = mole
Mole = 5/40 = 0.125
Q.25. Number of moles present in 100 Kg of lime stone is
(a) 10^{4}
(b) 10^{3}
(c) 10^{5}
(d) 10^{6}
Ans. (b)
Solution.
Mole = 100 × 10^{23}gm = 10^{3 }mole
lime stone CaCo_{3 }= Mw = 100 gm
Q.26. The number of molecules of CO_{2} in 4.4g of the gas at STP
(a) 6.02 × 10^{23}
(b) 5.02 × 10^{23 }
(c) 6.02 × 10^{24 }
(d) 6.02 × 10^{22}
Ans. (d)
Solution.
mole = 4.4/44 = 0.1 mole
1 mole pf Co_{2 }= N_{A} molecule
0.1 = ?
= 6.023 × 10^{23 }× 0.3
= 6 × 10^{22}molecules
Q.27. Weight of 6.02 ×10^{20} molecules of hydrogen is
(a) 0.002g
(b) 0.02g
(c) 2g
(d) 0.01g
Ans. (a)
Solution.
6.023 × 10^{23}
Moleculares = 2 gm of H_{2}
6 × 10^{20} = 6 × 10^{20}/6.023 × 10^{23}
Q.28. Which of the following pairs of gases contains the same number of molecules?
(a) 11 g of CO_{2} and 7 g of N_{2 }
(b) 44 g of CO_{2} and 14 g of N_{2 }
(c) 22 g of CO_{2 }and 28 g of N_{2 }
(d) All the above pairs of gases
Ans. (a)
Solution.
(a) CO_{2} = 11/44=0.25 × N_{A} molecule N_{2} = 7/28 = 1/4 = 0.2 × N_{A}
(b) CO_{2} = 44/44 = 1 × N_{A}, 14/28 = 0.5 × N_{A}
(c) 22/44 = 0.5 × N_{A,}or, 28/28 × N_{A} = 1× N_{A}
Q.29. 2.0 × 10^{22} atoms of an element weights 6 g. The atomic weight of the element is approximately
(a) 290
(b) 180
(c) 34.4
(d) 104
Ans. (b)
Solution.
2 × 10^{22 }atoms ⟶ 6 g
6 × 10^{23 }atom  6 × 6 × 10^{23}/2 × 10^{22 }
= 18 × 10
= 180 g
Q.30. A mixture of 2 mole of H_{2} and 1 mole of He occupies litres at NTP.
(a) 22.4
(b) 44.8
(c) 67.2
(d) 22400
Ans. (c)
Solution.
1 mole = 22.4 litre volume of any gas
2 + 1 = 44.8 + 22.4 volume
= 67.2 litre
Q.31. Which of the following will contain the same number of atoms as 20 gram of Calcium?
(a) 24 gram of Mg
(b) 12 gram of carbon
(c) 24 gram of carbon
(d) 12 gram of Mg
Ans. (d)
Solution.
20/40 = 0.5 mole
(a) 24/24 = 1 mole
(b) 12/12 = 1 mole
(c) 24/12 = 2 mole
Total no. of atoms mole × N_{A}
= 0.5 × 6.023 × 10^{23}
Q.32. Which of the following has the smallest number of molecules?
(a) 11.2 litre of SO_{2} gas at STP
(b) 1 mole of SO_{2} gas
(c) 1 × 10^{23} molecules of SO_{2} gas
(d) 3.2 gram of SO_{2} gas
Ans. (d)
Solution.
22.4 litre volume of any gas = 1 mole (stp)
(a) 11.2 litre = 0.5 mole
(b) So_{2} = 1 mole
(c) 6 × 10^{23} 1 mole
1 × 10^{23} 10^{23}
= 1/6 mole
(d) 32/640 = 1/20 mole
Total molecular = mole × N_{A}
=1/20 × 6.203 × 10^{23}
Q.33. The weight of a single molecule of a substance is 8.5 × 10^{}^{23} gram. The molecular weight of the substance is
(a) 51.2
(b) 30.1
(c) 60 × 1023
(d) 14.5
Ans. (a)
Solution.
1 molecular = 8.5 × 10^{}^{23}^{ }gm
N_{A} = ?
8.5 × 10^{23} × 6.023 × 10^{23}
= 51.2 gm
Q.34. The number of molecules present in 35.5 gram of chlorine is
(a) 3.0115 × 10^{15 }
(b) 3.0115 × 10^{23}
(c) 2.0115 × 10^{23}
(d) 6.023× 10^{23}
Ans. (b)
Solution.
35.5/35.5 = 1 × N_{A} total atom
= 1 × N_{A}/2 total no. of molecules
= 3 × 10^{23}
Q.35. 16 grams of a gas at STP occupies 11.2 litres. The molecular weight of the gas is
(a) 23
(b) 25
(c) 30
(d) 32
Ans. (d)
Solution.
22.4 litre of any gas occupied mole = 1 mole
11.2 litre =16 g
22.4 litre = 16 × 22.4/11.2 = 32g
Q.36. The total number of electrons present in 18 ml of water is
(a) 6.024 × 10^{}^{24}
(b) 7.240 × 10^{23 }
(c) 6.023 × 10^{24 }
(d) 6.023 × 10^{23}
Ans. (c)
Solution.
Density of H_{2}O = 1 gm/ml=1 kg/dm^{3}
1 ml = 1 gm
Mole H_{2}O_{ }= 18/18 = 1 mole
= 10e^{} = 1×10 × 6.033 × 10^{23}
= 6 × 10^{24}
Q.37. The ratio between the number of molecules in equal masses of nitrogen and oxygen is
(a) 7 : 8
(b) 1 : 9
(c) 9 : 1
(d) 8: 7
Ans. (d)
Solution.
⇒ 16 × N_{A}/14 × N_{A} = 8 ∶ 7
Q.38. Equal masses of oxygen, hydrogen and methane are kept under identical conditions.
The ratio of the volumes of the gases will be
(a) 2 : 16 : 2
(b) 2 : 16 : 1
(c) 1 : 16 : 2
(d) 1 : 1 : 1
Ans. (c)
Solution.
Equal massess
x/32 + x/2 + x/16
= (x + 16x + 2x)/32
= 1 : 16 : 2
⇒ mole ratio = volume ratio = molecular ratio
Q.39. 4.4 gram of CO_{2} and 2.24 litre of H_{2} at STP are mixed in a container. The total number of molecules present in the container will be
(a) 6.022 × 10^{23 }
(b) 1.2044 × 10^{23}
(c) 2 moles
(d) 6.023 × 10^{24}
Ans. (b)
Solution.
CO_{2} = 4.4/4.4 = 0.1 H_{2} = 2.24/22.4 = 0.1
= total number of moles = 0.2
= 0.2 × 6.023 × 10^{23}
= 1.2 × 10^{23}
Q.40. Out of the following the largest number of atoms are contained in
(a) 11 g of CO_{2}
(b) 4g of H_{2}
(c) 5 g of NH_{3}
(d) 8 g of SO_{2}
Ans. (b)
Solution.
(a) CO_{2} = 11/44 = 0.25 mole
(b) 4/2 = 2 mole
(c) 5/17 = 2 × N_{A} largest no. of atoms
(d) 8/64 = 1/8
Q.41. The percentage of nitrogen in urea, (NH_{2}CONH_{2)} is:
(a) 38.4
(b) 46.6
(c) 59.1
(d) 61.3
Ans. (b)
Solution.
⇒ 60 gm urea = 28 gN
100 g 28 × 100/60 = 46.6%
Q.42. The empirical formula of a compound is CH_{2}O, if vapour density is 90. Then the molecular formula is
(a) CH_{2}O
(b) C_{2}H_{4}O_{2}
(c) C_{3}H_{6}O_{3}
(d) C_{6}H_{12}O_{6}
Ans. (d)
Solution.
⇒ V. D × = mw
⇒ 90 × 2 = 180 gm
Empirical formula weight = 30 gm
Value of n = 180/30 = 6
C_{6}H_{12}O_{6}
Q.43. An organic compound having carbon and hydrogen has 80% carbon. The empirical formula of the hydrocarbon is
(a) CH_{4}
(b) CH_{3}
(c) CH_{2}
(d) CH
Ans. (b)
Solution.
C = 80%, h = 20%
%C = 80/12, %H = 20/1
= 6.66 = 20
= 6.66/6.66 = 20/6.66 = 3 {least Ratio}
1 : 3
C_{1}H_{3} = CH_{3}
Q.44. A compound contains 90% C and 10% H. The empirical formula of the compound is
(a) C_{8}H_{10}
(b) C_{15} H_{30}
(c) C_{3}H_{4}
(d) C_{15}H_{32}
Ans. (c)
Solution.
%C = 90/12, %H = 10/1
= 7.5/7.5 = 10/7.5 = 1.33
= 1
C_{3}H_{4}
Q.45. An organic compound contains C = 50% and H = 9.25%. Its empirical formula is
(a) C_{3}H_{6}
(b) C_{3}H_{7}O_{2}
(c) C_{2}H_{4}O
(d) C_{4}H_{8}O
Ans. (b)
Solution.
C% = 50/12 = 4.1666/2.546 = 1.636 × 2
%H = 9.25%/1 = 9.25/2.546 = 2.546/2.546 = 3.633 × 2
O% = 40.76/16 = 2.546/2.546 = 1 × 2
Ch_{3}H_{7}O_{2}
Q.46. A metal M having an atomic weight of 197 yields a chloride containing 35.1% chlorine.
The empirical formula of the compound is
(a) MCl_{3 }
(b) MCl
(c) MCl_{2}
(d) MCl_{4}
Ans. (a)
Solution.
Eq. weight of metal chloride = % of metal/% chloride × 35.5 = 64.9/35.1 × 35.5 = 64.9
Eq = 197/V_{F }
= 64.9 = 197/V_{f }= 197/64.9
V_{F} = 3
M^{+3 } Cl^{⊖}
MCL_{3}
Q.47. When 1 gram CaCO_{3} is dissolved in excess dilute acid the volume of CO_{2} evolved at STP will be
(a) One litre
(b) 224 ml
(c) 22.4 litre
(d) 2.24 litre
Ans. (b)
Solution.
CaCO_{3} ⟶ Cao + CO_{2}
1 gm  44 × 10^{}^{2 }gm
44 g  22.4 litres
44 × 10^{2 }g  = 0.224 litre = 224 ml
Q.48. 8 gram of sulphur is completely burnt in a large excess of oxygen, the volume in litres of SO_{2} formed as reduced to STP is
(a) 5.6
(b) 8.0
(c) 11.2
(d) 16.0
Ans. (a)
Solution.
S + O_{2} → SO_{2}
32 g  64 g = 1 mole = 22.4 liter
8g  64 x 8/32 = 16 g
mole = 16/64 = 1/4 mole
5.6 litre value
Q.49. 0.01 mole of iodoform (CHI_{3}) reacts with Ag powder to produce a gas whose volume at NTP is
(a) 224 ml
(b) 112 ml
(c) 336 ml
(d) 1120 ml
Ans. (b)
Solution.
2CHl_{3} + 6Ag ⟶ H − C ≡ C − H + 6 Agl
2  1 mole
10^{2 } ?
10^{2} × 1/2 = 0.5 × 10^{2}
= 5 × 10^{3}
∴ 1 mole  22.4 literes
5 × 10^{3 } 22.4 × 5 × 10^{3}
= 112 × 10^{3}
112 ml
Q.50. Molarity of liquid HCl with density equal to 1.17 g/ cc is
(a) 36.5
(b) 18.25
(c) 32.05
(d) 4.65
Ans. (c)
Solution.
d = M/v
1 cc = 1 ml
1000 ml = 1.17 × 1000 = 1170 g/36.5 = 32.05
Q.51. How many milliliters (mL) of 1 M H_{2}SO_{4} solution are required to neutralize 10 mL of 1 M NaOH solution?
(a) 2.5 mL
(b) 5.0 mL
(c) 10.0 mL
(d) 20.0 mL
Ans. (b)
Solution.
Neutralization
Acid = Base (milliequivalent)
N_{1}V_{1} = N_{2}V_{2}
⇒ 1 × V = 10 × 1 (N = m × VF)
V = 5ml
Q.52. The No. of moles of barium carbonate which contain 1.5 moles of oxygen atoms is
(a) 1.0 mole
(b) 1.5 mole
(c) 0.5 mole
(d) 2.0 mole
Ans. (c)
Solution.
Baco_{3} → 1 mole BaCo_{3 }
→ 1 mole Ba
→ 1 mole C
→ 3 mole (O)
3 mole oxygen  1 mole BaCo_{3}
1.5 mole oxygen  1.5/3 = 0.5 mole BaC0_{3}
Q.53. A molecule of Haemoglobin contains 0.33% of iron by weight. The molecular weight of Haemoglobin is 67200. The number of iron atoms (At. wt. = 56) present in one molecule of Haemoglobin is
(a) 1
(b) 2
(c) 3
(d) 4
Ans. (d)
Solution.
Mass of Fe in 1 mole of Hb = ?
100 gm Hb – 0.33 gm Fe
1 mole_{(hb) }= 67200g  0.33 × 67200/100
= 221.76 gm
Mole of Fe = 221.76/56 = ≅ 3.9 ≅ 4
= 4N_{A} atom of Fe
Na molecules of Hb – 4N_{A} atom of Fe
1 mole  4N_{A}/N_{A}
= 4 atoms of Fe
Q.54. 48 gram of Mg contains the same number of atoms as 160 gram of another element. The atomic mass of the element is
(a) 24
(b) 320
(c) 80
(d) 40
Ans. (c)
Solution.
mole of mg = 48/24 = 2 mole = 2 × N_{A}
(A) = 160/2 = 80
Q.55. The number of moles of oxygen in one litre of air (21% oxygen by volume) at STP would be
(a) 0.186 mole
(b) 0.21 mole
(c) 2.10 mole
(d) 0.0093 mole
Ans. (d)
Solution.
In 100 ml air  21 ml oxygen
1000 ml air  21 × 1000/100 = 210 ml
22400 ml – 1 mole oxygen
210 ml  210 × 1/22400 = 0.009375 mole oxygen
Q.56. What is correct for 10 g of CaCO_{3}?
(a) It contains 1 g atom of carbon
(b) It contains 0.3 g atoms of oxygen
(c) It contains 12 g of calcium
(d) It refers to 0.1 g equivalent of CaCO_{3}.
Ans. (b)
Solution.
10/100 = 0.1 mole CaCO_{3}
1 mole CaCO_{3} →1 mole Ca
→1 mole C
→3 mole
∴ 0.1 mole CaCO_{3} → 0.1 Ca
→0.1C
→0.3(oxygen)
Q.57. If isotopic distribution of C12 and C14 is 98% and 2% respectively, then the number of C14 atoms in 12 g of carbon is
(a) 1.032 × 10^{22}
(b) 3.01 × 10^{22}
(c) 5.88 × 10^{23}
(d) 6.02 × 10^{23}
Ans. (a)
Solution.
In 100 gram of carbon C14 is=2 gram
12 gm = 2 × 12/100 = 0.24 gram
Mole of C14 = 0.24/14 =o.017
Total no. of atoms C14 = 0.017 × NA
= 0.0017 × 6.023 × 10^{23}
= 0.102 × 10^{23}
= 1.02 × 10^{22} atom
Q.58. The number of moles of water present in 100g of water of 90% purity is
(a) 3
(b) 18
(c) 5
(d) 5.5
Ans. (c)
Solution.
100% purity = 100/18 mole H_{2}O
90% = 100 × 90/18 × 100
5 mole of H_{2}O
Q.59. Which of the gases contains the same number of molecules as that of 16 grams of oxygen
(a) 16g of O_{3}
(b) 32g of SO_{2}
(c) 16g of SO_{2}
(d) All
Ans. (b)
Solution.
16/16 = 1 mole N_{A }atom = N_{A}/2 = molecals
(a) 6/48 = 1/2 × N_{A}
(b) 32/64 = 1/2
(c) 16/4 = 1/4
Q.60. The correct arrangement of the following in order of increasing mass is
(i) N_{2} molecule
(ii) oxygen atom
(iii) 1 Avogram
(iv) 1 × 10^{10} gram atom of copper
(a) I > II > III > IV
(b) III < II < I < IV
(c) I > III > II > IV
(d) IV > I > II > III
Ans. (d)
Solution.
(a) N_{2 }mole = 28 gm
(b) O atom = 16 gm
(c) 1/6 × 10  23 gm
(d) 63.5
1 mole of Cu = 63.5 gm
1 × 10^{10 }mol = 63.5 × 10^{10 }gm
Q.61. 0.5 mole of a gas (Mol. Wt. 80) occupies 11.2 litres at STP. The volume occupied by 0.25 mole of a lighter gas (Mol. Wt. 20) at STP will be
(a) 11.2 lit
(b) 5.6 lit
(c) 8 lit
(d) 22.4 litre
Ans. (b)
Solution.
0.5 mole  11.2 litre
0.25 mole → 11.2 × 0.25/0.5 = 5.6 litre
Q.62. The molecular weight of an unknown substance is found to 24000. If it contains 0.2% Magnesium, then the number of Magnesium atoms that can be present in a molecule of it is
(a) 1
(b) 2
(c) 4
(d) 10
Ans. (b)
Solution.
100 gm2 gm Mg
24002 × 2400/100 = 48 g
Mole of mg = 48/24 = 2 mole
Total atom of Mg = 2 × N_{A}
NA molecules of unknown sub = 2N_{A}
0.1 mole = 2 × N_{A}/N_{A} = 2
Q.63. The percentage of magnesium in chlorophyll is 2.68%. The number of magnesium atoms in 2 gram of chlorophyll is
(a) 1.34 × 10^{21 }
(b) 1.34 × 10^{–21 }
(c) 1.35 × 10^{–24 }
(d) 1.35 × 10^{24}
Ans. (a)
Solution.
100 g – 2.68 g Mg
2g  2.68 × 2/100 = weight of Mg
Mole= 2.68 × 2/100 × 24 = 0.2 × 10^{2}
Total atom of Mg = 0.22 × 10^{2} × 6.023 × 10^{23}
= 1.34 × 10^{21}
Q.64. Ordinary water contains one part of heavy water per 6000 parts by weight. The number of heavy water molecules present in a drop of water of volume 0.01 ml is
(a) 2.5 × 10^{16 }
(b) 5 × 10^{17 }
(c) 5 × 10^{16}
(d) 7.5 × 10^{16}
Ans. (c)
Solution.
D_{2}O = Mw = 20
Mole of D_{2}O = 0.01/20 = 5 × 10^{4 }mole D_{2}O
6000 parts 5 × 10^{4 }mol D_{2}O
1 part 5 × 10^{4}/6000 × N_{A}
= 5 × 10^{16}
Q.65. The number of molecules present in a drop of water. If its volume is 0.05 ml are
(a) 1.66 × 10^{21}
(b) 1.60 × 10^{22}
(c) 1.66 × 10^{23 }
(d) 1.60 × 10^{24}
Ans. (a)
Solution.
H_{2}O density = gm/ml =1 ml=1 gm
⇒ mole H_{2}O = 0.05/18 × N_{A}
= 0.05/18 × 6.023 × 10^{23}
= 1.66 × 10^{21}
Q.66. Which of the following will not have a mass of 10 g?
(a) 0.1 mol CaCO_{3}
(b) 1.51 × 10^{23 }Ca^{2+} ions
(c) 0.16 mol of Cl^{}ions
(d) 7.525 × 10^{22} Br atom.
Ans. (c)
Solution.
(a) CaCO_{3} = 0.1 mol = 10g
(b) N_{A }Ca^{2+} 408 1.51 × 10^{23}  1.51 × 10^{23}/6 × 10^{23 }= 1/4 × 40 = 10 g
(c) 1 mole Cl =35.5 0.16 = ? 0.16 × 35.5 = 5.68 gm × 2 ≅ 10
(d) 7.525 × Br atom
1 mole – 6.023× 10^{23}
?  7.525 × 10^{22}
⟹ 7.525 × 10^{22}/6.2023× 10^{23} = 1.2 × 10^{1}
Q.67. x L of N_{2} at STP contains 3 × 10^{22 }molecules. The number of molecules in x/2 L of ozone at STP will be
(a) 3 × 10^{22 }
(b) 1.5 × 10^{22 }
(c) 1.5 × 10^{21 }
(d) 1.5 × 10^{11}
Ans. (b)
Solution.
xLN_{2} at stp → 3 x 10^{22} molecules
∴ volume occupied by gas at S.T.P doesn’t depend on nature of gas.
Q.68. 10 ml of a gaseous hydrocarbon combustion gives 40 ml of CO_{2} and 50 ml of H_{2}O vapour under the same condition. The hydrocarbon is
(a) C_{4}H_{6}
(b) C_{6}H_{10}
(c) C_{4}H_{8}
(d) C_{4}H_{10}
Ans. (d)
Solution.
_{X}H_{y} + 0_{2}[x + y/4] C0_{2} → + H_{2}O
Volume of CO_{2 }formed
10x = 40ml CO_{2}
X = 4
X = c = 4
C_{4}H_{10 }
volume of H_{2}O
10 y/2 = 50ml H_{2}O
y = 100/4 = 10
y = H = 1
OR
P.O.A.C method
P.O.A.C for carbon C atom
10x = 40
X = 4
C_{4}H_{10}
D.O.A.C for H atom
10y/2 = 50
y = 100/10
y = 10
Q.69. 15 ml hydrocarbon requires 45 ml of O_{2} for complete combustion and 30 ml of CO_{2} is formed. The formula of the hydrocarbon is
(a) C_{3}H_{6}
(b) C_{2}H_{6}
(c) C_{4}H_{10}
(d) C_{2}H_{4}
Ans. (d)
Solution.
C_{x}H_{y} + 0_{2} → C0_{2} + H_{2}O
x + Y/4 x y
(1) volume of CO_{2} formed
15x = 30 ml of CO_{2}
X = 2
(2) The volume of O_{2} required 15x + Y/4 = 45 ml of O_{2}
⟹ 30+15Y/4 = 45
⟹ 120+15y/4 = 45
⟹ 120+15y = 180
⟹ 15y = 180120
15 y = 60
Y = 60/15 = 4
C_{2}H_{4}
x = 2 y = 4
Q.70. Complete combustion of a sample of a hydrocarbon gives 0.66 2 grams of CO_{2 }and 0.362 grams of H_{2}O the formula of a compound is
(a) C_{3}H_{8}
(b) CH_{4}
(c) C_{2}H_{6 }
(d) C_{2}H_{4}
Ans. (a)
Solution.
hydrocarbon+ O_{2} → CO_{2} + H_{2}O
Q.71. The simplest formula of a compound containing 50% of element x (atomic weight = 10) and 50% of element y (atomic weight = 20) is
(a) xy
(b) x_{2}y
(c) xy_{2}
(d) x_{2}y_{3}
Ans. (b)
Solution.
%x = 50/10
x = 5/2.5
x = 2
y% = 50/20
y = 2.5/2,5
y = 1
x_{2}y
Q.72. The empirical formula of an organic compound is CH. 6.023 × 10^{22} molecules of same organic compound weigh 7.8 g. The molecular formula is
(a) C_{2}H_{2}
(b) C_{6}H_{6}
(c) C_{2}H_{4}
(d) None
Ans. (b)
Solution.
empirical formula weight = 12+1 = 13g
6 × 10^{22}− 7.8g
6 × 10^{23}  ?
= 7.8 × 6 × 10^{23}/6 × 10^{22}
= 78 gm
n = 78/13 = 6
C_{n}H_{n} = C_{6}H_{6}
Q.73. An organic compound contains C = 21.56%, H = 4.56% and Br = 73.36%. Its molecular weight is 109. Its molecular formula is
(a) C_{2}H_{5}Br
(b) C_{3}H_{7}Br
(c) C_{4}H_{8}Br
(d) C_{6}H_{6}Br
Ans. (a)
Solution.
C% = 21.56/12 = 1.7/0.917 = 1.853 ≅ 2
H% = 4.56/1 = 4.56/0917 = 4.97 ≅ 5
Br = 73.36/80 = 0.917/0.917 = 1
n = log/log = n
e.f = m.f
= C_{2}H_{5}Br
Q.74. 0.078 gram of hydrocarbon occupies 22.4 ml volume at STP. The empirical formula of hydrocarbon is CH. The molecular formula of hydrocarbon is
(a) C_{5}H_{5}
(b) C_{6}H_{6}
(c) C_{2}H_{2}
(d) C_{8}H_{8}
Ans. (b)
Solution.
22.4 ml 0.078 gm hydrocarbon
= 78 g
n = 78/13 = 6 CH = 13
12+1
C_{6}H_{6}
Q.75. Empirical formula of a compound is CH_{2}. The mass of one litre of this organic gas is exactly equal to that of one litre of nitrogen. Therefore the molecular formula of the organic gas is
(a) C_{3}H_{8}
(b) C_{2}H_{6}
(c) C_{2}H_{4}
(d) C_{3}H_{6}
Ans. (c)
Solution.
e.f = CH_{2} 12 + 2 = 14
n= 28/14 = 2
C_{2}H_{4}
Q.76. When 1.2 g of carbon is completely burnt in 6 litres of oxygen at STP, the remaining volume of oxygen is
(a) 3.76 lit
(b) 2.6 lit
(c) 5.8 lit
(d) 37.6 lit
Ans. (a)
Solution.
C + 1/2 O_{2} → (Incomplete combustion)
1C → (1 mole oxygen)
0.1C → 0.1 × 1 = 0.1 mole of oxygen
⟹ 22.4 litre → 1 mole O_{2}
1 C  22.4 litre
0.1  22.4 litre
6 − 2.24 = 3.76 liter
Q.77. 0.5 mole of H_{2}SO_{4} is mixed with 0.2 mole of Ca (OH)_{2}. The maximum number of moles of CaSO_{4 }formed is
(a) 0.5
(b) 0.2
(c) 0.4
(d) 0.25
Ans. (b)
Solution.
H_{2}SO_{4} + Ca(OH)_{2} → CaSO_{4 }+ 2H_{2}O
d = 0 0.5 0.2 0
t = ? 0.5  x 0.2  x x
x = 0.2
LR
L.R = 0.5/1
L.R = 0.2/1 = 0.2 = x
nx = aa
1 × = 0.2 × 100/100
x = 0.2
Q.78. 70 gram of a sample of magnesite on treatment with excess of HCl gave 11.2 litre of CO_{2} at STP. The percentage purify of the sample is
(a) 80
(b) 70
(c) 60
(d) 50
Ans. (d)
Solution.
Magnetic MgCO_{3}+ Fe]=140gm
Fe + MgCO_{3} → MgCl_{2} + H_{2}CO_{3}
↓
H_{2}O = CO_{2}
1 mole1CO_{2}
1 CO_{2}1 mol MgCO_{3}
0.50.5 mol
70/140 × 100
% purity = 50%
Q.79. When 100 gram of ethylene polymerises to polythene according to the equation nCH_{2} = CH_{2 }→_{ }(CH_{2}  CH_{2})_{n } the weight of polythene produced will be
(a) n/2 gram
(b) 100 gram
(c) 100/n gram
(d) 100n gram
Ans. (b)
Solution.
nCH_{2} = CH_{2} → (CH_{2}  CH_{2} )_{n}
100g 100g
n28gm  nx28g
Q.80. Air contains 20% by volume of oxygen. The volume of air required for the complete combustion of one litre of methane under the same conditions is
(a) 2 litre
(b) 4 litre
(c) 10 litre
(d) 0.4 litre
Ans. (c)
Solution.
CH_{4} +20_{2} → CO_{2} +2H_{2}O
1 litre 2 litre
2 liter O_{2} → 1 litre CH_{4}
20 litre  20/2 = 10 litre CH_{4}
Q.81. The hydrated Na_{2}SO_{4}nH_{2}O undergoes 56% loss in weight on heating and become anhydrous. The value of n will be
(a) 5
(b) 3
(c) 7
(d) 10
Ans. (d)
Solution.
Na_{2}SO_{4}H_{2}O → Na_{2}SO_{4}+ xH_{2}O
142 + 18x = 56/100%
Lose in weigh due to H_{2}O = 18x/142 + 18x = 56/100
⟹ 1800x = 142 × 56 + 18x × 56
⟹ 1800x = 7952 + 1000x
⟹ 792x = 795x
x = 10.004
x = 10
Q.82. 1.25 g of a solid dibasic acid is completely neutralized by 25 mL of 0.25 molar Ba (OH)_{2} solution . The molecular mass of the acid is
(a) 100
(b) 150
(c) 120
(d) 200
Ans. (d)
Solution.
Neutralisation rxn
Solid dibasic acid =
N_{1}V_{1} = N_{2}V_{2}
1000 × w/e = N_{1}V_{1}
mw = 160 × 1.25
mw = 200
Q.83. 0.126 g of acid requires 20 ml of 0.1 N NaOH for complete neutralization. The equivalent mass of the acid is
(a) 45
(b) 53
(c) 40
(d) 63
Ans. (d)
Solution.
w/e × 1000 = 20 × 0.1
⟹ 0.126/E × 1000 = 2
E = 0.126 × 1000/2
E = 63
Q.84. The mole fraction of the solute in one mole aqueous solution is
(a) 0.009
(b) 0.018
(c) 0.027
(d) 0.036
Ans. (b)
Solution.
X_{A} = 1/56.5 = 0.18
Q.85. How many grams of phosphoric acid would be needed to neutralize 100 g of magnesium hydroxide?
(a) 66.7 g
(b) 252 g
(c) 112 g
(d) 168 g
Ans. (c)
Solution.
⇒ w/E_{1} × 1000 = w_{2}/E_{2} × 1000
⇒ 3x/98 = 200/58
⇒ x = 200 × 98 / 58 × 3 = 19600/174
= 112g
Q.86. Normality of solution of FeSO_{4}. 7H_{2}O containing 5.56 g / 200 mL which converts to ferric from in a reaction is (Fe = 56, s = 32, O = 16, H = 1)
(a) 1
(b) 0.1
(c) 0.01
(d) 10
Ans. (b)
Solution.
Molarity of FeSO_{4}7H_{2}O
156 + 126 = 278g mw
Mole = 5.56/278g mol
2005.56/278
1000?
5.56 × 1000/278 × 200 × 10^{2}
= 20/200 = 0.1
Fe^{2+ }→ Fe^{3+} VF1
N = M_{X}V_{F}
N = 0.1 × 1
N = 0.1N
Q.87. 100 g of a sample of HCl solution of relative density 1.17 contains 31.2 g of HCl. What volume of this HCl solution will be required to neutralize exactly 5 litres of N/20 KOH solution?
(a) 25 ml
(b) 29.2 ml
(c) 34.2 ml
(d) 250 ml
Ans. (a)
Solution.
D = m/v
1.74 × g/m =100/V
V = 100/1.17
V = 85.4m volume of solution
⟹ 85.4  31.2/36.5 = 0.84
1000ml 31.2 × 1000/36.5 × 85.4
M = 10m
N= m × V_{F}
N=10N1
Molarity of HCl solution = 10m
Normality = molarity × V_{F}
= 10 ×1
= 10N
N_{1}V_{1} = N_{2}V_{2}
⟹ 10 × V_{litre }= 5/10
V_{litre }= 5/100 = 0.025 litre
= 25 ml
Q.88. 300 ml of 1 M HCI and 100 ml of 1 M NaOH are mixed. The chloride ion concentration in the resulting solution is
(a) 1 M
(b) 0.5 M
(c) 0.75 M
(d) 0.25 M
Ans. (c)
Solution.
HCL ⇋ H^{+} + Cl^{}
0.3 0.3 0.3
NaOH ⇋ Na^{+} + OH
0.1 0.1 0.1
Cl^{} = 0.3/0.4 = 0.75M
Cl^{} = c0ncentration of Cl^{1}/total volume
Q.89. 200 ml of 1 M H_{2}SO_{4}, 300 ml 3 M HCI and 100 ml of 2 M HCI is mixed and made up to 1 litre. The proton concentration in the resulting solution is
(a) 1.25 M
(b) 1.5 M
(c) 2.5 M
(d) 0.75 M
Ans. (b)
Solution.
All are acids so it shows additive properties
H^{+} = (N_{1}V_{1}+N_{2}V_{2}+N_{3}V_{3})/(V_{1}+V_{2}+V_{3})
400+900+200/1000 = 1500/1000 = 1.5
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195 videos337 docs190 tests

Solved Examples: Some Basic Concepts of Chemistry 2 Doc  14 pages 
Solved Examples: Mole Concept Video  10:08 min 
Dalton's Atomic Theory Video  07:06 min 

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