1)According to the Stock system of nomenclature UO2SO4 is
a) dioxo uranium (VI) sulphate
b) dioxo uranium (IV) sulphate
c) dioxo uranium (II) sulphate
d) dioxo uranium (III) sulphate
2)The best method to identify trace elements is
a) neutron activation analysis
b) chromatogrphic method
c) n.m.r spectroscopy
d) radiocarbon dating
3)The number of significant figures in 0.0200 are
4)The number of significant figures in 0.0045 are
5)The greatest number of significant figures in 10500 are
d) can be any of these
6)The number of significant figures in are
d) can be any of these
7)The correctly reported answer of the addition of 40523, 2.3 and 6.24 will have significant figures
8)The correctly reported answer of the addition of 154.21, 6.142 and 23 will be
9)The correctly reported difference of 16.4215 and 6.01 will have significant figures equal to
10)After rounding off to three digits 1.235 and 1.225 we will have their answers respectively as
11)The actual product of 4.327 and 2.8 is 12.1156. The correctly reported answer will be
12)On dividing 0.46 by 15.374, the actual answer is 0.029236. The correct reported answer will be
13)Two students X and Y report the weight of the same substance as 5.0 g and 5.00 g respectively. Which of the following statements is correct?
a) Both are equally accurate
b) X is more accurate than Y
c) Y is more accurate than x
d) Both are inaccurate scientifically
14)Which of the following is incorrect about S.I units?
b) Force in Newtons
c) Pressure in pascals
15)Which of the following is correct?
16)The relationship between picometer (pm) and nanometer (nm) is
b) 1pm= 1000 nm
c) 1 pm = 100 nm
d) 1 pm = 100 pm
17)The nuclear cross section is measured in . It is equal to
18)In which of the following numbers are all zeros significant?
19)Planck’s constant has the dimensions of
c) angular momentum
20)The units , nanometer, fermi, angstrom and attometre, arranged in decreasing order are expressed as
a) angstrom, nanometer, fermi, attometre
b) fermi, attometre, angstrom, nanometer
c) nanometer, angstrom, fermi, attometre
d) attometre, angstrom, fermi, nanometer
21)The value of Planck’s constant is . The numbers of significant figures in it is
d) thirty four
22)81.4 g sample of ethyl alcohol contains 0.002 g of water. The amount of pure ethyl alcohol (to the proper number of significant figures ) is
23)One torr is equal to
a) 1 cm of Hg
b) 1 atm pressure
c) 1 mm of Hg
d) 1 m of Hg
24)The Metric system was first introduced by
c) The British Academy of Science
d) The French Academy of Science
25)The number of significant figures in are
b) infinite number
26) Rydberg’s constant is . It can be expressed to three significant figures as
27) is the unit of
28)The mass of a piece of paper is g and the mass of a solid substance along with the same piece of paper is 20.036 g. If the volume of the solid is , its density to the proper number of significant digits will be
29)Which of the following is not an element ?
30)Which one of the following statements is correct?
a) all elements are homogenous
b) compounds made up of a number of elements are heterogeneous
c) a mixture is not always heterogeneous
d) Air is a heterogeneous mixture
31)The most abundant element in the earth’s crust is
32)Which one of the following processes results in the formation of a new chemical compound?
a) dissolving common salt in water
b) heating water
c) heating a platinum rod
d) heating an iron rod
33)Which of the following statements is false ?
a) An element of a substance contains only one kind of atoms
b) A compound can be decomposed into its constituents
c) Milk is a homogenous mixture
d) All homogenous mixture are called solutions
34)Which one of the following is not a mixture ?
a) Iodized table salt
c) Liquefied petroleum gas ( L.P.G. )
d) Distilled water
35)Which one of the following mixtures cannot be separated by fractional distillation ?
a) Benzene + toluene
b) Acetone + Water
c) Acetone + Methyl alcohol
d) Glucose +Sucrose
36)A mixture of sand and naphthalene can be separated by
a) Washing with water
b) Magnetic separation
37)Magnesium sulphate containing calcium sulphate as impurity can be separated by
d) fractional crystallisation
38)A mixture of can be separated by
d) adding acetic acid
39)A mixture that can be separated by sublimation method is
40)Chlorophyll extracted from green leaves can be separated into its components by
41)10 g on heating leaves behind a residue weighing 5.6g. Carbon dioxide released into the atmosphere at STP will be
a) 2.24 L
b) 4.48 L
c) 1.12 L
d) 0.56 L
42)The atomic mass of an element is
a) The actual mass of one atom of the element.
b) The relative mass of an atom of the element
c) The weighted average relative mass of different isotopes of the element.
d) The same as the atomic number of the element
43)Nitrogen forms five stable oxides with oxygen of the formula . The formation of these oxides explains fully the
a) Law of definite proportions
b) Law of partial pressures
c) Law of multiple proportions
d) Law of reciprocal proportions.
44)Two metallic oxides contain 27.6% and 30.0 % oxygen respectively. If the formula of the first oxide is , that of the second will be
45)1L of combines with 3L of to form 2L of under the same conditions. This illustrates the
a) Law of constant composition
b) Law of multiple proportions
c) Law of reciprocal proportions
46)12 g carbon combine with 64 g sulphur to form . 12 g carbon also combine with 32 g oxygen to form . 10 g sulphur combine with 10 g oxygen to form . These data illustrate the
a) law of multiple proportions
b) law of definite proportions
c) law of reciprocal proportions
d) law of gaseous volumes
47)Oxygen combines with two isotopes of carbon and to form two samples of carbon dioxide. The data illustrates
a) Law of conservation of mass
b) Law of multiple proportions
c) Law of reciprocal proportions
d) None of these
48)Which of the following does not represent 32 g of the substance?
a) one mole of oxygen atoms
b) one gram atom of sulphur
d) Oxygen present in 2 moles of water
49)The isotopes of chlorine with mass numbers 35 and 37 exist in the ratio of
50)The correct value of Avogadro’s number is
51)Which one of the following statements is incorrect ?
52)5.6 litres of a gas N.T.P are found to have a mass of 11 g. The molecular mass of the gas is
53)The number of atoms of oxygen present in 10.6 g of will be
54)Which of the following contains the maximum number of molecules ?
55)The number of molecules present in a drop of water weighing 0.06g is approximately
56)Which of the following pairs contains an equal number of atoms ?
a) 11.2 cc of nitrogen and 0.015 g of nitric oxide
b) 22.4 litre of nitrous oxide and 22.4 litres of nitric acid
57)Which of the following has the maximum mass?
a) 0.1 g atom of carbon
b) 0.1 mol of ammonia
d) 1120 cc of carbon dioxide
58)The total number of atoms present in 17 g of is
59)The mass of one atom of hydrogen is approximately
a) 1 g
b) 0.5 g
60)The mass of one molecule of water is approximately
61)Which of the following contains the same number of atoms as are present in 12 g of carbon ?
a) 28 g of iron (at mass of Fe = 56 )
b) 48 g of magnesium (at mass of Mg = 24)
d) 14 g of carbon monoxide
62)The number of molecules present in of water is
63)The volume at N.T.P of 0.22 g of is not the same as that of
a) 0.01 g of hydrogen
d) 0.16 g of oxygen
64)0.6 g of carbon was burnt in the air to form . The number of molecules of introduced into the air will be
65)An electric bulb was filled with 100 cc of argon gas( At. Mass = 40) at and 1 atm pressure
66)An evacuated observation balloon after filling with helium gas had an increase in weight equal to 2 g. The volume of the balloon at and 1 atm pressure must be, approximately
a) 2.8 L
b) 5.6 L
c) 11.2 L
d) 22.4 L
67)Equal volumes of different gases at any definite temperature and pressure have
a) Equal densities
b) Equal masses
c) Equal number of atoms
d) Equal number of molecules
68)The crystalline salt, on heating loses 59.6% of its weight. The formula of the crystalline salt is
d) None of these
69)Which of the following represents the formula of a substance which contains about 26% nitrogen and 74% oxygen ?
70)How many atoms are present in a mole of ?
71)Which one of the following statements is incorrect?
a) Atoms of the same element may have different atomic weights.
b) Atoms can be created or destroyed
c) Half of an atom can also take part in a reaction
d) Elements can exist as atoms or molecules but compounds exist only as molecules.
72)An oxide of iodine (I= 127) contains 25.4 grams of iodine and 8 grams of oxygen. Its formula could be
73)One litre of a gas is at a pressure of mm of Hg at . How many molecules are present in the vessel ?
74)The normality of a in 60 mL solution is
75)The milliequivalents of in 60 mL solution is
76)A compound contains 69.5% oxygen and 30.5 % nitrogen and its molecular weight is 92 . The formula of the compound is
77)The number of molecules in one litre of a gas at S.T. P. is known as
a) Avogadro Number
b) Berzelius number
d) Loschmidt number
78)280 ml of sulphur vapour at NTP weighs 3.2 g .The Species present in the sulphur vapour is
79)A person adds 1.71 gram of sugar () in order to sweeten his tea. The number of carbon atoms added are (mol. mass of sugar = 342)
80)Which pair of species have the same percentage of carbon ?
81)The number of potassium atoms in 2 moles of potassium ferrocyanide is
82)The product of the atomic weight and specific heat of any element is constant which is approximately 6.4. This is known as
83)The approximate atomic weight of a metal having specific heat 0.32 is
84)0.1 g of a metal on reaction with dil. acid at S.T.P.liberates 34.2 ml hydrogen gas. The equivalent weight of the metal is
85)In the chemical scale , the relative mass of the isotopic mixture of oxygen atoms is assumed to be equal to
86)The weight of two elements which combine with one another are in the ratio of their
a) atomic weight
b) molecular weight
c) gram mole
d) equivalent weight
87)The vapor density of a metal chloride is 66. Its oxide contains 53% metal. The atomic weight of the metal is
88)One gram of hydrogen is found to combine with 80 g of bromine. One gram of a metal (valency=2) combines with 4 gm of bromine. The equivalent weight of the metal is
89)When 100 ml of 1 M NaOH solution and 10 ml of solution are mixed together, the resulting solution will be
c) strongly acidic
90)The percentage of copper and oxygen in samples of CuO obtained by different methods were found to be the same. This illustrates the law of
a) constant proportions
b) conservation of mass
c) multiple proportions
d) reciprocal proportions
91)If molecules are removed from 200 mg of , then the number of moles of left are
92)A compound contains 38.8% C, 16.0% H and 45.2% N. The formula of the compound would be
93)What is the normality of 1M solution of ?
94)How much water should be added to 200 c.c. of semi normal solution of NaOH to make it exactly deci normal?
95)The number of molecules in 100 ml each of , and
c) the same
96)Two samples of lead oxide were separately reduced to metallic lead by heating in a current of hydrogen. The weight of lead from one sample was half that obtained from the other. The data illustrates the
a) Law of reciprocal proportions.
b) Law of constant proportions
c) Law of multiple proportions
d) Law of equivalent proportions.
97)What is the concentration of nitrate ions, if equal volume of and 0.1 M NaCl are mixed together?
a) 0.1 M
b) 0.2 M
c) 0.05 M
d) 0.25 M
98)One part of an element A combines with two parts of another element B. Six parts of the element C combine with four parts of the element B. If A and C combine together the ratio of their weights will be governed by the
a) law of definite proportions
b) law of multiple proportions
c) law of reciprocal proportions
d) law of conservation of mass.
99)1.520 g of the hydroxide of a metal on ignition gave 0.995 g of oxide. The equivalent weight of the metal is
100)A solution of was titrated with solution in an acidic medium. The amount of used will be <I
a) 5 ml of 0.1 M
b) 10 ml of 0.1 M
c) 10 ml of 0.5 M
d) 10 ml of 0.02 M
Ans 1) a
Ans Desc 1) In Stock system, the oxidation number is given in Roman numerals in parenthesis at the ends of name of metals to represent the charge on the monatomic Cations. Eg: VO2+ is named as Oxo Vanadium(IV) ion. In the given compound UO2SO4, oxidation number of Uranium is six
Ans 2) a
Ans Desc 2)
Ans 3) c
Ans Desc 3)
The significant figures in 0.0200 are three because zeros to the left of the first non-zero digit are not significant, while zeros at the end but to be right of the decimal point is significant.
Ans 4) a
Ans Desc 4)
The number of significant figures in 0.0045 is two because zeros to the left of the first non-zero digit are not significant, while zeros at the end but to the right of the decimal point is significant.
Ans 5) c
Ans Desc 5) The number 10500 ends in zeros but these zeros are not to the right of a decimal point. These zeros may or may not be significant. It depends how it is expressed i.e. which have significant figures 3,4 and 5 respectively.
Ans 6) b
Ans Desc 6) The no. of significant figures in are 4.
Ans 7) b
Ans Desc 7) 4.523+2.3+6.24=13.063. As 2.3 has the least no. of decimal places i.e. one, therefore, the sum should be reported to one decimal place only. After rounding off, the reported sum = 13.1 which has three significant figures.
Ans 8) d
Ans Desc 8) 154.21+6.142+23=183.352. As 23 has no decimal places. i.e. it is a whole number , the reported sum = 183.
Ans 9) b
Ans Desc 9) 16.4215-6.01=10.4115. As 6.01 has the least no. of decimal places, viz. 2, therefore, the reported difference = 10.41 which has four significant figures.
Ans 10) d
Ans Desc 10) If the digit just next to the last digit to be retained is 5, the last significant figure is left unchanged if it is even and is increased by 1 if it is odd. Hence 1.235 = 1.24 and 1.225 = 1.22
Ans 11) a
Ans Desc 11) The product of 4.327 and 2.8 should have 2 significant figures because the least precise term viz. 2.8 has only two significant figures. After rounding off to two significant figures , 12.1156 = 12
Ans 12) b
Ans Desc 12) As the least precise term involved has two significant figures (viz., 0.46), therefore the reported answer =0.029.
Ans 13) c
Ans Desc 13) 5.00 is more accurate than 5.0 because the former has three significant figures while the latter has two.
Ans 14) d
Ans Desc 14) Amount of the substance is expressed in moles and not moles L-1
Ans 15) a
Ans Desc 15)
Ans 16) a
Ans Desc 16)
Hence 1 nm =1000 pm
Ans 17) c
Ans Desc 17) Self explanatory
Ans 18) c
Ans Desc 18) In 20,000 , all zeros are significant.
Ans 19) c
Ans Desc 19)
Ans 20) c
Ans Desc 20)
Ans 21) a
Ans Desc 21) Self explanatory
Ans 22) c
Ans Desc 22) Pure ethyl alcohol = 81.4-0.002=81.398, as 81.4 has the least number of decimal places viz., one, the reported answer after rounding off = 81.4
Ans 23) c
Ans Desc 23) Self explanatory
Ans 24) d
Ans Desc 24) Self explanatory
Ans 25) b
Ans Desc 25) hence significant figures are infinite.
Ans 26) c
Ans Desc 26) After rounding off, to express to three significant figures will be
Ans 27) d
Ans Desc 27) Force = mass acceleration. Hence
Ans 28) a
Ans Desc 28) Mass = 20.036 – 0.02 =20.016=20.02
(reported to two decimal places)
( as it should contain three significant figures)
Ans 29) c
Ans Desc 29) Silica is
Ans 30) c
Ans Desc 30) A compound is always heterogeneous.
Ans 31) c
Ans Desc 31) Self explanatory
Ans 32) d
Ans Desc 32) On heating iron rod, it changes into its oxide.
Ans 33) c
Ans Desc 33) Milk is a heterogeneous mixture, two phases being fats and water.
Ans 34) d
Ans Desc 34) Distilled water is a pure compound.
Ans 35) d
Ans Desc 35) Acetone (b.pt. = ) and methyl alcohol (b.pt. = ) have a small difference in their boiling points.
Ans 36) d
Ans Desc 36) Naphthalene sublimes, while sand does not.
Ans 37) b
Ans Desc 37) being insoluble in water can be separated by filtration. The filtrate containing can be subjected to crystallization.
Ans 38) b
Ans Desc 38) is insoluble while is soluble in water. Hence, they can be separated by crystallization
Ans 39) b
Ans Desc 39) sublimes while NaCl does not.
Ans 40) c
Ans Desc 40) Red and blue ink is absorbed to different extents. Hence they can be separated by chromatography.
Ans 41) a
Ans Desc 41) Thus 100 g give 22.4 L CO2 of STP. Hence 10 g will give 2.24 L at STP.
Ans 42) c
Ans Desc 42) Self explanatory
Ans 43) c
Ans Desc 43) The weights of oxygen which combine with the fixed weight of nitrogen (= 28 g) in are 16, 32, 48, 64 and 80 respectively. They are in the ratio 1:2:3:4:5. This proves the law of multiple proportions.
Ans 44) d
Ans Desc 44)
In the 1st Oxide , oxygen = 27.6 parts.
Metal = 100- 27.6=72.4 parts
In the 2nd oxide, oxygen = 30 parts
Metal = 100-30=70 parts.
As 1st oxide is parts of M= 3 atoms of M and 27.6 parts of O= 4 atoms of O
70 parts of M= atoms of M
= 2.9 atoms of M
30 parts of atoms of O
= 4.35 atoms of O
Ratio of M:O in the 2nd oxide = 2.9: 4.35
Hence the formula is
Ans 45) d
Ans Desc 45) The ratio of volumes of is 1:3:2 which is simple ratio. This proves Gay Lussacs law of gaseous volumes.
Ans 46) c
Ans Desc 46) Ratio of the weights of S and O combining with fixed weights of C= 64:32 = 2:1, Ratio of the weights of S and O combining directly 10:10=1:1. The two ratios are simple multiple of each other. This proves law of reciprocal proportions.
Ans 47) d
Ans Desc 47) Isotopes have nothing to do with laws of chemical combination.
Ans 48) a
Ans Desc 48)
1 mol O – atoms = 16 g, 1 g atom of S= 32 g, 22.4 L at STP= 1 mol = 32g
1 mol of molecules =
Ans 49) c
Ans Desc 49) Self explanatory
Ans 50) c
Ans Desc 50) Self explanatory
Ans 51) c
Ans Desc 51) Should be – one mole of hydrogen atoms contains number of atoms.
Ans 52) b
Ans Desc 52) Molecular mass is the mass of 22.4 L of the gas at NTP.
Ans 53) c
Ans Desc 53) 10.6 g of formula units
Ans 54) b
Ans Desc 54) 22400cc of any gas at STP contain number of molecules . Thus greater the volume of the gas greater is the number of molecules.
Ans 55) b
Ans Desc 55) 1 mol of
Ans 56) a
Ans Desc 56) 22400 cc of atoms,
11.2 cc of
Ans 57) d
Ans Desc 57) 0.1 g atom of C=1.2 g, 0.1 mol of
Ans 58) d
Ans Desc 58)
Ans 59) c
Ans Desc 59) Mass of 1 atom of
Ans 60) a
Ans Desc 60) Mass of molecule of
Ans 61) c
Ans Desc 61)
Ans 62) b
Ans Desc 62)
Ans 63) d
Ans Desc 63)
Ans 64) d
Ans Desc 64) 12 g C give molecules. Hence 0.6 g of C will give
Ans 65) b
Ans Desc 65)
Number of atoms present
Ans 66) c
Ans Desc 66) 2g He = 0.5 mole = 11.2 L at STP ( which is nearly the same at )
Ans 67) d
Ans Desc 67) Is based on definition of hypothesis.
Ans 68) c
Ans Desc 68) 100 g of crystalline salt contain
Anhydrous salt = 10 -59.6 = 40.4g
40.4 g of anhydrous salt have water = 59.6 g
122 g of anhydrous salt has water
Ans 69) d
Ans Desc 69) % of N in
Ans 70) d
Ans Desc 70) 1 mol of
Ans 71) c
Ans Desc 71) Self explanatory
Ans 72) c
Ans Desc 72) 48 g oxygen are combined with 254 g iodine.
8 g oxygen are combined with iodine 16 g of O are combined with I= 254 g. Hence 8 g O are combined with I = 127 g 80 g O are combined with I = 254 g. Hence 8 g O are combined with I = 25.4 g
Ans 73) c
Ans Desc 73)
No. of molecules
Ans 74) b
Ans Desc 74)
Ans 75) b
Ans Desc 75) milliequivalents . Hence 60 mL will contain.
Ans 76) c
Ans Desc 76)
, Hence mol. Formula
Ans 77) d
Ans Desc 77) Self explanatory
Ans 78) d
Ans Desc 78) 22400 ml at STP will weigh
Molecular mass of
Hence x = 256/32 = 8. So formula is
Ans 79) a
Ans Desc 79) moles
3.6 x 1022
Ans 80) a
Ans Desc 80) % of C in
% of C in
Ans 81) c
Ans Desc 81) 2moles of
molecules atoms of K.
= 8 x 6.02 x 1023
Ans 82) d
Ans Desc 82) Self explanatory
Ans 83) b
Ans Desc 83) At.wt. sp. Heat = 64 approx.(Dulong and law)
Ans 84) a
Ans Desc 84)
Ans 85) b
Ans Desc 85) Relative mass (atomic mass) of O on the chemical scale = 16
Ans 86) d
Ans Desc 86) The elements combine in the ratio of their equivalent weights.
Ans 87) c
Ans Desc 87) Eq. Wt. Of metal
Valency of metal (x)
Ans 88) b
Ans Desc 88) As 1 g combines with 80 g , eq.wt. of Br=80
4 g combine with 1 g Ca
80 f combine with Ca=20 g
Hence Eq.wt. of Ca=20
Ans 89) d
Ans Desc 89) 100 ml of 1 M NaOH = 100 ml of 1 N NaOH
10 ml of 10 N = 100 ml of 1 N
Ans 90) a
Ans Desc 90) Self explanatory
Ans 91) a
Ans Desc 91) 200 mg
No of moles left
Ans 92) a
Ans Desc 92)
Ans 93) d
Ans Desc 93)
Ans 94) c
Ans Desc 94)
200 cc of
Hence volume of water to be added
= 1000-200=800 cc
Ans 95) c
Ans Desc 95) Equal volumes under similar conditions will contain equal number of molecules.
Ans 96) c
Ans Desc 96) This shows that the weights of lead combining with fixed weight of oxygen are in the ratio which is in accordance with law of multiple proportions.
Ans 97) c
Ans Desc 97) will react with 0.1 M NaCl to form 0.1 M . But as the volume is doubled, conc. of
Ans 98) c
Ans Desc 98) Self explanatory
Ans 99) d
Ans Desc 99)
Ans 100) d
Ans Desc 100)