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Matter and its nature, Dalton’s atomic theory, Concept of atom, Molecule, Element, Compound, Precision, Significant figures, SI units, Derived units, Laws of chemical combinations, Mole, Mass, Molecular mass, Equivalent mass, Chemical equation, Stoichiometry of chemical equation and various levels of multiple-choice questions.
MATTER
Any species having mass and occupying space is known as matter. It can exist in the three physical states, namely, solid, liquid and gas.
Matter
Homogeneous mixtures
Heterogeneous mixtures
Elements Compounds
Mixtures^ Pure^ substances
Figure 1.1 Classification of Matter
Pencil, air, water, justify the physical states and are all composed of matter.
- At the bulk level or macroscopic level, we can fur- ther classify matter as mixtures or pure substances.
Mixture
A mixture is composed of two or more substances which are known as its components or constituents (in any ratio).
The components of the mixture can be separated with the help of physical separation methods like filtration, crystal- lization, distillation.
- A mixture is further classified into two categories— homogeneous and heterogeneous.
- In a homogeneous mixture all the components under- go complete mixing forming a uniform composition as, air or sugar solution.
- In a heterogeneous mixture the composition formed due to the mixing of components is not entirely uniform like in the case of grains mixed with dust etc.
Pure Substance
Pure substances have fixed compositions and their constit- uents cannot be separated by using simple physical meth- ods of separation.
- A pure substance can be further classified into an ele- ment or a compound.
- An element is composed of one type of particle which could either be atoms or molecules. Na, Cu, Ag have only one type of atoms.
- A compound is formed by the combination of two or more atoms or different elements. For example, H 2 O, CO 2.
Chapter Highlights
Basics of Chemistry
CHAPTER
BASIC CONCEPTS
1.2 Chapter 1
Dalton’s Atomic Theory
An atom is the smallest particle of an element which is neutral in nature, retains all the properties of the element and takes part in a chemical reaction. The word atom was introduced by Dalton ( alamos means undivided). The Dalton’s atomic theory was proposed by Dalton on the basis of laws of chemical combination.
Main assumptions
- Matter (of any type) is composed of atoms.
- An atom is the smallest, fundamental, undivided particle. (Building block material)
- An atom can neither be created nor destroyed.
- Atoms of an element have similar size, energy and properties while atoms of different element differ in these aspects.
- Atoms combine in whole number ratios to form a molecule, therefore, a molecule is the smallest iden- tity that exists individually.
Modern view about atom
According to the modem view:
- An atom is divisible into other smaller particles which are known as subatomic particles. It can also combine in non-whole number ratio as in the case of non-stoichiometric compounds (Berthollide com- pounds) like Fe0.93O.
- Atoms of same element also differ in mass and mass related properties as in the case of isotopes.
- Chemical reactions involve rearrangement of atoms.
Molecule
The term molecule was introduced by Avogadro. It is the smallest particle (identity) of matter that can exist inde- pendently and retains all the properties of the substance. Normally the diameter of the molecules is in the range of 4–20 Å and the molecular mass is between 2–1000.
- In case of macromolecules, the diameter is in the range of 50–250 Å and the molecular weight may be in lakhs.
Berzelius Hypothesis
According to the Berzelius hypothesis, “Equal volumes of all the gases contain same number of atoms under the similar conditions of temperature and pressure.” This hypothesis on application to law of combining volume confirms that atoms are divisible which is in con- trary to Dalton’s theory.
PHYSICAL QUANTITIES AND THEIR
MEASUREMENTS
In order to describe and interpret the behaviour of chem- ical species, we require not only chemical properties but also few physical properties. Physical properties are mass, length, temperature time, electric current etc. Further, to express the measurement of any physical quantity we require its numerical value as well as its unit. Hence, the magnitude of a physical quantity can be given as Magnitude of physical quantity 5 Its numerical value 3 Unit.
Precision and Accuracy
- The measurements are considered accurate when the aver- age value of different measurements is closer to the actual value. An individual measurement is considered more accurate when it differs slightly from the actual value.
- When the values of different measurements are close to each other as well as to the average value, such measurements are called precise.
- In fact, precision is simply the measurement of reproductability of an experiment.
Significant Figures
These are some uncertainties in values during measure- ment of matter. In order to make accurate measurements we use these figures. The total number of digits in a number including the last digit with uncertain value is known as the number of significant figures, for example, 14.3256 6 0.0001 has six significant figures.
Rules to determine significant numbers
- All non-zero digits as well as the zeros present between the non-zero digits are significant, for exam- ple, 6003 has four significant figures.
- Zeros to the LHS of the first non-zero digit in a given number are not significant figures, for example, 0.00336 has only three significant figures.
- In a number ending with zeros, if the zeros are pres- ent at right of the decimal point then these zeros are also significant figures, for example, 33.600 has five significant figures.
- Zeros at the end of a number without a decimal are not counted as significant figures, for example, 12600 has just three significant figures.
- The result of division or multiplication must be reported to the same number of significant figures as possessed by the least precise term, for example, 3.331 3 0.011 5 0.036641 ≈ 0.037.
1.4 Chapter 1
1 Exa (E) 5 1018 1 Zetta (Z) 5 1021 1 Yotta (Y) 5 1024
1 litre 5 1023 m^3 5 1 dm^3 1 atmosphere 5 760 mm or torr 5 101.325 Pa or Nm^22 1 bar 5 105 Nm^22 5 105 Pa 1 calorie 5 4.184 joule 1 eV (electron volt) 5 1.602 3 10219 joule 1 joule 5 1027 erg So, 1 eV 5 1.602 3 10212 erg 1 cal. 1 J. 1 erg. 1 eV
‘Barn’ is a unit of area to measure the cross section of nucleus. 1 Barn 5 10228 m^2 ≈ 10^224 cm^2
LAWS OF CHEMICAL COMBINATIONS
Law of Conservation of Mass
- Law of conservation of mass was proposed by Lavoisier in 1774.
- It was verified by Landolt.
- According to this law, “In a chemical change the total mass of the products is equal to the total mass of the reactants, that is, mass is neither created nor destroyed.” For example, when a solution with calcu- lated weight of AgNO 3 and NaCl is mixed, white pre- cipitates of AgCl are formed while NaNO 3 remains in solution. The weight of the solution remains the same before and after this experiment.
- It is not applicable to nuclear reactions.
Law of Constant Composition or Law of
Definite Proportion
- Law of constant composition was proposed by Proust in 1779.
- It was verified by Star and Richards.
- According to this law, “A chemical compound always contains same elements combined together in same proportion by mass.” For example, NaCl extracted from sea water or achieved from deposits will have 23 g Na and 35.5 g of chlorine in its one mole.
- It is not applicable to non-stoichiometric compounds like Fe0.93 O.
Law of Multiple Proportion
- Law of multiple proportion was proposed by Dalton in 1804.
- It was verified by Berzilius.
- According to this law, “Different weights of an ele- ment that combine with a fixed weight of another element bear a simple whole number ratio.” For example, in case of CO, and CO 2 weight of oxygen which combines with 12 g of carbon is in 1 : 2 ratio.
- It is applicable when same compound is prepared from different isotopes of an element. For example, H 2 O, D 2 O.
Law of Reciprocal Proportion
- Law of reciprocal proportion was proposed by Richter in 1792.
- It was verified by Star.
- According to this law, “When two different elements undergo combination with same weight of a third element, the ratio in which they combine will either be same or some simple multiple of the ratio in which they combine with each other.”
- It is also known as Law of equivalent proportion which states “Elements always combine in terms of their equivalent weight.”
Law of Combining Volume
- Law of combining volume was proposed by Gay-Lussac.
- It applies to gases.
- According to this law, “When gases react with each other they bear a simple whole number ratio with one another as well as the product under conditions of same temperature and pressure.”
AVOGADRO’S LAW
- Avogadro’s law explains law of combining volumes.
- According to this law “Under similar conditions of temperature and pressure equal volume of gases contain equal number of molecules.”
- It is used in:
- Deriving molecular formula of a gas
- Determining atomicity of a gas
- Deriving a relation Molecular mass 5 2 3 Vapour Density (M 5 2 3 V.D.)
- Deriving the gram molecular volume
Basics of Chemistry 1.
- Avogadro number (N 0 or NA) 5 6.023 3 1023.
- Avogadro number of gas molecules occupies 22.4 litre or 22400 mL or cm^3 volume at STP.
- The number of molecules in 1 cm^3 of a gas at STP is equal to Loschmidt number that is, 2.68 3 1019.
- Reciprocal of Avogadro number is known as avogram.
MOLE
- Mole is a unit which represents 6.023 3 1023 par- ticles, atoms, molecules or ions etc., irrespective of their nature.
- Mole is related to the mass of substance, the volume of gaseous substance and the number of particles.
- Volume of one mole of any gas is equal to 22.4 litres or 22.4 dm^3 at STP. It is known as molar volume.
- Mole 5
W
M
_____________________Wt of substance in g Molar mass of substance (g.m.m.)
Here, g.m.m. 5 Gram molecular mass
Mole 5 ____________________Vol. of substance in litre22.4 L
Mole Concept: An Example A mole of any substance (like N 2 ) stands for:
- 6.023 31023 molecules of N 2
- 2 3 6.023 31023 atoms of nitrogen
- 28 g of nitrogen
- 22.4 litre of N 2 at STP
To find total number of identities
⇒ Total Number of Molecules = mole (n) × NA ⇒ Total Number of Atoms in = mole (n) × NA × No. of atoms present in one molecule ⇒ Total Number of Electrons = mole (n) × NA × No. of electrons in one molecule ⇒ Total Charge on Any Ion = mole (n) × NA × charge on one ion × 1.6 × 10–19C
MASS
Mass can be expressed in terms of atoms or molecules as follows:
Atomic Mass
Atomic mass is the relative mass of an atom which shows the number of times an atom is heavier than __ 121 mass of C-12.
- The atomic mass of any element expressed in grams is called g.a.m. (gram atomic mass) or gram atom.
- A gram atom has number of atoms of the element. Atomic mass 5 E 3 V Here, E 5 Equivalent weight V 5 Valency
- Atomic mass 5 6.
____________________
Specific heat in calories This is known as Dulong Petit’s Law.
Atomic mass unit
The quantity of
mass of an atom of C-12 represents it and it
is abbreviated as a.m.u
1 A.m.u =
11 × (^) = × 24 − −
Atomic mass Mass of one atom of an element 1 A.m.u.
Here 1. 99 × 10–23^ g is wt. of one C–12 atom
Average atomic mass
(At. mass)AV.^
m a m b m c a b c
= 1 ×^ +^2 ×^ +^2 ×
Here m 1 , m 2 , m 3 are masses of isotopes and a, b, c are their percentage ratio.
Molecular Mass
Molecular mass represents the total mass of a molecule,
that is, number of times a molecule is heavier than
weight of C–12 atom or
weight of an oxygen atom
Determination of molecular mass Vapour density method Mol. mass 5 2 3 V.D.
V.D. 5
___________________W^3
Volume at STP (in mL) Here, W 5 Weight of substance in g V.D. = Vapour Density
Basics of Chemistry 1.
(c) Chloride formation method
E 5 ____________Wt of chlorideWt of metal 3 35.
Weight of chloride 5 Weight of metal chloride 2 Weight of metal (d) Double decomposition method
Eq. wt. of salt taken Eq. Wt. of salt ppt.
Wt. of salt taken Wt. of salt ppt.
(e) Metal displacement method E___ 1 E 25
___W 1
W 2
MOLE FRACTION
- Mole fraction is t he ratio of moles of one component to the total number of moles present in the solution. It is expressed by X, for example, for a binary solu- tion of two components A and B.
XA 5 _______nA nA 1 nB
XB 5 _______nB nA 1 nB
XA 1 XB 5 1
- Mole fraction of solute (X 2 ) 5 _______n 2 n 1 1 n 2 Here, n 1 and n 2 represent number of moles of solvent and solute respectively.
- Mole fraction does not depend upon temperature as both the solute and the solvent are expressed by weight.
CHEMICAL EQUATION AND
STOICHIOMETRY OF CHEMICAL
REACTIONS
1. A balanced chemical reaction represents a stoichio- metric equation. 2. In a stoichiometric equation, the coefficient of reactants and products represents their stoichiometric amounts. 3. The reactant which is completely used up during an irre- versible reaction is called the limiting reagent while the reactant left is called the excess reagent , for example, 20 g of calcium is burnt in 32 g of O 2 , then Ca is the limiting reagent while O 2 is the excess reagent. 4. Stoichiometric calculations help in finding whether the production of a particular substance is economi- cally feasible or not. 5. These stoichiometric calculations are of following four types:
(a) Calculations based on weight–weight relationship (b) Calculations based on weight–volume relationships (c) Calculations based on volume–volume relationships (d) Calculations based on weight–volume–energy relationships
6. If the amount of the reactant in a particular reaction is known, then the amount of the other substance needed in the reaction or the amount of the product formed in the reaction can be calculated. 7. For stoichiometric calculations the following steps must be considered: (a) A balanced chemical equation using chemi- cal formulas of reactants and products must be written. (b) Here, the coefficients of balanced chemical equa- tion provide the mole ratio of the reactants and products. (c) This mole ratio is convertible into weight– weight (w/w) ratio, weight–volume (w/v) ratio or volume–volume (v/v) ratio. These are called percentage by weight, percentage by volume and percentage by strength respectively.
SOLVED EXAMPLES
Mole Concept
1. If 1 Faraday was to be 60230 coulombs instead of 96500 coulombs, what will be the charge on an electron?
Solution One mole electron carries 1 Faraday charge. 6.023 3 1023 electrons carry 5 60230 C
1 electron carries 5
___________60230 C
51 3 10219 C.
2. If a piece of copper weighs 0.635 g, how many atoms does it contain?
Solution Number of moles of Cu in 0.635 g
5 ___________0.635 g 63.5 g mol^21 5 1022 mol
1 mole Cu contains 6.023 3 1023 atoms of Cu 1022 mole Cu contains 6.023 3 1023 3 1022 5 6.023 3 1021 atoms of Cu.
1.8 Chapter 1
The volume of H 2 at NTP given by Zn
(1.67______________ 2 A) 22.
65.4 L^ (ii) From (i) and (ii) (^3) ____________ 3 22.4 3 A 54 1
(1.67______________ 2 A) 22.
142.2 3 A 5 176.
A 5 1.248 g
7. Find the equivalent mass of H 3 PO 4 in the reaction: Ca(OH) 2 1 H 3 PO 4 CaHPO 4 1 2H 2 O Solution As in this reaction only two hydrogen atoms are replaced so its equivalent mass will be given by the following expression: Equivalent mass of H 3 PO 4
5 _____________________Molecular mass of H 3 PO 4 2 5
___^98
8. How many years would it take to spend Avogadro num- ber of rupees at the rate of 10 lac rupees per second?
Solution Avogadro number 5 6.023 3 1023 Total rupees 5 6.023 3 1023 Rs Rate of spending 5 10 lac rupees/s 5 106 Rs/s Number of years to spend all the rupees
5 6.023^3 ______________________________^23 Rs 1036 603 603 243 365 Rs/year 5 1.90988 3 1010 years
9. Oxygen is present in a one litre flask at a pressure of 7.6 3 10210 mm of Hg. Calculate the number of oxy- gen molecules in the flask at 0 °C.
Solution Since, PV 5 nRT ___________________7.6 3 10210 atm 3 1 L 760 5 n 3 0.0821 L atm K^21 mol^21 3 273 K
n 5 7.6^3 __________________________________^210 L atm 7603 0.0821 L atm K^21 mol^21 3 273 K
5 10 _________^212 mol
Number of molecules 5 (6.02 3 1023 mol^21 ) 3 10 _________^212 mol
5 2.68 3 1010.
3. Calculate the number of atoms of oxygen present in 88 g of CO 2. What would be the mass of CO having the same number of oxygen atoms?
Solution
Number of moles of CO 2 5 __________88 g 44 g mol^21 5 2 moles 1 mole of CO 2 contains 2 moles of oxygen atoms, 2 moles of CO 2 will contain 4 moles of oxygen atoms. Number of oxygen atoms 5 4 3 6.023 3 1023 5 2.5092 3 1024 1 mole oxygen atom is present in 1 mole of CO, 4 moles oxygen atoms are present in 4 moles of CO Its mass is 4 (12 1 16) 5 112 g.
4. Calculate the total number of electrons present in 1. gram of methane.
Solution Molecular mass of methane 5 16 g mol^21 16 g CH 4 contains 6.02 3 1023 molecules of CH 4 1.6 g CH 4 contains 6.02 3 1022 molecules of CH 4 As one molecule of CH 4 contains (6 1 4) 5 10 ele- ctrons, 6.02 3 1022 molecules of CH 4 will have 10 3 6.02 3 1022 5 6.02 3 1023 electrons.
5. How many atoms of carbon has a young man given to his bride-to-be if the engagement ring contains 0. carat diamond? (1 carat 5 200 mg)
Solution Mass of diamond (C) 5 0.5 3 200 mg 5 100 mg 5 100 3 1023 g 5 0.1 g
Number of mole of C 5 __________0.1 g 12 g mol^21 5 1/120 mole Number of C atoms 5 ____ 1201 3 6.023 3 1023 5 5.02 3 1021.
6. A mixture of aluminium and zinc weighing 1.67 grams was completely dissolved in acid and the evolved 1. litres of hydrogen gas was measured at 273 K and one atmosphere pressure. What was the mass of alumin- ium in the original mixture?
Solution Let the mass of aluminium in the sample be ‘A’ g. The mass of Zn 5 (1.67 2 A) g The volume of H 2 at NTP given by Al (i)
5
____________^3 3 22.4^3 A
2 3 27 L
1.10 Chapter 1
16. Calculate the volume of water to be added to a 100 mL of 5N solution to make it 0.01 N.
Solution According to normality equation, N 1 V 1 5 N 2 V 2 0.01 3 V 1 5 5 3 100
V 1 5
5 _______ 3 100
0.01 5 50000 mL So, volume of water to be added 5 50000 2 100 5 49900 mL 5 49.9 L.
17. A small amount of CaCO 3 completely neutralizes 525 mL of 0.1 N HCl and no acid is left in the end. After converting all calcium chloride to CaSO 4 , how much plaster of paris can be obtained?
Solution 525 mL of 0.1 N HCl 5 525 mL of 0.1 N CaCl 2 5 525 mL of 0.1 N plaster of paris Molecular mass of plaster of paris 5 145 Equivalent mass of plaster of paris 5
145 ____
Mass of plaster of paris in 525 mL of 0.1 N solution 5
N__________ 3 E 3 V
5 0.1_______________^3 72.5 10003 525
5 3.806 g.
Calculations Based on Reactions
18. Metallic tin in the presence of HCl is oxidized by K 2 Cr 2 O 7 solution to stannic chloride. What volume of decinorrnal dichromate solution would be reduced by 1 g of Sn?
Solution 3Sn 1 2K 2 Cr 2 O 7 1 28 HCl 3SnCl 4 1 4KCl 1 33 118.7 (^2 3 294) 4CrCl g g 3 1 14H 2 O K 2 Cr 2 O 7 required for 1 g of Sn 5 _________ 32 3 3 118.7^294 5 1.65 g.
19. How many grams of CaO are required to neutralize 852 g of P 4 O 10?
Solution The reaction is as follows: 6CaO 1 P 4 O 10 2Ca 3 (PO 4 ) 2 852 g P 4 O 10 3 mol P 4 O 10
1 mole of P 4 O 10 neutralizes 6 moles of CaO 3 moles of P 4 O 10 will neutralize 18 moles of CaO Mass of CaO 5 18 3 56 5 1008 g.
20. Find the weight of iron which will be converted into its oxide by the action of 18 g of steam. Solution The reaction is 3Fe 1 4H 2 O Fe 3 O 4 1 4H 2 4 moles steam reacts with 3 moles Fe 1 mole (18 g) steam reacts with 3/4 moles Fe 5
3 __
4 mole^3 56 g mol
21
5 42 g Fe.
21. The mineral haematite is Fe 2 O 3. Haematite ore con- tains unwanted material called gangue in addition of Fe 2 O 3. If 5.0 kg of ore contains 2.78 kg of Fe, what per cent of ore is Fe 2 O 3? Solution 2Fe Fe 2 O 3 2 55.85 g 3 159.7 g 2 55.85 g Fe is present in 159.7 g Fe 3 2 O 3 2.78 kg Fe is present in
=
159.7 g_______________ 3 2.78 kg 32 55.85 g 5 3.97 kg Fe 2 O 3 As 5 kg ore contains 5 3.97 kg Fe 2 O 3
So, 100 kg ore contains 5
3.97__________ 3 100
5 79.4 kg Fe 2 O 3 Thus, percentage of Fe 2 O 3 in ore 5 79.4%.
22. What should be the weight of NaNO 3 to make 50 mL of an aqueous solution so that it contains 70 mg Na mL^21? Solution Molecular mass of NaNO 3 5 23 1 14 1 3 3 16 5 85 g mol^21 23 mg Na is present in 85 mg of NaNO 3
70 mg Na is present in 5 85 _______^323
5 258.7 mg NaNO 3 1 mL solution contains 258.7 mg NaNO 3 50 mL solution contains 258.7 mg________________ 3 50 mL 1 mL = 13935 mg = 13.935 g.
Basics of Chemistry 1.
CONCEPTS AT A GLANCE
Giorgi introduced MKS system. π has infinite number of significant numbers. 1 mole of H 2 O ≠ 2240 0 mL or cc of H 2 O (since it is liquid) 1 mole of H 2 O = 18 cc of H 2 O (as density of H 2 O = 1g/cc) Mass of one mole of e^2 = mass of one e^2 × NA = 9.1 × 10231 × 6.02 × 1023 = 0.55 mg 20 carat gold is a mixture having 20 parts by weight of gold and 4 parts by weight of copper. Some substances like CuSO 4 .5H 2 O, Na 2 CO 3 .10H 2 O have a tendency to lose water in air. These are called efflorescent substances and this tendency is called efflorescence. Some solid substances like NaOH, KOH, which have a tendency to absorb moisture greatly from air and to get wet are called deliquescent and this tendency is called deliquescence. Hygroscopic substances like quicklime (CaO) anhydrous P 2 O 5 etc., absorb moisture from air. Compounds having similar chemical composition in the same crystalline form are called isomorphs.
For example, all alums [M 2 SO 4. M 2 (SO 4 ) 3. 24H 2 O] Here, M = Monovalent (K) M = Trivalent (Al) FeSO 4 .7H 2 O (Green vitriol) and ZnSO 4 .7H 2 O Different crystalline forms of a substance are called polymorphs and this phenomenon is called polymorphism. For example, ZnS → Zinc blende ↓ Wurtzite To find equivalent weight of an acid, Silver Salt Formation Method is used Eq. wt of RCOOAg_________________ 108 =^
Wt of RCOOAg______________ Wt of Ag Eq. wt of R–COOH = Eq. wt of RCOOAg – 107 Equivalent weight (E) = Weight deposited by 96500 coulombs or 1 Faraday.
TOOL BOX AND SUMMARY OF IMPORTANT RELATIONS
Mole (n) 5 ___W M 5
______V 22.4 L 5
___N N 0 Here, W 5 Weight M 5 Molecular weight N 5 Number of atoms/molcules V 5 Volume in litre Molar mass (M) 5 2 3 V.D. (Vapour density)
V.D. 5 ___________________W^3 Volume at STP (in mL) Here, W 5 Weight of substance in g
r r
= M M
1 2
2 1 Here rl, r 2 are rates of diffusion for two species while M 1 , M 2 are their molecular masses respectively.
p V 5 ___WM RT Here, p = Osmotic pressure in atm V = Volume in litre W = Weight in gram R = Universal gas constant T = Given temperature
Equivalent weight (E)
=
=
=
E (^) + − M H or OH or Charge
E M Change in oxidation number
E = M Number of replaceable hydrogen atoms E E
W W
1 2
1 2
Specifi gravity =
Mass of Liquid vol. of liquid
(^) M = % by mass^ ×^ d^ ×^10 ×^ × M. wt. of solute ; N = % mass d 10 Eq. wt. of solute
× × − ×
m = M 1000 (1000 d M M.wt.)
(Here ‘d’ is density
of solution in g cm–3, M is molarity, N is normality and ‘m’ is molality)
; ,
, (^) ;
; ,
-F
, (^) ;
Basics of Chemistry 1.
19. Normality of 0.04 M H 2 SO 4 is
(a) 0.02 N (b) 0.01 N (c) 0.04 N (d) 0.08 N
20. Which among the following is the heaviest?
(a) one mole of oxygen (b) one molecule of sulphur trioxide (c) 100 amu of uranium (d) 44 g of carbon dioxide
21. The empirical formula of a commercial ion exchange resin is C 8 H 7 SO 3 Na. The resin can be used to soften water according to the reaction Ca+2^ + 2C 8 H 7 SO 3 Na → (C 8 H 7 SO 3 ) 2 Ca + 2Na+. What would be the maximum uptake of Ca+2^ by the resin expressed in mole/g resin? (a) 0.0024 (b) 0. (c) 0.246 (d) 24. 22. A boy drinks 500 mL of 9% glucose solution. The number of glucose molecules he has consumed are [mol. wt of glucose 5 180] (a) 0.5 31023 (b) 1.0 31023 (c) 1.5 31023 (d) 2.0 31023 23. The pollution of SO 2 in air is 10 ppm by volume. The volume of SO 2 per litre of air is (a) 10^22 mL (b) 10^23 mL (c) 10^24 mL (d) 10^26 mL 24. The molarity of pure water is
(a) 55.56 M (b) 5.56 M (c) 1.0 M (d) 0.01 M
25. The number of grams of a dibasic acid (molecular weight 200) present is 100 mL of its aqueous solution to give decinormal strength is (a) 1 g (b) 2 g (c) 3 g (d) 4 g 26. Normality of 0.3 M H 3 PO 4 solution is
(a) 0.1 N (b) 0.45 N (c) 0.6 N (d) 0.9 N
27. 2 g of O 2 at NTP occupies the volume
(a) 1.4 L (b) 2.8 L (c) 11.4 L (d) 3.2 L
28. The molecular weight of O 2 and SO 2 are 32 and 64 respectively. At 15 8 C and 150 mm Hg pres- sure, one litre of O 2 contains ‘N’ molecules. The number of molecules in two litres of SO 2 under the same conditions of temperature and pressure will be
(a) N (b) N__ 5 (c) 4N (d) 2N
29. 7.5 gram of a gas occupies 5.6 litres as STP. The gas is (a) CO (b) NO (c) CO 2 (d) N 2 O 30. 50 gram of calcium carbonate was completely burnt in air. What is the weight (in grams) of the residue? (a) 28 (b) 2. (c) 46 (d) 4. 31. At STP the density of a gas (mol. wt 5 45) in g/L is (a) 11.2 (b) 1000 (c) 2 (d) 224 32. The amount of NH 3 formed by the combustion of 2 L of N 2 and 2 L of H 2 is? (a) 2 L (b) 1 L (c) 0.66 L (d) 1.33 L 33. How many moles of acidified FeSO 4 can be completely oxidized by one mole of KMnO 4? (a) 20 (b) 10 (c) 5 (d) 0. 34. A compound possess 8% sulphur by mass. The least molecular mass is (a) 200 (b) 400 (c) 155 (d) 355 35. The vapour density of ozone is (a) 24 (b) 16 (c) 48 (d) 72 36. The weight of one molecule of a compound C 60 H 122 is (a) 1.3 310220 g (b) 5.01 3 10221 g (c) 3.72 31023 g (d) 1.4 3 10221 g 37. 1000 g calcium carbonate solution contains 10 g car- bonate. The concentration of solution is (a) 10 ppm (b) 100 ppm (c) 1000 ppm (d) 10,000 ppm 38. One mole of CH 4 contains (a) 4.0 g atoms of hydrogen (b) 3.0 g atom of carbon (c) 6.02 31023 atoms of hydrogen (d) 1.81 31023 molecules of CH 4 39. The amount of O 2 for me a at N.T.P by the complete combastion of 1 kg coal is? (a) 22.4 L (b) 2240 L (c) 1866 L (d) 100 L
1.14 Chapter 1
40. The maximum number of molecules is present in (a) 15 L of H 2 gas at STP (b) 5 L of N 2 gas at STP (c) 1.5 g of H 2 gas (d) 5 g of O 2 gas 41. Molarity of liquid HCl if density of solution is 1.17 g/ cc, is (a) 32.05 (b) 12. (c) 3.05 (d) 22. 42. The incorrect statement for 14 g of CO is (a) it occupies 2.24 litre at NTP (b) it corresponds to 0.5 mol of CO (c) it corresponds to same mol of CO and N 2 (d) it corresponds to 3.01 3 1023 molecules of CO 43. How many moles are present in 2.5 litre of 0. MH 2 SO 4? (a) 0.25 (b) 0. (c) 0.75 (d) 0. 43. Area of nuclear cross section is measured in ‘Barn’. It is equal to (a) 10^228 m^2 (b) 10^218 m^2 (c) 10^28 m^2 (d) 10^234 m^2 44. Which of the following statement is correct? (a) 1 mole of electrons weighs 5.4 mg (b) 1 mole of electrons weighs 5.4 kg (c) 1 mole of electrons weighs 0.54 mg (d) 1 mole of electrons has 1.6 3 10219 C of charge 45. Which of the following pairs of gases contain equal number of molecules? (a) CO 2 and NO 2 (b) CO and (CN) 2 (c) NO and CO (d) N 2 O and CO 2 46. The samples of NaCl are produced when Na combines separately with two isotopes of chlorine Cl^35 and Cl^37. Which law is illustrated? (a) Law of constant volume (b) Law of multiple proportions (c) Law of reciprocal proportions (d) None of these 47. Which of the following is the odd one with regard to mass? (a) 1 g atom of sulphur (b) 0.5 moles of CO 2 (c) 1 mole of O 2 (d) 3 31023 molecules of SO 2 48. A breakfast cereal in advertised to contain 110 mg of sodium per 100 g of the cereal. The per cent of sodium in the cereal is (a) 0.110 % (b) 0.01 10 % (c) 11.0 % (d) 0.22 % 49. Express 145.6 L of chlorine in terms of gram moles. (a) 6.5 g moles (b) 4.5 g moles (c) 0.65 g moles (d) 9.5 g moles 50. The number of significant figures in 306.45 and 40440 are respectively (a) 4, 5 (b) 5, 5 (c) 5, 4 (d) 4, 6 51. The quantity of ____KPV BT
represents the (a) molar mass of a gas (b) number of molecules in a gas (c) mass of gas (d) number of moles of a gas
52. Which is the correct order of micro, nano, femto and pico here? (a) micro < nano < pico < femto (b) pico < femto < nano < micro (c) femto < pico < nano < micro (d) femto < nano < micro < pico 53. Find the number of atoms present in 0.016 g of methane. (a) 0.5 N 0 (b) 0.05 N 0 (c) N 0 (d) 1.6 N 0 54. 15 litre atmosphere is equal to (a) 1.515 3108 erg (b) 15.15 3 109 erg (c) 1.515 31010 erg (d) 15.15 3 1012 erg 55. If equal moles of water and urea are taken in a ves- sel what will be the mass percentage of urea in the solution? (a) 22.086 (b) 11. 536 (c) 46.146 (d) 23. 56. Mixture X 5 0.02 mol of [Co(NH 3 ) 5 SO 4 ]Br and [Co(NH 3 ) 5 Br]SO 4 was prepared in 2 litre of solution. 1 litre of mixture X 1 excess AgNO 3 Y 1 litre of mixture X 1 excess BaCl 2 Z Number of moles of Y and Z are (a) 0.02, 0.01 (b) 0.01, 0. (c) 0.01, 0.02 (d) 0.02, 0. 57. At STP the density of CCl 4 vapour in g/L will be nearest to (a) 8.67 (b) 6. (c) 5.67 (d) 4.
1.16 Chapter 1
76. In the standardization of Na 2 S 2 O 3 using K 2 Cr 2 O 7 by iodometry, the equivalent weight of K 2 Cr 2 O 7 is
(a) same as mol. wt (b)
mol. wt 2
(c) mol. wt 4
(d)
mol. wt 6
77. The number of molecules in 4.25 g of ammonia is (a) 1.5 31023 (b) 2.5 31023 (c) 3.5 31023 (d) 15 31023 78. The volume in litres of CO 2 liberated at STP, when log of 90% pure limestone is heated completely is (a) 2.24 (b) 22. (c) 2.016 (d) 20. 79. The weight of a single atom of oxygen is (a) 5.057 31023 g (b) 1.556 3 1023 g (c) 2.656 310223 g (d) 4.538 3 10223 g 80. From the complete decomposition of 20 g CaCO 3 at STP the volume of CO 2 obtained is (a) 2.24 L (b) 4.48 L (c) 44.8 L (d) 48.4 L 81. 5 g of CH 3 COOH is dissolved in one litre of ethanol. Suppose there is no reaction between them. If the density of ethanol is 0.789 g/mL then the molality of resulting solution is (a) 0.0256 (b) 0. (c) 1.1288 (d) 0. 82. 800 g of a 40% solution by weight was cooled. 100 g of solute precipitated. The percentage composition of remaining solution is (a) 31.4% (b) 57.6% (c) 45.8% (d) 41.4% 83. 0.25 mol of P 4 molecules contains_______atoms. (a) 1.764 31023 (b) 6.02 31019 (c) 6.023 31023 (d) 8.086 31023 84. How many grams of CH 3 OH would have to be added to water to prepare 150 mL of a solution that is 2.0 M CH 3 OH? (a) 9.6 g (b) 906 g (c) 4.3 3102 g (d) 9.6 3 103 g 85. The oxide of an element contains 67.67% of oxygen and the vapour density of its volatile chloride is 79. Equivalent weight of the element is (a) 2.46 (b) 3. (c) 4.36 (d) 4. 86. The molar concentration of 20 g of NaOH present in 5 litre of solution is (a) 0.1 mol/L (b) 0.2 mol/L (c) v1.0 mol/L (d) 2.0 mol/L 87. Volume of a gas at NTP is 1.12 3 1027 cc. The number of molecules in it is (a) 3.01 31012 (b) 3.01 31018 (c) 3.01 31024 (d) 3.01 31030 88. Maximum number of molecules will be in (a) 1 g of H 2 (b) 10 g of H 2 (c) 22 g of O 2 (d) 44 g of CO 2 89. Haemoglobin contains 0.33% of iron by weight. The molecular weight of haemoglobin is approximately 67200. The number of iron atom (at. wt of Fe is 56) present in one molecule of haemoglobin are (a) 1 (b) 6 (c) 4 (d) 2 90. The equivalent weight of phosphoric acid (H 3 PO 4 ) in the reaction: NaOH 1 H 3 PO 4 NaH 2 PO 4 1 H 2 O is (a) 89 (b) 98 (c) 59 (d) 29 91. 2.76 g of silver carbonate (At. mass of Ag 5 108) on being heated strongly yields a reduce weigh (a) 2.32 g (b) 3.32 g (c) 1.36 g (d) 0. 92. 4 g caustic soda is dissolved in 100 cc of solution. The normality of solution is (a) 1 (b) 0. (c) 0.6 (d) 0. 93. When a mixture of Na 2 CO 3 and NaH CO 3 was heated at 423K 112 ml of CO 2 was formed only. What is the % of Na 2 CO 3 here in the mixture? (a) 84% (b) 16% (c) 32% (d) 68% 94. The mass of BaCO 3 formed where excess of CO 2 is passed through a solution having 0.205 mole of Ba (OH) 2 is? (a) 40.5 gm (b) 20.25 gm (c) 81 gm (d) 4.05 gm 95. 120 gram of urea is present in 5 litre of solution. The active mass of urea is (a) 0.06 (b) 0. (c) 0.4 (d) 1.
Basics of Chemistry 1.
96. The normality of orthophosphoric acid having purity of 70 % be weight and specific gravity 1.54 is (a) 11 N (b) 22 N (c) 33 N (d) 44 N 97. 1021 molecules are removed from 200 mg of CO 2. The moles of CO 2 left are (a) 2.88 31023 (b) 28.8 31023 (c) 288 31023 (d) 28.8 3103 98. What is the volume (in litres) of oxygen at STP required for complete combustion of 32 g of CH 4? (mol. wt of CH 4 5 16) (a) 89.6 (b) 189. (c) 98.4 (d) 169. 99. Two grams of sulphur is completely burnt in oxygen to form SO 2 , In this reaction, what is the volume (in litres) of oxygen consumed at STP? (At. wt of sulphur and oxygen are 32 and 16 respectively)
(a)
______22.
16 (b)^
_______ 16
(c) ______32.414 18 (d) 42.414______ 16
100. How many water molecules are there in one drop of water (volume 5 0.0018 mL) at room temperature? (a) 4.86 31017 (b) 6.023 31024 (c) 2.584 31019 (d) 6.023 31019 101. ‘X’ litres of carbon monoxide is present at STP. It is completely oxidized to CO 2. The volume of CO 2 formed is 11.207 litres at STP. What is the value of ‘X’ in litres? (a) 32.2 (b) 21. (c) 10.2 (d) 11. 102. Which has maximum number of molecules?
(a) 7 g N 2 (b) 2 g H 2 (c) 18 g NO 2 (d) 16 g O 2
103. One mole of fluorine is reacted with two mole of hot and concentrated KOH. The products formed are KF, H 2 O and O 2. The molar ratio of KF, H 2 O and O 2 respectively is (a) 1 : 2 : 1 (b) 1 : 2 : 2 (c) 0.5 : 1 : 2 (d) 2 : 1 : 0. 104. The total number of protons in 10 g of calcium carbon- ate is (N 0 5 6.023 3 1023 ) (a) 3.01 31024 (b) 4.06 31024 (c) 30.1 31024 (d) 3.01 31023 105. What is the volume (in litre) of oxygen required at STP to completely convert 1.5 moles of sulphur to sul- phur dioxide? (a) 33.6 (b) 43. (c) 11.2 (d) 23. 106. In acidic medium, dichromate ion oxidize ferrous ion to ferric ion. If the gram molecular weight of potas- sium dichromate is 294 g, its equivalent weight is (a) 19 (b) 49 (c) 99 (d) 294 107. 10 g of CaCO 3 is completely decomposed to X and CaO. X is passed into an aqueous solution containing one mole of sodium carbonate. What is the number of moles of sodium bicarbonate formed? (Mol. wt of CaCO 3 5 100, Na 2 CO 3 5 106, NaHCO 3 5 84) (a) 0.010 (b) 0. (c) 0.4 (d) 10 108. The mass of carbon anode consumed (giving only CO 2 ) in the production of 270 kg of aluminium metal from bauxite by the Hall process is (atomic mass of Al 5 27) (a) 180 kg (b) 270 kg (c) 145 kg (d) 90 kg 109. The number of moles of KMnO 4 reduced by one mole of KI in alkaline medium is (a) 2 (b) 1 (c) 5 (d) 6 110. How many grams of dibasic acid (mol. wt 200) should be present in 100 mL of the aqueous solution to give 0.1 normality? (a) 1 g (b) 1.5 g (c) 0.5 g (d) 20 g 111. ‘X’ gram of calcium carbonate was completely burnt in air. The weight of solid residue formed is 28 g. What is the value of ‘X’ (in grams)? (a) 50 (b) 100 (c) 150 (d) 200 112. One mole of acidified K 2 Cr 2 O 7 on reaction with excess KI will liberate …… moles (s) of I 2. (a) 2 (b) 3 (c) 6 (d) 7 113. 0.59 g of the silver salt of an organic acid (molar mass 210) on ignition gave 0.36 g of pure silver. The basic- ity of the acid is (a) 2 (b) 3 (c) 4 (d) 5
Basics of Chemistry 1.
129. 500 mL of NH 3 contains 6.0 3 1023 molecules at STP. How many molecules are present in 100 mL of CO 2 at STP? (a) 6 31023 (b) 1.5 3 1023 (c) 1.2 31023 (d) none of these 130. In the reaction, 4NH 3 1 5O 2 4NO 1 6H 2 O, when one mole of ammonia and one mole of oxygen are made to react to completion, then (a) 1.0 mol of H 2 O is produced (b) all the oxygen is consumed (c) 1.5 mol of NO is formed (d) all the ammonia is consumed 131. The number of gram molecules of oxygen in 6.02 3 1024 CO molecules is (a) 10 g molecules (b) 5 g molecules (c) 1 g molecules (d) 0.5 g molecules 132. The number of oxalic acid molecules in 100 ml of 0. N oxalic acid solution is (a) 6.023 31022 (b) 10^23 (c) 6.022 31020 (d) none of these 133. In the reaction
4NH 3 (g) 1 5O 2 (g) 4NO (g) 1 6H 2 O (l) when 1 mol of ammonia and 1 mol of O 2 are made to react to completion then (a) 1.0 mol of H 2 O is produced (b) 1.0 mol of NO will be produced (c) all the ammonia will be consumed (d) all the oxygen will be consumed
134. Pressure in a mixture of 4 g of O 2 and 2g of H 2 con- fined in a container of 1 litre capacity at 0 °C is (a) 25.2 atm (b) 35.6 atm (c) 15.4 atm (d) 48.2 atm 135. What is the volume (in litres) of CO 2 liberated at STP, when 2.12 gram of sodium carbonate (mol. wt 5 106) is treated with excess dilute HCl? (a) 11.2 (b) 2. (c) 0.448 (d) 4. 136. Percentage of Se in peroxidase anhydrous enzyme is 0.5% by weight (at. wt 5 78.4) then minimum molecu- lar weight of peroxidase anhydrous enzymes is (a) 1.568 3 103 (b) 1.568 3 104 (c) 25.68 (d) 4.316 3 104 137. For the formation of 3.65 gm of HCl, what volume of H 2 , and Cl 2 are needed at N.T.P? (a) 1.12 L, 1. 12 L (b) 1.12 L, 2.24 L (c) 3.65 L, 1.83 L (d) 1 L, 1 L 138. Study the following table:
Table 1. Compound (mol. wt)
Wt of compound (in g) taken
- CO 2 (44) 4.
- NO 2 (46) 2.
- H 2 O 2 (34) 6.
- SO 2 (64) 1.
Which two compounds have least weight of oxygen? (a) 1 and 2 (b) 1 and 3 (c) 4 and 4 (d) 3 and 4
139. The decomposition of a certain mass of CaCO 3 gave 11.2 dm^3 of CO 2 gas at STP. The mass of KOH required to completely neutralize the gas is (a) 56 g (b) 28 g (c) 42 g (d) 20 g
More than One Option Correct Type
140. If 1 mole of H 3 PO 4 is reacted with 1 mole of X(OH) 2 as: H 3 PO 4 + X(OH) 2 XHPO 4 + 2H 2 O then
(a) The equivalent weight of H 3 PO 4 is 98 3
(b) The equivalent weight of base is Molecular mass 2 (c) 1 mole of X(OH) 2 more is required for complete neutralization of XHPO 4.
(d) The resulting solution required 1 mole NaOH for complete neutralization.
141. Which of the following contain the same number of molecules? (a) 0.1 mole of CO 2 (b) 3.2 g of O 2 (c) 0.1 g atom of Helium gas (d) 11.2 L of SO 2 at S.T.P
1.20 Chapter 1
142. 2 moles of CO 2 is required to prepare:
(a ) 168 g of NaHCO 3 (b) 148 gm Li 2 CO 3 (c) 162 g of Ca(HCO 3 ) 2 (d) 462 g of Ca(HCO 3 ) 2
143. 5.3 % (W/V) Na 2 CO 3 solution and 6.3% (W/V) H 2 C 2 O 4 .2H 2 O solution have same (a) molarity (b) molality (c) normality (d) mole fraction 144. 0.1 mole of MnO 4 −^ in acidic medium can oxidize
(a) 0.60 mole of Cr 2 O 7 2− (b) 0.25 mole of C 2 O 4 2− (c) 0.167 mole of FeC 2 O 4 (d) 0.5 mole of Fe+
145. Which of the following statements are true?
(a) 392 g of ferrous ammonium sulphate is required to reduce 31.6 g KMnO 4 in aqueous acidic medium. (b) Molality of a solution does not vary with temperature. (c) KBrO 3 can quantitatively convert Br–^ to Br 2. (d) The oxidation state of S in CNS–^ is zero.
14 6. Which of the following statement is/are correct?
(a) CrI 3 + KOH + Cl 2 → K 2 CrO 4 + KCl + KIO 4 + H 2 O So, in balanced chemical reaction coefficient of KOH is 64.
(b) If Zn + KMnO 4 + H 2 SO 4 → ZnSO 4 + K 2 SO 4 + MnSO 4 + H 2 ; then equivalents of Zn = equivalents of Zn = equivalents of KMnO 4 + equivalents of H 2 SO 4. (c) (NH 4 ) 2 Cr 2 O 7
N 2 + Cr 2 O 3 + H 2 O, n-factor of (NH 4 ) 2 Cr 2 O 7 is 12. (d) [Fe(CN) 6 ]3−^ → Fe3+^ + CO 2 + NO 3 −, n-factor for [Fe(CN) 6 ]3−^ is 60.
147. The strength of “20 volume” H 2 O 2 is equal to (a) 3.58 N (b) 3.035 % (c) 60.86 g/L (d) 1.79 M 148. 100 ml of 0.06 M Ca(NO 3 ) 2 is added to 50 mL of 0. M Na 2 C 2 O 4. After the reaction is complete (a) 0.003 M of excess of Ca2+^ will remain in excess. (b) 0.003 moles of calcium oxalate will get precipitated (c) Ca(NO 3 ) 2 is excess reagent. (d) Na 2 C 2 O 4 is limiting reagent. 149. Which of the following contains same number of molecules? (a) 3.2 g of O 2 (b) 0.1 mole of NO 2 (c) 0.1 g atom of Ar gas (d) 11.2 L of CO 2 at S.T.P.
Passage Based Questions
Passage –I
Oleum is considered as a solution of SO 3 in H 2 SO 4 , which is obtained by passing SO 3 in concentrated H 2 SO 4. When 100 g sample of oleum is diluted with desired weight of H 2 O, then the total mass of H 2 SO 4 obtained after dilution is known as % labeling in oleum.
For example, an oleum labeled as ‘109% H 2 SO 4 ’ means the 109 g total mass of pure H 2 SO 4 will be formed when 100 g of oleum is diluted by 9 g of H 2 O which combines with all the free SO 3 present in oleum to form H 2 SO 4 as SO 3 + H 2 O → H 2 SO 4.
150. What is the % of free SO 3 is an oleum that is labeled as ‘104.5 % H 2 SO 4 ’? (a) 30 (b) 10 (c) 20 (d) 40 151. 9.0 g water is added into oleum sample labeled as “112%” H 2 SO 4 then the volume of free SO 3 remaining in the solution at1 atm pressure and 0 oC is (a) 7.46 L (b) 3.73 L (c) 11.2 L (d) 7.34 L 152. 1 g of oleum sample is diluted with water. The solution required 54 mL of 0.4 N NaOH for complete neutral- ization. The % of free SO 3 in the sample is (a) 20 (b) 52 (c) 26 (d) none of these
