XI MCQs Chapter 1 ………… Stoichiometry
1. If the volume occupied by oxygen gas (O2) at STP is 44.8dm3, the number molecules of O2 in the vessels are:
(a) 3.01 x 1023
(b) 6.02 x 1023
(c) 12.04 x 1023
(d) 24.08 x 1023
Explanation; (Answer, c)
22.4 dm3 (1 mole) of any gas at STP contains 6.02 x 1023 molecules. Hence 44.8 dm3 (2 mole) of any gas at STP would contain twice of 6.02 x 1023 molecules i.e. 12.04 x 1023 molecules.
No. of molecules = Vg/Vm x NA = 44.8/22.4 x 6.02 x 1023 = 12.04 x 1023 or 1.204 x 1024 molecules.
2. The number of carbon atoms in 1 mole of sugar (C12H22O11) are approximately:
(a) 6 x 1023
(b) 24 x 1023
(c) 60 x 1023
(d) 72 x 1023
Explanation; (Answer, d)
1 mole of sugar (C12H22O11) = 12 moles of carbon atoms
12 moles of carbon atoms = n x NA = 12 x 6.02 x 1023 = 72 x 1023 or 7.2 x 1024 carbon atoms.
3. In the reaction 2Na + 2H2O → 2NaOH + H2, if 23g of Na reacts with excess of water, the volume of hydrogen gas (H2) liberated at STP should be
(a) 11.2dm3
(b) 22.4dm3
(c) 33.6dm3
(d) 44.8dm3
Explanation; (Answer, a)
From stoichiometric ratio in the given equation:
46g (2 mole) Na liberates 22.4 dm3 of hydrogen gas at STP
23g (1 mole) Na liberates 22.4/46 x 23 dm3 of hydrogen gas at STP = 11.2 dm3 of hydrogen gas
4. Which of the following sample of substances contains the same number of atoms as that of 20g calcium:
(a) 16g S
(b) 20g C
(c) 19 g K
(d) 24g Mg
Explanation; (Answer, a)
Species containing same number of moles will have same number of particles (atoms, molecules or formula units). 20 g Ca and 16 g S constitute same number of moles (0.5 moles), hence they have same number of atoms.
No. of moles in 20 g Ca = mass/molar mass = 20/40 = 0.5 mol
No. of moles in 16 g S = mass/molar mass = 16/32 = 0.5 mol
5. The minimum number of moles are present in:
(a) 1 dm3 methane gas at STP
(b) 5 dm3 of helium gas at STP
(c) 10 dm3 of hydrogen gas at STP
(d) 22.4 dm3 of chlorine gas at STP
Explanation; (Answer, a)
For gases at STP, minimum number of moles are present in that gas which has least volume. Here 1 dm3 methane gas at STP has least number of moles as it has least volume.
6. Number of atoms in 60g carbon are:
(a) 3.01x 1023
(b) 3.01 x 1024
(c) 6.02 x 1023
(d) 6.02 x 1024
Explanation; (Answer, b)
No. of atoms = (Mass/molar mass) x NA = 60/12 x 6.02 x 1023 = 3.01 x 1024
7. Maximum number of molecules present in the following sample of gas:
(a) 100g O2
(b) 100g CH4
(c) 100g CO2
(d) 100g Cl2
Explanation; (Answer, b)
The species containing greatest moles will have greatest number of molecules.
For species having same mass, the specie having least molar mass would have greatest number of moles. Here CH4 has least molar mass of 16, so its 100 g would contain maximum number of moles and also maximum number of molecules.
8. Which of the following statement is incorrect?
(a) The mass of 1 mole Cl2 gas is 35.
(b) One mole of H2 gas contains 6.02 x 1023 molecules of H2
(c) Number of atoms in 23g Na and 24 g Mg are equal
(d) One moles of O2 at S.T.P occupy 22.4 dm3 volume
Explanation; (Answer, a)
The statement a is incorrect as the mass of 1 mole of Cl2 is 71 g (not 35).
9. For Avogadro's number, this statement is incorrect?
(a) It is the no. of particles in one mole of any substance
(b) Its numerical value is 6.02x 1023
(c) Its value change if temperature increases
(d) Its value change if number of moles increases.
Explanation; (Answer, c)
Avogadro’s number is independent of temperature and pressure. Its value only depends upon the number of moles and volume of gas at STP.
10. Generally actual yield is:
(a) Greater than theoretical yield
(b) Less than theoretical yield
(c) Equal to the theoretical yield
(d) Some times greater & some times less than theoretical yield
Explanation; (Answer, c)
Generally actual yield is Less than theoretical yield mainly due to side reactions and mechanical loss of product.
XI MCQs Chapter 2 …… Atomic Structure
1. Bohr's theory is not applicable to which of the following
(a) H
(b) H+
(c) He1+
(d) Li2+
Explanation; (Answer, b)
One of the limitations of Bohr's atomic model is that it does not explain the spectra of multi-electron atoms. Bohr's theory is applicable to hydrogen like atoms or hydrogenic ions (single electron system).
All the given species like H, He+ and Li2+ are isoelectronic and have only one electron. Their electronic configurations are same and so their spectra is explained by Bohr's atomic model. But H+ has no electron at all and hence cannot form spectrum.
2. Nitrogen has the electronic configuration 1s2 2s2 2px1 2py1 2pz1 and 1s2 2s2 2px2 2py1.This is determined by
(a) Aufbau principle
(b) Pauli’s rule
(c) Hund's rule
(d) n+l rule
Explanation; (Answer, c)
The filling of electrons in degenerate orbitals like p, d and f is governed by Hund’s rule of maximum electron multiplicity accordingly electrons in degenerate orbitals tend to remain singly with same spin until all of the degenerate orbitals become half filled then pairing of electrons are allowed in them.
3. Quantum number values for 3s orbital are
(a) n=0, l=1
(b) n=1, l=0
(c) n=3, l=1
(d) n=3, l=0
Explanation; (Answer, d)
In the notation 3s (nl), 3 stands for n while s represent l.
For 3s orbital, n =3 and l=0.
4. The radius of first orbit of hydrogen atom is .....
(a) 529Å
(b) 52.9 Å
(c) 5.29 Å
(d) 0.529 Å
Explanation; (Answer, d)
Where ao is known as Bohr’s constant or Bohr radius and it the radius of first orbital of hydrogen atom. and its value is 0.529 x 10−10 m or 0.529 x 10−8 cm or 0.529 Å or 0.0529nm or 52.9 pm. This is the radius of the first orbit of H. This equation is used for the determination of nth orbit of hydrogen atom and hydrogen like ions like He+, Li2+ etc.
5. Line spectrum is used as a tool for the identification of ....
(a) Colors
(c) Elements
(b) Electrons
(d) Molecules
Explanation; (Answer, c)
Samples of same element always produces same characteristic line spectrum. Each element emit of light of specific wavelength therefore the number of lines and the distance between them depends upon the nature of element, so line spectra is used as “Finger Print” for the identification of elements. For example, line spectrum of sodium contains two yellow coloured lines separated by a definite distance.
6. The shape of orbital for which l = 0 is
(a) Spherical
(b) Dumbbell
(c) Double dumbbell
(d) Complicated
Explanation; (Answer, a)
Its values show shape of orbitals. It has values l = 0 to (n – 1). e.g.
l = 0, used for s-orbital, spherical in shape
l = 1, used for p-orbital, dumb-bell in shape
l = 2, used for d-orbital, double dumb-bell in shape
l = 3, used for f-orbital, complicated shape
7. When 4d orbital is filled, the next electron enter into
(a) 5s
(b) 5p
(c) 5d
(d) 6s
Explanation; (Answer, b)
According to n+ l rule, electrons are filled in various orbitals in the increasing order of (n+l) value. Those orbitals which have lower value of (n+l) value filled first. In case, if two orbitals which have same (n+l) value, then orbital having lower ‘n’ value will be filled first.
The n+l value of 4d orbital is 4+2 =6. The next orbital to be filled must have n+l value either equal to 6 or greater than 6. The n+l value of 5d orbital is 5+2 =7, for 5s it is 5+0=5, for 5p it is 5+1=6 and for 6s it is 6+0=6. Since 5p and 6s orbitals have same n + l value, hence the electron first go to that orbital which has least n value i.e. 5p.
8. Which of the following is not an iso electronic pair?
(a) Na+ and Ne
(b) Na+ and F−
(c) Na and Ca
(d) Na+ and Mg2+
Explanation; (Answer, b)
The species having same number of total electrons and hence have same electronic configuration are called isoelectronic pair. Na+ and F− are isoelectronic pair as both of them contains 10 electrons.
9. Balmer series appears in the hydrogen spectrum if electron jumps from any appropriate higher energy orbit to
(a) Second orbit
(b) Third orbit
(c) Fourth orbit
(d) Fifth orbit
Explanation; (Answer, a)
Balmer noted first that the Hydrogen spectrum consisted of some well defined discrete lines in the visible region (i.e. having l between 4000 to 7000Å) of spectrum. This series is obtained by the transition of electrons from any higher orbit (n2 = 3, 4, 5, 6, 7 …. µ) to 2nd orbit (n1 = 2).
10. In 1935 A.D. James Chadwick was awarded Nobel Prize because ...
(a) He discovered proton
(b) He discovered neutron
(c) He determined the radius of hydrogen atom
(d) He gave the rules for electronic configuration
Explanation; (Answer, b)
James Chadwick won the 1935 Nobel Prize in Physics for the discovery of the neutron.
XI MCQs Chapter 3 … Chemical Bonding
1. If the bond angle is AB2 type molecule is 104.5o, it geometry should be;
(d) Planar Trigonal
Explanation; (Answer, c)
The bond angle of 104.5o is the characteristic of AB2 type of molecules with bent or V-shaped or angular geometry.
2. The highest bond energy in the following is:
(d) H–N
Explanation; (Answer, b)
Bond energy is directly proportional to the bond polarity or partial ionic character which in turn is proportional to the difference in electronegativity of the bonded atoms. Greater the ΔEN, greater is the bond energy. H – F bond has the greatest ΔEN (1.9) and hence it has the highest bond energy (in fact highest among all diatomic molecules containing single bond; 570 kJmol−1).
The second highest bond energy is for H – O bond with second highest ΔEN (1.4). The next highest bond energy is for H – N bond with third highest ΔEN (0.9). Cl–Cl being non-polar has the least bond energy.
3. The molecule which has zero dipole moment is:
(d) CCl4
Explanation; (Answer, d)
Symmetrical molecules with linear, trigonal or tetrahedral geometry show zero dipole moment. CCl4 has symmetrical tetrahedral structure in which all four bond moments for C – Cl bonds being directed in opposite directions are cancel out giving net zero dipole moment.
4. The molecule which has maximum bond angle
(d) BF3
Explanation; (Answer, a)
Linear molecules like CS2, CO2, BeCl2, C2H2, HgCl2, ZnCl2 etc. with AB2 formula has the largest bond angle of 180o.
5. The shape and hybridization of BCl3 molecule is:
(b) Linear and sp
(d) Angular and sp3
Explanation; (Answer, c)
Being AB3 molecule with no non-bonding orbital, BCl3 has sp2 hybridization with planar trigonal shape.
6. Amongst the following molecules which one has trigonal pyramidal shape?
(c) NH3
(d) C2H4
Explanation; (Answer, c)
The ÄB3 molecules like NH3, PH3, PCl3, NCl3 etc. with one lone pair and 3 bond pairs has trigonal pyramidal shape
7. A simple covalent molecule possesses two bond pairs and two lone pairs around the central atom, its shape should be:
(d) Tetrahedral
Explanation; (Answer, c)
The presence of two bond pairs and two lone pairs around the central atom gives rise to angular shape to the molecules. Such molecules have :ÄB2 type general formula.
8. The correct relation between debye and coulomb meter is:
(b) 1D = 1.6x10−19Cm
(d) 1D = 1.23x10−8Cm
Explanation; (Answer, a)
The SI unit of dipole moment C-M and cgs unit debye is inter related as
1D = 3.33x10−30Cm
9. The bond order of N2 molecule is:
(d) 3
Explanation; (Answer, d)
The bond order of N2 molecule is 3 showing that it has triple bond between two nitrogen atoms.
10. The number of sigma and pi bonds in C2H4 molecules are respectively:
(d) 4 and 2
Explanation; (Answer, c)
There are total 5 sigma and one pi bond in in C2H4 molecule.
XI MCQs Chapter 4 State of Matter, Gas
1. According to Graham’s Law of diffusion, the ratio of diffusion of H2 and O2 are respectively:
(d) 4:1
Explanation; (Answer, d)
r1/r2= √M2/√M2 ⇒ r1/r2= √32/√2 ⇒ r1/r2= √16 ⇒ r1/r2= 4:1
2. Collection of gas over water is an example of:
(b) Dalton’s law
(d) Gay-Lussac law
Explanation; (Answer, b)
Collection of gas over water is an example of Dalton’s law. The pressure of dry gas is calculated by using Dalton’s law.
Pdry gas = Pmoist gas – Pwater vapours
3. The molar volume of oxygen gas is maximum at:
(b) 0°C and 2 atm
(d) 25°C and 2 atm
Explanation; (Answer, c)
The molar volume of a gas varies directly with temperature and varies inversely with pressure. The molar volume of a gas is maximum at highest temperature and least pressure. In option c and d, temperature is highest hence deciding factor will be pressure which is least in option c.
4. The volume of gas would be theoretically zero at:
(d) 273°C
Explanation; (Answer, b)
The volume of gas would be theoretically zero at 0 K (-273°C).
5. If the Kelvin temperature of ideal gas is increase to double and pressure is reduced to one half, the volume of gas will:
(b) Double
(d) Four times
Explanation; (Answer, d)
PV=nRT ⇒ PV = T (R and n = constant) ⇒ V = T/P ⇒ V = 2/ ½ = 2 x 2 = 4
Thus, doubling the Kelvin temperature and halving the pressure quadruples the volume.
6. The molar volume of oxygen gas is 22.4 dm3 at:
(a) 0°C and 1 atm
(b) 25°C and 0.5 atm
(d) 25 K and 0.5 atm
Explanation; (Answer, a)
The molar volume of a gas is 22.4 dm3 at STP i.e. at 0C and 1 atm.
7. Under similar condition CH4 gas diffuses........ times faster than SO2 gas:
(d) 16 times
Explanation; (Answer, b)
r1/r2= √M2/√M2 ⇒ r1/r2= √64/√16 ⇒ r1/r2= √4 ⇒ r1/r2= 2
8. Which one of the following statement is incorrect about the gas molecules?
(a) They have large spaces
(b) They possess kinetic energy
(c) Their collision is elastic
(d) Their molar mass depends upon temperature
Explanation; (Answer, d)
The molar mass of gas is independent of temperature.
9. The diffusion rate of C3H8 and CO2 are same because:
(a) Both are poly atomic gases
(b) Both are denser than air
(d) Both contains carbon atoms
Explanation; (Answer, c)
The rate of diffusion depends upon molar masses of gases. The gases with same molar masses would have same rate of diffusion. The given gases ethane (C3H8) and CO2 have same molar mass of 44 gmol−1.
10. Real gas reaches the ideal behavior at:
(a) Low temperature and low pressure
(b) High temperature and high pressure
(c) Low temperature and high pressure
(d) High temperature and low pressure
Explanation; (Answer, d)
High temperature and low pressure make the ideal.
XI MCQs Chapter 5 - Liquid State
1. Hydrogen bond is not found in:
(d) HF
1. Explanation (Answer; b)
Hydrogen bonding is found in those polar compounds which have polarized hydrogen. Since CH4 is non-polar, it cannot form hydrogen bond.
2. The boiling points of different liquids may be different at the same external pressure due to;
(c) Surface area
(d) Viscosities
2. Explanation (Answer; b)
The boiling points of liquids depend upon the strength of intermolecular forces. Greater the strength of IMF, greater is the boiling point. Since different liquids have different strength of IMF, therefore, different liquids have different boiling points at the same external pressure.
3. Which of the following possesses weakest London dispersion forces:
(d) I2
3. Explanation (Answer; b)
The strength of London dispersion forces depends upon the size of atom or molecules. Large size atom or molecules have greater number of electrons therefore more distortion of electrons is possible, which increases the strength of London forces. In this questions given options belongs to halogens of group VIIA in which atomic size increases down the group. Hence iodine has the largest size and has the strongest dispersion forces while fluorine has the smallest size and hence has the weakest dispersion forces.
4. A non-polar molecule with bigger size will experience:
(c) Hydrogen bonding
(d) All of these
4. Explanation (Answer; a)
Non-polar molecules have only London dispersion forces. The strength of London dispersion forces depends upon the Size of atom or molecules. Large size atom or molecules have greater number of electrons therefore more distortion of electrons is possible, which increases the strength of London forces.
5. Which of the following liquids show maximum surface tension?
(d) Gasoline
5. Explanation (Answer; b)
Surface tension of liquids depends upon the strength of IMF. Greater is the strength of IMF, greater is the surface tension. Mercury has strong interatomic forces in the form strong metallic bond. Hence mercury has the highest surface tension among all liquids.
6.The boiling point of water (H2O) is 100oC where as that of hydrogen sulphide (H2S) is 42oC, This can be attributed to:
(a) Smaller bond angle of H2S than H2O
(b) Smaller radii of oxygen than sulphur
(c) High I.P of oxygen than sulphur
(d) Tendency of water to form hydrogen bond
6. Explanation (Answer; d)
The boiling points of liquids depend upon the strength of intermolecular forces. Greater the strength of IMF, greater is the boiling point. Water being a polar liquid forms strong hydrogen bonds with neighbouring water molecules while hydrogen sulphide being non-polar cannot do so. Hence due to hydrogen bonding water shows high boiling point than hydrogen sulphide.
7. Liquids can form convex meniscus in a narrow glass tube when:
(a) Cohesive forces are stronger than adhesive forces
(b) Adhesive forces are stronger than cohesive forces
(c) Cohesive and adhesive forces are equal in strength
(d) None of these
7. Explanation (Answer; a)
The formation of meniscus by liquids in a narrow tube depends the strength of cohesive and adhesive forces in that liquids. Liquids with strong cohesive forces than adhesive forces form convex meniscus while liquids with strong adhesive forces than cohesive forces forms convex meniscus.
8. Which of the following statement is incorrect?
(a) Viscosity is the resistance against flow of liquid
(b) Liquid possesses definite volume
(c) One feels sense of cooling after bath due to condensation
(d) Sublimation is endothermic process
8. Explanation (Answer; d)
Cooling sensation after taking bath is due to evaporation not due to condensation. Evaporation is a cooling process.
9. Which statement is incorrect about evaporation?
(a) It is an exothermic process
(b) It is a reverse process of condensation
(c) It occurs at all temperatures and pressure
(d) It causes cooling effect
9. Explanation (Answer; a)
Evaporation is an endothermic process.
10. Cooking time is reduced in a pressure cooker because:
(a) Boiling point of water increases
(b) Boiling point of water decreases
(c) Vapor pressure of liquid is reduced
(d) Heat is uniformly distributed
10. Explanation (Answer; a)
In pressure cooker, boiling point of water increases which reduces cooking time.
XI MCQs Chapter 6 - Solid State
Coming Soon
XI MCQs Chapter 7 -Chemical Equilibrium
1. In the equilibrium system of PCl5(g) ⇌ PCl3(g)+ Cl2(g), the correct relationship between Kc and Kp is:
(a) Kp> Kc
(b) Kp< Kc
(c) Kp = Kc
(d) Kp/Kc=1
Explanation; (Answer, a)
2. In the reaction A(g)+ B(g) ⇌2C(g), the equilibrium constants Kp=Kc because:
(a) ∆n > 1
(b) ∆n < 1
(c) ∆n = 1
(d) ∆n = 0
Explanation; (Answer, d)
3. What happens to the value of Kc when a catalyst is added to a chemical system at equilibrium?
(a) It decreases
(b) it increases
(c) It becomes zero
(d) it remains unchanged
Explanation; (Answer, d)
4. The correct Ksp expression of a sparingly soluble salt Li2C2O4 among the following is
(a) Ksp=[Li+][C2O42−]
(b) Ksp=[Li+]2[C2O42−]
(c) Ksp=[2Li+][ C2O42−]
(d) Ksp=[2Li+]2[C2O42−]2
Explanation; (Answer, b)
5. If the equilibrium expression of a reversible reaction is
Kc = [C]2/[A][B]
The balanced equilibrium equation should be:
(a) C ⇌ A+B
(b) A+B ⇌ C
(c) 2C ⇌ A+ B
(d) A+B ⇌ 2C
Explanation; (Answer, d)
6. The term active mass use in law of mass action means:
(a) No. of mole
(b) No. of molecules
(c) mole per dm³
(d) gram per dm³
Explanation; (Answer, c)
7. The equilibrium of N2(g) +O2(g) ⇌ 2NO(g) (∆H = +ve) is affected by change in:
(a) Temperature only
(b) Pressure only
(c) Both temperature & pressure
(d) Neither temperature nor pressure
Explanation; (Answer, a)
8. The equilibrium of which of the following reaction would not be affected by an increase in pressure:
(a) PCl5(g) ⇌ PCl3(g)+ Cl2(g)
(b) 2NO(g) + Cl2(g) ⇌ 2NOCl(g)
(c) N2(g) + O2(g) ⇌ 2NO(g)
(d) 2SO2(g) + O2(g) ⇌ 2SO3(g)
Explanation; (Answer, b)
9. NaCl when added to an aqueous silver chloride solution:
(a) Decreases the solubility of AgCI
(b) Increases the solubility of AgCl
(c) Forms a clear solution
(d) Does not effect
Explanation; (Answer, a)
10. Some reactions are nearly to completion in the forward direction and identified by their:
(a) Very high value of Kc
(b) Very low value of Kc
(c) Very high value of ∆H
(d) Very low value of ∆H
Explanation; (Answer, a)
XI MCQs Chapter 8 - Acids, bases and Salts
1. H2SO4 is stronger acid than CH3COOH because:
(a) It gives two H+ ion per molecule
(b) Its boiling point is high
(c) Its degree of ionization is high
(d) It is highly corrosive
Explanation (Answer; c)
The strength of an acid depends upon its degree of ionization and acid ionization constant (Ka). Greater the value of degree of ionization (equal or greater than 30%) and Ka (more than 10−3), stronger is the acid. The degree of ionization of H2SO4 is quite high about 60% (first step) and its Ka is also quite high about 103 (Ka2 for HSO4− = 1.3 x 10−2) showing that it is a very strong acid. Contrarily, The degree of ionization of CH3COOH is quite low about 1.4% and its Ka is also quite low about 1.7 x 10−5 showing that it is a weaker acid than H2SO4.
2. Which of the following statements is not correct about the bases?
(a) They have bitter tastes (b) They have high pH value
(c) They react with acids to form salts (d) They turn blue litmus red
Explanation (Answer; d)
Bases turns red litmus paper blue.
3. Al2O3 is amphoteric oxide because it reacts with:
(a) Acids
(b) neither acid nor base
(c) Both acids and base
(d) Base
Explanation (Answer; c)
amphoteric oxides have dual characteristics of acid and base and hence they react both with acid and base through neutralization forming salt and water.
The oxides of amphoteric metals like Be, Al, Sn, Pb, Zn, Cr etc are amphoteric . These oxides react with both acids and base via neutralization to form salt and water. Al2O3, ZnO, BeO, SnO, SnO2, PbO, PbO2, Cr2O3 are amphoteric oxides.
4. Which of the following is not a buffer solution?
(i) Na2CO3 /NaHCO3
(b) CH3COOH/CH3COONa
(c) NH4OH/NH4Cl
(d) NaOH/HCl
Explanation (Answer; d)
Buffer solution is a mixture of weak acid and its salt with strong base or weak base and its salt with strong base. In option ‘d’, NaOH and HCl are both base and acid.
5. Which oxide is amphoteric in nature:
(a) K2O
(b) CO2
(c) CaO
(d) Al2O3
Explanation (Answer; d)
The oxides of amphoteric metals like Be, Al, Sn, Pb, Zn, Cr etc are amphoteric . These oxides react with both acids and base via neutralization to form salt and water. Al2O3, ZnO, BeO, SnO, SnO2, PbO, PbO2, Cr2O3 are amphoteric oxides.
The oxides of metals like K2O, Li2O, Na2O, MgO, CaO, BaO etc. are basic.
The oxides of non-metals like CO2, SO2, SO3, NO2, N2O5, P2O3, P2O5, Cl2O, Cl2O7 etc. are acidic. (Some non-metallic oxides like H2O, N2O, NO and CO are neutral).
6. Which of the following does not alter the pH of a solution?
(a) NH4Cl
(b) Na2CO3
(c) NaCl
(d) Mg(OH)Cl
Explanation (Answer; c)
Neutral salts formed from strong acids and strong bases like NaCl, KBr, KCl, BaCl2, Na2SO4, KNO3 etc. do not change the pH of solution.
7. Conjugated acids of NH3 is:
(a) NH4+
(b) NH2-
(c) NH2
(d) NH
Explanation (Answer; a)
To form conjugate acid from a compound or ion, just add H into it while to form conjugate base of a compound or ion just remove H from it. Don’t forget to balance charge either adding or subtracting H. Addition of H increases one unit + charge while removing of H decreases one + charge.
NH3 (adding add H into it will introduce + charge) …………… NH4+ (conjugate acid)
NH3 (removing H from it will introduce – charge) ……………. NH2− (conjugate base)
8. Salt which is formed by the neutralization of weak acid and strong base is:
(a) NaNO3
(b) NH4Cl
(c) Na2CO3
(d) NH4CN
Explanation (Answer; c)
Strong bases contain Na+, K+, Ba2+ as cation while weak acids contains anions other than Cl−, Br−, I−, NO3−, SO4−, ClO4−, and ClO3−. A salt formed from weak acid and strong base via neutralization contains any one of the cation like Na+, K+, Ba2+ and anion other than Cl−, Br−, I−, NO3−, SO4−, ClO4−, and ClO3−. Sodium carbonate (Na2CO3) is a salt derived from the neutralization of weak acid (H2CO3 containing CO32− anion) and strong base (NaOH).
9. A conjugate acid base pair has the difference of only:
(a) One electron
(b) One proton
(c) One electron pair
(d) One proton pair
Explanation (Answer; b)
conjugate acid-base pair differs from one another only by a single proton (H+). To identify a conjugate acid base pair, we look at the specie that donates a proton and the specie that is formed from it by the loss of proton. In the given reaction, NH3 accepts a proton acting as a Bronsted base to form NH4+ which is its conjugate acid. Similarly, HCl donates a proton acting as a Bronsted acid to form Cl− which is its conjugate base.
10. Salt formed by neutralization of weak acid and weak base is:
(a) NH4Cl
(b) Na2CO3
(c) NH4CN
(d) Na2SO4
Explanation (Answer; c)
Strong bases contain Na+, K+, Ba2+ as cation while weak acids contains anions other than Cl−, Br−, I−, NO3−, SO4−, ClO4−, and ClO3−. A salt formed from weak acid and weak base via neutralization contains any cation other Na+, K+, Ba2+ and any anion other than Cl−, Br−, I−, NO3−, SO4−, ClO4−, and ClO3−. Ammonium cyanide (NH4CN) is a salt derived from the neutralization of weak acid (HCN containing CN− anion) and weak base (NH4OH).
XI MCQs Chapter 9 - Chemical Kinetics
1. For a hypothetical reaction x+y → z, if the conc. of x is double, the rate increases by square and if the conc. of y is doubled the rate increases by twice. The experimental rate law of this reaction is:
(a) R=K [x]1 [y]1
(b) R= K [x][y]
(c) R= K [x]2 [y]1
(d) R=K [x][y]
Explanation (Answer; c)
For second order reaction, doubling the concentration of reactant (x) quadruples the rate (i.e. increases by square).
For first order reaction, doubling the concentration of reactant (y) doubles the rate (i.e. increases by twice).
Hence for given reaction, order of reaction with respect to x is 2 and that of y is 1. The experimental rate law of this reaction is R= K [x]2 [y]1
2.The unit of rate constant (K) for the first order reaction is:
(a) s−1
(b) conc. s−1
(c) conc−1. s
(d) conc−1. s−1
Explanation (Answer; a)
3. Rate constant of a reaction is affected by:
(a)Conc. of reactants
(b) Conc. of products
(c) Temperature
(d) Reaction time
Explanation (Answer; c)
Rate constant of a reaction is affected by temperature only.
4. The purpose of using of catalyst in a chemical reaction is to change:
(a) Equilibrium constant
(b) Enthalpy of reaction
(c) Activation energy
(d) Nature of reaction
Explanation (Answer; c)
The main function of catalyst is to change the activation energy.
5. The overall order of reaction to which the rate law is R= K:
(a) Zero order
(b) First order
(c) Second order
(d) Third order
Explanation (Answer; a)
For zero order reaction, concentration of reactants are taken as 1, so for such reactions the rate law becomes R = K
Rate of reaction = K [A]x[B]y ⇒ Rate of reaction = K [A]0[B]0 ⇒ Rate of reaction = K x 1 x 1⇒ Rate=K
6. Ionic reactions are classified into
(a) Reversible reaction
(b) Moderate reactions
(c) Fast reactions
(d) Slow reaction
Explanation (Answer; c)
Ionic reactions are classified into fast reactions as they have low activation energy due to involvement of ions.
7. The decomposition of H2O2 is inhibited by:
(a) 2% ethanol
(b) Glycerin
(c) MnO₂
(d) V₂Os
Explanation (Answer; b)
Prevention of decomposition of H2O2 is done by using glycerin as a negative catalyst.
8. The best alternative term for the velocity of reaction is:
(a) Rate of appearance
(b) Rate of disappearance
(c) Average rate
(d) Instantaneous rate
Explanation (Answer; d)
The velocity of reaction is called Instantaneous rate when time interval is approached to zero.
9. Order of reaction is the power to which concentration of reactant is:
(a) Lowered
(b) Raised
(c) Stopped In the reaction
(d) Constant
Explanation (Answer; b)
Order of reaction is the power to which concentration of reactant is raised.
10. In the reaction A→B, the rate of disappearance is written as:
(a) dA/dt
(b) -dA/dt
(c) dB/dt
(d) -dB/dt
Explanation (Answer; b)
The rate of disappearance of reactant is written as -dA/dt
XI MCQs Chapter 10 - Solution and Colloids
1.Only one pair of liquid in the following set does not obey Raoult’s law, identify it:
(a) Methanol and Ethanol
(b) Benzene and toluene
(c) n-Hexane and n-heptane
(d) Ethanol and Acetone
Explanation (Answer; d)
The solutions which obey Raoult’s law over the entire range of concentration are known as ideal solutions. Some examples of ideal solutions are methanol and ethanol, benzene, and toluene, n-hexane and n-heptane, bromoethane and chloroethane etc.
Methanol and ethanol are polar in nature with similar intermolecular forces, hence they form an ideal solution.
Both benzene and toluene and n-hexane and n-heptane are non-polar in nature with similar intermolecular forces. Hence, they form an ideal solution.
Ethanol and Acetone
When Acetone and ethanol are mixed, the weakening of hydrogen bonds in ethanol takes place. Hence, an ideal solution is not formed.
2. Identify the incorrect statement about colloidal solution:
(a) It shows Tyndall effect
(b) Its particles movement is Brownian type
(c) Its particle size is less than 1 nm
(d) Its physical appearance is translucent.
Explanation (Answer; c)
In colloidal solution, the particles is in between 1 nm to 1000 nm.
4. Which is not a colligative property?
(a) Lowering in vapours pressure
(b) Elevation in boiling point
(c) Depression in freezing point
(d) Atmospheric pressure
Explanation (Answer; d)
Atmospheric pressure is a colligative property.
3. Effect of pressure change play significant role in the solubility of:
(a) Solid into liquid
(b) Liquid into liquid
(c) Gas into liquid
(d) All of them
Explanation (Answer; c)
The solubility of gas into liquid is significantly changed by the change in pressure obeying Henry’s law accordingly the solubility of gas in liquid solvent is directly proportional to the pressure.
5. According to Raoult law the relative lowering of vapour pressure is equal to:
(a) Molality
(b) Mole fraction of solute
(c) Mole fraction of solvent
(d) Molarity
Explanation (Answer; b)
According to the third form of Raoult’s law, the relative lowering of vapour pressure is equal to the mole fraction of solute.
∆P/Po = X2---------------- [Here ∆P/Po is referred as relative lowering in vapours pressure]
6. The sum of mole fractions of components of a solution is equal to:
(a) 0.0
(b) 1.0
(c) 10
(d) 100
Explanation (Answer; b)
The sum of mole fractions of components of a solution is equal to 1.0.
7. How many mole of NaOH are present in 2dm3 of 1 molar aqueous solution of it.
(a) 0.5 mole
(b) 1 mole
(c) 1.5 mole
(d) 2 mole
Explanation (Answer; d)
1 molar aqueous solution contains 1 mole of solute per 1 dm3. Hence 1 molar aqueous solution contains 2 mole per 2 dm3.
8. A colloidal solution of liquid into liquid is known as:
(a) Gel
(b) Foam
(c) Sol
(d) Emulsion
Explanation (Answer; d)
A colloidal solution of liquid into liquid is known as Emulsion
9. An example of completely immiscible liquid pair is:
(a) Benzene to toluene
(b) Water and phenol
(c) Water and Benzene
(d) Water & methanol
Explanation (Answer; b)
Water and phenol is an example of completely immiscible liquid pair
10. A 15% W/W KOH solution can be prepared by mixing 15g KOH in:
(a) 15g water
(b) 85g water
(c) 100g water
(d) 115g water
Explanation (Answer; b)
The numerical value of percent concentration indicates amount of solute. The amount of solvent can be calculated by subtracting amount of solute from 100 (only in case of W/W % or V/V%). Hence 15% W/W KOH solution can be prepared by mixing 15g KOH in 85 g (100 – 15) water.
XI MCQs Chapter 11 - Thermochemistry
1. Least entropy found in which of the following state of water:
(a) Steam at 100°C
(b) Liquid water at 25°C
(c) Liquid water at 4°C
(d) Ice at 0°C
Explanation (Answer; d)
Entropy, S, is a thermodynamic quantity and it is defined as the measure of the disorder or randomness in a system. The more “messed up” or “disorderly” or “messy” a system becomes, we say the entropy of the system increases.
The entropy of substances with different states are different since the arrangement of the molecules are different. Solids have the lowest entropy value among the other states of matter since the molecules in solids are stationary. These molecules only vibrate but do not move due to the strong interactions. Their molecules are therefore more ordered than those in other states of matter. Thus among three states of matter, the randomness and also entropy follows the order:
Gas > liquid > solid
Sgas > Sliquid > Ssolid
2. The energy corresponds to the given thermochemical process is labeled as:
Li+(g) + Cl−(g) →LiCl(s)
(a) Ionization energy
(b) Enthalpy of atomization
(c) Enthalpy of combustion
(d) Lattice energy
Explanation (Answer; d)
The given equation for thermochemical process represents the lattice formation of ionic solid. The energy released during the formation of 1 mole solid lattice of ionic compound from its consistent gaseous ions under standard conditions is called lattice energy. It is negative.
3. Which of the following change is not an endothermic reaction
(a) Cracking of alkanes
(b) Decomposition of lime
(c) Combustion of butane
(d) Photosynthesis
Explanation (Answer; c)
Combustion is always exothermic releasing heat called heat of combustion.
4. Heat transfer cannot be feasible across the boundary of a
(a) Open system
(b) Thermo permeable system
(c) Isolated system
(d) Close system
Explanation (Answer; c)
Isolated system is an ideal thermodynamic system that does not allow either matter or heat transfer across the boundary.
5. In a thermochemical process, no work is done if the system is kept at:
(a) Constant pressure
(b) Constant temperature
(c) Constant volume
(d) Constant mass
Explanation (Answer; c)
When a reaction is carried out at constant volume (ΔV = 0), no work is done on or by the system.
6. Standard enthalpy of formation of all of the following elements at 25°C and 1 atm pressure are zero except:
(a) C(diamond))
(b) C(graphite)
(c) O₂
(d) N₂
Explanation (Answer; a)
Conventionally ΔHf° of most the elements (H2, O2, N2, Na, graphite, rhombic sulphur etc.) are taken as zero because no further change is needed to bring them to standard state conditions. However, for those elements which exists in more than one allotropic forms, the most stable allotropic form is conventionally assumed to have ΔHf° = 0.
For example
(i) ΔHf° of graphite is taken as zero as it is thermodynamically more stable allotrope of carbon than diamond at 25°C & 1 atm
(ii) O2 is more stable allotropic form than O3 (ozone) and therefore, ΔHf° of O2 is assumed to be zero.
(iii) Rhombic sulphur has ΔHf° = 0 as it is the most stable allotrope of sulphur.
7. In the equation of First law of thermodynamics (∆E=q + w), the property(s) which depends upon initial and final state is (are):
(a) ΔE
(b) q
(c) W
(d) Both q and W
Explanation (Answer; a)
Only ΔE is a state function which depends upon initial and final state. The other two variables of the first law of thermodynamics i.e. q and W both are path functions that do not depend upon initial and final state.
8. Volume is a:
(a) State function
(b) Colligative property
(c) Intensive properties
(d) Path function
Explanation (Answer; a)
Volume is a state function as it depends upon initial and final states.
9. Which statement is incorrect?
(a) For constant pressure process, ∆H = ∆E + P∆V
(b) For constant volume process, ∆E = q
(c) For exothermic reactions, ∆H > 0
(d) For Hess law Σ∆H°(cycle) = 0
Explanation (Answer; c)
For exothermic changes, ∆H < 0
10. Which of the following enthalpy change is always negative:
(a) Enthalpy of formation
(b) Enthalpy of decomposition
(c) Enthalpy of combustion
(d) Enthalpy of reaction
Explanation (Answer; c)
Enthalpy of decomposition is always negative as decomposition is always endothermic.
Enthalpy of formation and Enthalpy of reaction may be positive or negative. Enthalpy of combustion is always negative.
XI MCQs Chapter 12 - Electrochemistry
1. The outer body of dry cell serves as anode, it is made up of
(a) Copper
(b) Zinc
(c) Lead
(d) Iron
1. Explanation (Answer; b)
Composition and Construction of Dry Cell
1. Zinc vessel (Outer body………….……; Acts as anode
2. carbon or graphite rod ……………...; acts as cathode (Centrally placed)
3. manganese dioxide (40%) ………….; Acts as cathode, helps to depolarize electrodes to produce steady current
4. Ammonium Chloride(8%) and Zinc chloride (8%) ………….; Acts as electrolyte giving necessary ions for conduction
5. Glycerin or Starch (4%) ……………..; serves to retains moisture
6. Carbon powder (40%) ……………….; serves to increase the surface area of graphite cathode
7. copper cap ………………………………..; for conduction of electricity
8. Zinc container internally lined with porous paper works as a separator.
Electrolyte
The region between cathode and anode is filled a paste of MnO2, carbon, NH4Cl and ZnCl2.
2. The conduction of electricity through an electrolytic solution is due to the flow of
(a) Electrons
(b) ions
(c) Atoms
(d) Molecules
2. Explanation (Answer; b)
The conduction of electricity through an electrolytic solution is called electrolytic conduction which is due to the movement of free floating ions of electrolyte. In contrast, the electronic or metallic conduction of electricity through metal or graphite is due to free moving valence electrons.
3. During electrolysis, the reaction that takes place at anode is
(a) Oxidation
(b) Simultaneous oxidation and reduction
(c) Hydrolysis
(d) Reduction
3. Explanation (Answer; a)
In electrochemical cells (whether electrolytic or Galvanic cell), oxidation always takes place at anode while reduction occurs at cathode.
4.The strongest oxidizing agent in the electro chemical series is:
(a) Li
(b) H₂
(c) Cu
(d) F
4. Explanation (Answer; d)
The strongest oxidizing must have highest reduction potential. In ECS, F has the highest reduction potential, therefore fluorine is the strongest oxidizing agent.
5. Galvanized rod of iron is coated with:
(a) Nickel
(b) Zinc
(c) Chromium
(d) Carbon
5. Explanation (Answer; b)
Galvanized iron is coated with zinc. Galvanizing is the type of electroplating involving coating of zinc on baser metals like iron.
6. Oxidation number of Cr in Na2Cr2O7, is:
(a) +3
(b)+6
(c) +8
(d) +12
6. Explanation (Answer; b)
Oxidation number of Na = +1
Oxidation number of O = −2
In a neutral compound, sum of oxidation numbers of all atoms is equal to zero.
2(Na) + 2Cr + 7(O) = 0
2(+1) + 2Cr + 7(−2) = 0
2 + 2Cr + (−14) = 0
2 + 2Cr −14 = 0
2Cr −12 = 0
2Cr = +12
Cr = +12/2 = +6
7. Which of the following half cell reaction show oxidation?
(a) Fe3+ → Fe2+
(b) Cl₂→ 2Cl−
(c) SO42− → SO32−
(d) Zn → Zn2+
7. Explanation (Answer; d)
Oxidation means increase in oxidation number of atom during a reaction. The conversion of Zn (ON =0) to ion (ON = +2) involves loss of 2 electrons which is manifested by the 2 units increase in oxidation number.
In reaction (a) where ferric ion (ON = +3) changes into ferrous ion (ON = +2) is reduction as oxidation number of Fe has been decreased.
In reaction (b) where chlorine (ON = 0) changes into chloride ion (ON = −1) is reduction as oxidation number of Cl has been decreased.
In reaction (c) where sulphate ion (ON = ++6) changes into sulphite ion (ON = +4) is reduction as oxidation number of S has been decreased.
8. Fuel cell is a typical Galvanic cell which is based on the reaction between:
(a) Nitrogen and oxygen
(b) Hydrogen and oxygen
(c) Methane and oxygen
(d) Hydrogen & zinc
6. Explanation (Answer; b)
A fuel cell is an electrochemical cell that generates electrical energy from fuel via an electrochemical reaction. A hydrogen oxygen fuel cell is a unique type of Galvanic cell based on the reaction between hydrogen and oxygen to produce water and the heat released in the reaction is used to produced electricity.
Cathode ………….. Platinum coated porous graphite
Anode ……………. Platinum coated porous graphite
Electrolyte ……... Concentrated alkaline electrolyte like KOH
Hydrogen gas …. fed to anode side (where it reacts with four OH- to produce 4 water molecules)
Oxygen gas ……… Fed to the cathode side (where it reacts with water to produced four OH- ions)
9. In Zn-SHE voltaic cell, the half reaction occurs at anode is
(a) Zn2+ + 2e− → Zn
(b) Zn → Zn2+ +2e−
(c) 2H+ + 2e− → H₂
(d) H₂ →2H2+ + 2e−
6. Explanation (Answer; b)
In Zn-SHE voltaic cell, zinc oxidizes to zinc ion (Zn2+) at anode (while SHE or hydrogen ion reduces to H2 gas at cathode).
10. This statement is not correct for lead storage battery:
(a) It can be recharged
(b) It is a primary battery
(c) Anode is made up of lead
(d) Cathode is made up of lead oxide
10. Explanation (Answer; b)
Lead storage battery is a secondary cell.
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