Question Set D

cbse 12th Standard Chemistry Chemical Kinetics Important Questions And Answers Study Material - QB365

Download CBSE Class 12th Standard CBSE Chemistry question papers, sample papers, important questions, and previous year solved papers in PDF format. Get free study materials, NCERT solutions, and exam preparation resources for Class 12th Standard CBSE Chemistry

Questions:

Generic 2

6 x 2 = 12

Distinguish between the terms molality and molarity. Under what conditions are the molarity and molality of a solution nearly the same?

The reaction:
\({ SO }_{ 2 }{ Cl }_{ 2 }\longrightarrow { SO }_{ 2 }+{ Cl }_{ 2 }\) is a first order reaction. The half life period of this reaction is 60 minutes. calculate the rate constant of this reaction.

Write an expression to relate molar conductivity of an electrolyte to its degree of dissociation.

Predict which of the following will be coloured in aqueous solution. Ti³⁺, V³⁺, Cu⁺, Sc³⁺, Mn²⁺, Fe³⁺ and Co^2+. Give reason for each.

Reactivity of transition elements decreases almost regularly from Sc to Cu. Explain.

In permanganate ion, all the bonds formed between Mn and O are covalent. Give reasons.

Generic 5

5 x 5 = 25

(i) Give name:
(a) the complex used as oxygen carrier in the blood.
(b) the coordination compound of magnesium, which is responsible for photosynthesis.
(ii) Discuss some applications of complex compounds.

(i) Two liquids A and B boil at 155°Cand 190°C, respectively. Which of them has a higher vapour pressure at 80°C?
(ii) Heptane and octane form ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2kPa and 46.8 kPa, respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35.0 g of octane?
(iii) The vapour pressure of water is 12.3 kPa at 300 K. Calculate the vapour pressure of one molal solution of non -volatile, non - ionic solute in water.

Calculate the standard electrode potential of Ni²⁺ II Ni electrode if emf of the cell, Ni (s)1 Ni²⁺(0.01M) IICu²⁺(0.1 M) ICu(s) is 0.059 V. (Given \(E^{ 0 }_{ cu2+/cu }=+0.34V)\)

10.

(a) Out of Ag₂SO₄, CuF₂, MgF₂ and CuCI, which compound will be coloured and why?
(b) Explain :
(i) \({ CrO }_{ 4 }^{ 2- }\) is a strong oxidizing agent while \({ MnO }_{ 4 }^{ 2- }\) is not.
(ii) Zr and Hf have identical sizes.
(iii) The lowest oxidation state of manganese is basic while the highest is acidic.
(iv) Mn (II) shows maximum paramagnetic character amongst the divalent ions of the first transition series.

11.

(a) Define the following:
(i) Order of a reaction
(ii) Elementary step in a reaction
(b) A first order reaction has a rate constant value of 0.00510 min^-1. If we begin with 0.10 M concentration of the reactant, how much of the reactant will remain after 3.0 hours?

Generic 3

8 x 3 = 24

12.

Transition metals and their compounds show catalytic activities.

13.

Zn, Cd and Hg are soft and have low melting points. Why?

14.

A 1st order reaction is 50% complete in 30minutes at 27⁰ C and in 10 min at 47⁰C. Calculate
(i) rate constant for the reaction at 27⁰C and 47⁰C
(ii) energy of activation for the reaction.

15.

The decomposition of N₂O₅ in carbon tetrachloride solution was stuied N₂O₅(solution)\(\longrightarrow\)2NO₂(solution)+\(1\over2\)O₂(g)
The reaction has been found to be first order and rate constant is found to be 4.2\(\times\)10^-4s-1^.Calculate the rate of reaction when
(a) [N₂O₂] = 1.25mol L^-1 and
(b) [N₂O₅] = 0.25mol L_-1
(c) What concentration of N₂O₅ would give a rate of 2.4\(\times\)10^-3mol L^-1 s^-1?

16.

Calculate the equilibrium constant for the reaction, Zn + Cd²⁺\(\rightleftharpoons \) Zn²⁺ + Cd, if E⁰ _Cd²⁺/Cd =-0.403 V and E⁰ _Zn²⁺/Zn = 0.763 V.

17.

3.9 g of benzoic acid dissolved in 49 g of benzene shows a depression in freezing point of 1.62 K. Calculate the van't Hoff factor and predict the nature of solute (associated or dissociated). (Given : Molar mass of benzoic acid = 122 g mol^-1 , K_f for benzene = 4.9 K kg mol^-1)

18.

The molar conductivity of acetic acid at infinite dilution is 387\(\Omega\)^-1 cm² mol^-1. At the same temperature, but at a concentration of acetic acid at this temperature.

19.

Name the following coordination entities and draw the structures of their entities and draw the structures of their streoisomers:
(i) [Co(en)₂Cl₂]⁺ (en = ethane-1, 2-diamine)
(ii) [Cr(C₂O₄)₃]^3-
(iii) [Co(NH₃)₃Cl₃]
(Atomic numbers Cr = 24, Co = 27)

Multiple Choice Question

15 x 1 = 15

20.

Isotonic solutions have the same

(a)

density

(b)

refractive index

(c)

osmotic pressure

(d)

volume

21.

Choose the appropriate option about the misch metal.

(a)

It is an alloy which consists of a lanthanoid metal (-95%) and iron (-5%) and traces of S, C, Ca and AI

(b)

Used in Mg based alloy to produce bullets, shell and lighter flint.

(c)

It fmds application in making aeroplane body

(d)

Both (a) and (b)

22.

The factors affecting the rate of a reaction are

(a)

temperature

(b)

pressure

(c)

concentration of reactant or product

(d)

catalyst

23.

Which of the following reactions are disproportionation reactions?
\((i)\ { Cu }^{ + }\longrightarrow { Cu }^{ 2+ }+{ Cu }\)
\((ii)\ { 3MnO }_{ 4 }^{ - }+{ 4H }^{ + }\longrightarrow { 2MnO }_{ 4 }^{ - }+{ MnO }_{ 2 }+{ 2H }_{ 2 }O\)
\((iii)\ { 2KMnO }_{ 4 }\longrightarrow { K }_{ 2 }{ MnO }_{ 4 }+{ MnO }_{ 2 }+{ O }_{ 2 }\)
\((iv)\ { 2MnO }_{ 4 }^{ - }+{ 3Mn }^{ 2+ }+{ 2H }_{ 2 }O\longrightarrow { 5MnO }_{ 2 }+{ 4H }^{ + }\)

(a)

(i),(ii)

(b)

(i),(ii),(iii)

(c)

(ii),(iii),(iv)

(d)

(i),(iv)

24.

Which of the following shall from an octahedral complex?

(a)

d⁴(low spin)

(b)

d⁸(high spin)

(c)

d⁶(low spin)

(d)

all of these

25.

Given
\({ E }_{ { Cr }^{ 3+ }/{ Cr } }^{ 0 }=-0.74V,\\ { E }_{ { MnO }_{ 4 }^{ - }/{ Mn }^{ 2+ } }^{ 0 }=1.51V\)
\({ E }_{ { CrO }_{ 7 }^{ - }/{ Cr }^{ 3+ } }^{ 0 }=1.33V,{ E }_{ Cl/{ Cl }^{ - } }^{ 0 }=1.36V\)
Based on the data given above, strongest oxidizing agent will be

(a)

MnO₄^-

(b)

Cl^-

(c)

Cr³⁺

(d)

Mn²⁺

26.

Which has maximum potential for the half cell reaction : \(2{ H }^{ + }+2{ e }^{ - }\longrightarrow { H }_{ 2 }\) ?

(a)

1.0 M HCl

(b)

1.0 M NaOH

(c)

Pure water

(d)

A solution with pH = 4

27.

Oxidation number and coordination number of silver in Tollen's reagent respectively are

(a)

1, 1

(b)

2, 1

(c)

2, 2

(d)

1, 2

28.

IUPAC name of [Pt(NH₃)₃Br(NO₂)Cl]Cl is

(a)

Triamminechloridobromidonitroplatinum (IV) chloride

(b)

Triamminebromidonitrochloridoplatinum (IV) chloride

(c)

Triamminebromidochloridonitroplatinum (IV) chloride

(d)

Triamminenitrochloridobromidoplatinum (IV) chloride

29.

General electronic configuration of d - block elements is

(a)

(n - 1) d¹^-10ns^1-2

(b)

ns²np^1-6

(c)

(n - 2) f^0-14(n - 1)^1-2ns²

(d)

(n-1) d^1-5ns^1-2

30.

The van't Hoff factor for 0.1 M Ba(NO₃)₂ solution is 2.74. The degree of dissociation is

(a)

91.3%

(b)

87%

(c)

100%

(d)

74%

31.

A person is considered to be suffering from lead poisoning if its concentration in him is more than 15 micrograms of lead per decilitre of blood. Concentration in parts per billion parts is

(a)

(b)

(c)

100

(d)

1000

32.

In the reaction BrO^-₃ (aq) + 5 Br ^- (aq) + 6 H⁺ \(\longrightarrow\) 3 Br₂(I) + 3 H₂O (l), the rate of apperance of bromine (BHr₂) is related to the disapearance of bromide uions as follows :

(a)

\(\frac { d[{ Br }_{ 2 }] }{ dt } =-\frac { 5 }{ 3 } \frac { d[{ Br }^{ - }] }{ dt } \)

(b)

\(\frac { d[{ Br }_{ 2 }] }{ dt } =\frac { 5 }{ 3 } \frac { d[{ Br }^{ - }] }{ dt } \)

(c)

\(\frac { d[{ Br }_{ 2 }] }{ dt } =\frac {3 }{ 5} \frac { d[{ Br }^{ - }] }{ dt } \)

(d)

\(\frac { d[{ Br }_{ 2 }] }{ dt } =-\frac { 3 }{ 5 } \frac { d[{ Br }^{ - }] }{ dt } \)

33.

K_H value for Ar(g), CO₂(g), HCHO (g) and CH₄ (g) are 4.39,1.67,1.83 x 10^-5 and 0.413 respectively. Arrange these gases in the order of their increasing solubility.

(a)

\( \mathrm{HCHO}<\mathrm{CH}_4<\mathrm{CO}_2<\mathrm{Ar}\)

(b)

\( \mathrm{HCHO}<\mathrm{CO}_2<\mathrm{CH}_4<\mathrm{Ar} \)

(c)

\( \mathrm{Ar}<\mathrm{CO}_2<\mathrm{CH}_4<\mathrm{HCHO} \)

(d)

\( \mathrm{Ar}<\mathrm{CH}_4<\mathrm{CO}_2<\mathrm{HCHO} \)

34.

Camphor is often used in molecular mass determination because

(a)

it is readily available

(b)

it has a very high cryoscopic constant

(c)

it is volatile

(d)

it is solvent for organic substances

Assertion and reason

5 x 1 = 5

35.

Assertion (A) :In the coordination compound [Co(H2NCH2CH2NH2)3]2, ethane-1,2-diammine is a neutral molecule.
Reason (R): Oxidation number of Co in the complex ion is +3.
Codes:
(a) If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
(b) If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
(c) If the Assertion is correct but Reason is incorrect.
(d) If both the Assertion and Reason are incorrect.

36.

Assertion (A) : Order and molecularity of a reaction are always same.
Reason (R) : Complex reactions involve a sequence of elementary reactions and the slowest step is rate determining.
(a) Both Assertion and Reason are correct, Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are correct, Reason is not the correct explanation of Assertion.
(c) Assertion is correct; Reason is incorrect.
(d) Assertion is incorrect; Reason is correct.

37.

Assertion : If \(\lambda^{0}{ }_{\mathrm{Na}^{+}}\) and \(\lambda^{0}{ }_{\mathrm{Cl}^{-}}\) are molar limiting conductivities of the sodium and chloride ions respectively, then the limiting molar conductivity for sodium chloride is given by the equation, \(\Lambda_{\mathrm{NaCl}}^{\circ}=\lambda_{\mathrm{Na}^{+}}^{\circ}+\lambda_{\mathrm{Cl}^{-}}^{\circ}\)
Reason : This is according to Kohlrausch law of independent migration of ions.
Codes :
(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
(c) Assertion is correct statement but reason is wrong statement.
(d) Assertion is wrong statement but reason is correct statement.

38.

In the following questions an Assertion (A) is followed by a corresponding Reason (R) Use the following keys to choose the appropriate answer.
Assertion (A) Transition metals form substitutional alloys.
Reason (R) Alloys are made to develop some useful properties which are absent in the constitutent elements.
Codes:
(a) Both (A) and (R) are correct, (R) is the correct explanation of (A).
(b) Both (A) and (R) are correct, (R) is not the correct explanation of (A).
(c) (A) is correct; (R) is incorrect.
(d) (A) is incorrect; (R) is correct.

39.

In the following questions. an Assertion (A) is followed by a corresponding Reason (R) Use the following keys to choose the appropriate answer.
Assertion (A) Ethanol and acetone show positive deviation from Raoult's law.
Reason (R) Pure ethanol molecule show hydrogen bond and on adding acetone hydrogen bond between ethanol molecules breaks
(a) Both (A) and (R) are correct, (R) is the correct explanation of (A).
(b) Both (A) and (R) are correct, (R) is not the correct explanation of (A).
(c) (A) is correct; (R) is incorrect.
(d) (A) is incorrect; (R) is correct.

Case Study Questions

2 x 4 = 8

40.

KMnO₄ and K₂Cr₂O₇are most important chemicals which are used as oxidising agents and disinfectants. K2MnO4 is prepared by fusing MnO₂ with KOH in presence of O₂.K₂MnO ₄ is electrolysed to get purple coloured KMnO₄ . Na₂CO₄ is prepared by heating chromite ore with Na₂CO₃ in presence of O₂. Na₂CrO₄ is converted into Na₂Cr₂O₇ by reacting with concentrated H₂ SO₄ , Na₂Cr₂O₇ is reacted with KCI to get K₂Cr₂O₇ orange coloured solid, soluble in water, changes to yellow coloured \(\mathrm{CrO}_{4}^{2-}\) in basic medium, KMnO₄ acts as oxidising agent in acidic, neutral as well tasic medium. In acidic medium, it converts Fe²⁺ to Fe^{3 +}, Sn²⁺ to Sn⁴⁺, to CO₂. In asic medium it converts I^- to \(\mathrm{IO}_{3}^{-} \cdot \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) acts as oxidising agent only in acidic medium, converts H₂S to S, SO₂ to \(\mathrm{SO}_{4}^{2-}, \mathbf{I}^{-}\) to I₂. Lanthanoids and actinoids belong to f-block elements with general electronic configuration \((n-2) f^{1 \text { to } 14}(n-1) d^{0-2} n s^{2}\) . All actinoids are radio active. Both show contraction in atomic and ionic radii but actinoid contraction is more than lanthanoid contraction. Lanthanoid show +3 oxidation state, few elements show +2 and +4 oxidation states also. Actinoids show +3, +4, +5, +6, +7 oxidation states.
(a) Which lanthanoid shows +4 oxidation state and why?
(b) Give two similarity between lanthanoids and actinoids.
(c) Complete the equation and balance:
\(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{Fe}^{2+}+\mathrm{H}^{+} \longrightarrow\)
(d) Convert sodium chromate to sodium dichromate. Give chemical equation.
(e) Complete the following:
\(\mathbf{3} \mathbf{M n} \mathbf{O}_{4}^{2-}+\mathbf{4} \mathbf{H}^{+} \longrightarrow\)

41.

Read the passage given below and answer the following questions:
Molar conductivity of ions are given as product of charge on ions to their ionic mobilities and Faraday's constant.
\(\lambda_{A^{n+}}=n \mu_{A^{n+}} F\) (here \(\mu\) is the ionic mobility of Aⁿ⁺).
For electrolytes say A_XB_y, molar conductivity is given by
\(\lambda_{m\left(A_{x} B_{y}\right)}=x_{n} \mu_{A^{n+}} F+y_{m} \lambda_{A^{m}-F}\)

Ions	Ionic mobility
K⁺	7.616 x 10^{- 4}
Ca²⁺	12.33 x 10^-4
Br^-	8.09 x 10^{- 4}
\(\mathrm{SO}_{4}^{2-}\)	16.58 x 10^{- 4}

The following questions are multiple choice questions. Choose the most appropriate answer
(i) At infinite dilution, the equivalent conductance of CaSO₄ is

(a) 256 x 10^-4

(b) 279

(c) 23.7

(d) 2.0 x 10^{- 8}

(ii) If the degree of dissociation of CaSO₄ solution is 10% then equivalent conductance of CaSO₄ is

(a) 3.59

(b) 36.9

(c) 27.9

(d) 30.6

(iii) What is the unit of equivalent conductivity?

(a) ohm^-1 cm² eq^-1	(b) ohm cm²eq^-1
(c) ohm^-1 cm eq^-1	(d) ohm cm² eq^-1

(iv) If the molar conductance value of Ca²⁺ and Cl^-at infinite dilution are 118.88 x 10^-4m² mho mol^-1 and 77.33 x 10^-4 m²mho mol^-1respectively then the molar conductance of CaCl₂ (in m² mho mol^-1) will be

(a) 120.18 x 10^{- 4}

(b) 135 x 10^-4

(c) 273.54 x 10^-4

(d) 192.1 x 10^-4

Visit QB365-SLMS - https://www.qb365.in/slms/

Answers Key:

Generic 2

6 x 2 = 12

Molarity	Molality
The number of moles of solute present in per liter of solution	The number of moles of solute present in 1kg of solvent.
It depends on volume.	It depends on mass.
Volume changes with temperature, hence it is temperature dependent.	It depends on mass which is independent of temperature.

\({ t }_{ 1/2 }=\frac { 0.693 }{ k } for \ a \ first \ order \ reaction\)
\(\\ or \ k=\frac { 0.693 }{ { t }_{ 1/2 } } =\frac { 0.693 }{ 60\times 60 } =1.9\times { 10 }^{ -4 }{ s }^{ -1 }\)

\(\alpha=\frac{\lambda_m^c}{\lambda_m^0} \text { where } \lambda_m^c=\) molar conductivity at given concentration, c and ∧⁰_m = molar conductivity at infinite dilution.

Ti³⁺, V³⁺, Mn²⁺, Fe³⁺ and Co²⁺ are coloured due to presence of unpaired electrons, they can undergo d-d transitions others Cu⁺, Sc³⁺ are colourless due to absence of unpaired electrons.

It is due to increase in ionisation enthalpy, tendency to lose electron decreases, therefore, reactivity decreases.

In MnO^-_4, Mn shows +7 oxidation state. Mn cannot lose 7 electrons because very high energy is required to remove 7 electrons, therefore it can form 7 covalent bonds.

Generic 5

5 x 5 = 25

(i) (a) Haemoglobin
(b) Chlorophyll
(ii) Applications of complex compounds.
(a) They are used in photography, Le. AgBr forms soluble complex with sodium thiosulphate rn photography.
(b) K[Ag(CN)₂] is used for electroplating of silver, K[Au(CN)₂] is used for gold plating.
(c) Some of ligands oxidise Co²⁺ to Co³⁺ ion.
(d) EDTA is used for estimation of Cla²⁺ and Mg²⁺ in hard water.
(e) Silver and gold are extracted by treating zinc with their cyanide complexes.
(f) Ni²⁺ is tested and estimated by DMG (dimethylglyoxime).

(i) A is more volatile, therefore has higher vapour pressure.
(ii) Given that, \(p_{\text {heptane }}^{\circ}=105.2 \mathrm{kPa},\)
\(p_{\text {octane }}^{\circ}=46.8 \mathrm{kPa}\)
Molar mass of heptane
\(\left(\mathrm{C}_{7} \mathrm{H}_{16}\right)=100 \mathrm{~g} \mathrm{~mol}^{-1}\\ (As \left.C_{7} H_{16}=7 \times 12+16=84+16=100\right)\)
Molar mass of octane \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)=114 \mathrm{~g} \mathrm{~mol}^{-1}\)
\((As \left.C_{8} H_{18}=8 \times 12+18=96+18=114\right)\)
Number of moles of 26.0 g heptane
\(=\frac{26.0 \mathrm{~g}}{100 \mathrm{~g} \mathrm{~mol}^{-1}}=0.26 \mathrm{~mol} \)
Number of moles of 35.0 g heptane
\(=\frac{35.0 \mathrm{~g}}{114 \mathrm{~g} \mathrm{~mol}^{-1}}=0.31 \mathrm{~mol}\)
\(\chi_{\text {heptane }}=\frac{0.26 \mathrm{~g}}{0.26+0.31}=0.456\)
\(\chi_{\text {octane }}=1-0.456=0.544\)
\( p_{\text {heptane }}=0.456 \times 105.2 \mathrm{kPa}=47.97 \mathrm{kPa}\)
\(p_{\text {octane }} =0.544 \times 46.8 \mathrm{kPa}=25.46 \mathrm{kPa} \)
\(p_{\text {total }} =47.97+25.46=73.43 \mathrm{kPa}\)
(iii) As solution is 1 molal, it means that 1 mole of solute is dissolved in 1000 g of solvent (water).
No. of moles of solute = 1 mol, \(p_{\text {solvent }}^{\circ}=12.3 \mathrm{kPa}\)
No. of moles of water \(=\frac{1000}{18}=55.55 \)
Total moles = 56.55 mol
Mole fraction of water \(=\frac{\chi_{1} \text { (solvent) }}{\chi_{1}+\chi_{2} \text { (solution) }}=\frac{55.55}{56.55} \)
\(p_{\text {solution }}=\)\(p_{\text {solvent }}^{\circ}\)x \(\chi\)_H2O
\(p_{\text {solution }}=12.3 \times \frac{55.55}{56.55}\)
= 12.08 kPa

(i) First, find E⁰_cell
(ii) Then find \(E^{ 0 }_{ anode }\) by using the formula
\(E^{ 0 }_{ cell }=E^{ 0 }-_{ cathode }E^{ 0 }_{ anode }\)
Given E_cell = 0.059V; E⁰_cu2+/cu = +0.34V
[Ni²⁺] = 0.01M and [cu²⁺] = 0.1M
E_cell = E^₀_cell - \(\frac { 0.059 }{ 2 } log\frac { \left[ Ni^{ 2+ }(aq) \right] }{ \left[ Cu^{ 2+ }(aq) \right] } \)
\(0.059=E^{ 0 }_{ cell }-\frac { 0.059 }{ 2 } log\frac { \left( 0.01 \right) }{ 0.1 } \left( \because n=2 \right) \)
\(0.059=E^{ 0 }_{ cell }-\frac { 0.059 }{ 2 } log\frac { 1 }{ 10 } \left[ \because log10^{ -1 }=-1 \right] \)
\(\because 0.059=E^{ 0 }_{ cell }+0.0295 \times 1\)
\(E^{ 0 }_{ cell }=0.059-0.0295\)
= 0.0295 V = 0.03 V
\(E^{ 0 }_{ cell }=E^{ 0 }_{ cathode }E^{ 0 }_{ anode }\)
0.03 = 0.34 - E⁰_anode

or\(E^{ 0 }_{ anode }=E^{ 0 }_{ Ni2+/Ni }=0.34-0.03\)
_{= 0.31 V}

10.

(i) (a) CuF2 (b) (i) Cr in crO²₄- is in the highest oxidation state, i.e., +6 while Mn in MnO is in oxidation state +6 and its most stable oxidation state is +7
(ii) due to lanthanoid contraction
(iii) [Mn²⁺ has 3d 5 configuration]

11.

(a) (i) It is sum of powers to which cone. terms are raised in rate law or rate equation.
(ii) Each step of complex reaction (which takes place in more than one step) is called elementary, step in a reaction.
\((b) \ k=0.00510 \ { min }^{ -1 }\)
\(t=\frac { 2.303 }{ k } \log { \frac { { \left[ R \right] }_{ 0 } }{ \left[ R \right] } }\)
\(3\times 60\times 60=\frac { 2.303 }{ 0.00510 } \log { \frac { 0.1 }{ \left[ R \right] } }\)
\(\log { \frac { 0.1 }{ \left[ R \right] } } =\frac { 10800\times 0.00510 }{ 2.303 } \)
\(=23.94\)
\(\frac { 0.1 }{ \left[ R \right] } =Antilog \ 23.94\)
\(\frac { 0.1 }{ \left[ R \right] } =8.71\times { 10 }^{ 23 }\)
\(\left[ R \right] =\frac { 0.1 }{ 8.71\times { 10 }^{ 23 } }\)
\(\left[ R \right] =0.1148\times { 10 }^{ -24 }\)
\(\left[ R \right] =1.148\times { 10 }^{ -25 }M\)

Generic 3

8 x 3 = 24

12.

Transition metals and their compounds show catalytic activities because of Variable or multiple oxidation states, ability to form complexes, they provide large surface area for adsorption.

13.

Zn, Cd and Hg are generally soft and have low melting point because all the electrons in their d-orbital are paired.
Hence, the metallic bonds which are present in them are weak

14.

k_27⁰C = 2.31\(\times10^{-2}\) min^-1,
k₄_7⁰C = 6.931\(\times10^{-2}\) min^-1,
E_a = 43.85 kJ mol^-1.

15.

(a) 7.75\(\times\)10^-4mol L^-1s^-1
(b) 1.55\(\times\)10^-4mol L^-1s^-1
(c) conc.= 3.87 mol L^-1.

16.

1.52 \(\times\) 10¹².

17.

Given, W₂ = 3.9 s. WI = 49 g; \(\Delta T_f\)= 1.62 K
M₂(benzoic acid) = 122 g mol^-1
K_f(benzene) = 4.9 K kg mol^-1
\(\Delta T_f=\frac{i\times K_f \times W_2 \times 1000}{M_2 \times W_1}\)
\(i=\frac{\Delta T_f\times M_2 \times W_1}{K_f \times W_2 \times 1000}\)
\(i=\frac{1.62 K \times 122 g mol^{-1}\times 49 g}{4.9 K kg mol^{-1}\times 3.9 g \times 1000}\)
i = 0.50
Since, i < I, therefore solute (benzoic acid) undergoes association.

18.

14.21 %

19.

(i) [Co(en)₂Cl₂]⁺ (en = ethane-1, 2-diamine)
(ii) [Cr(C₂O₄)₃]^3-
trioxalatochromate (III) ion
(iii) [Co(NH₃)₃Cl₃]
Triamminetrichloridocobalt (Ill) ion. It will show geometrical isomerism.

Multiple Choice Question

15 x 1 = 15

20.

21.

22.

(d)
catalyst

23.

(a)
(i),(ii)

24.

25.

(a)
MnO₄^-

26.

(a)
1.0 M HCl

27.

(d)
1, 2

28.

(a)
Triamminechloridobromidonitroplatinum (IV) chloride

29.

(a)
(n - 1) d¹^-10ns^1-2

30.

(b)
87%

31.

32.

(d) \(\frac{1}{3} \frac { d[{ Br }_{ 2 }] }{ dt } =-\frac { 1 }{ 5 } \frac { d[{ Br }^{ - }] }{ dt } \) or \(\frac { d[{ Br }_{ 2 }] }{ dt } =-\frac {3 }{ 5 } \frac { d[{ Br }^{ - }] }{ dt } \)

33.

34.

Camphor has a very high cryoscopic constant (=39.7^o). Hence, It gives a large depression in melting point when an organic solute is dissolved in it.

Assertion and reason

5 x 1 = 5

35.

(a) If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
Explanation:
Ethan-1,2-diammine is a neutral molecule.as it carries no charge. Oxidation number of Co in the complex ion is +3.

36.

(d) Assertion is incorrect; Reason is correct.
Order and molecularity differs in case of complex reactions.
A is incorrect, but R is correct.

37.

(a) : According to Kohlrausch law, "limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte".

38.

(b) Transition metals form substitutional alloys, since they have nearly the same size, they can substitute one another in the crystal lattice. Thus, Both (A) and (R) are correct but (R) is not the correct explanation of (A).

39.

(a) Ethanol molecules show hydrogen bonding. On adding acetone, its molecules get in between the host molecule and break some of hydrogen bonds between them. Due to weakening of interaction, the mixture of ethanol and acetone shows the positive deviation from Raoult's law. Hence, both (A) and (R) are correct and (R) is the correct explanation of (A).

Case Study Questions

2 x 4 = 8

40.

(a) 'Ce' shows +4 oxidation state because if has stable electronic configuration.
(b) (i) Both show contraction, lanthanoid and actinoid contraction.
(ii) Both form coloured ions and undergo f· f transition
(c) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+6 \mathrm{Fe}^{2+}+14 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}+6 \mathrm{Fe}^{3+}\)
(d) \(2 \mathrm{Na}_{2} \mathrm{CrO}_{4}+\mathrm{H}_{2} \mathrm{SO}_{4}(\text { conc. }) \longrightarrow\) \(\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+\mathrm{Na}_{2} \mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{O}\)
(e) \(3 \mathrm{MnO}_{4}^{2-}+4 \mathrm{H}^{+} \longrightarrow 2 \mathrm{MnO}_{4}^{-}+\mathrm{MnO}_{2}+2 \mathrm{H}_{2} \mathrm{O}\)

41.

(i) (b) : Equivalent conductance of CaSO₄ :
\(\Lambda_{\mathrm{CaSO}_{4}}^{\infty}=\lambda_{\mathrm{Ca}^{2+}}^{\infty}+\lambda_{\mathrm{SO}_{4}^{2-}}^{\infty}\)
\(\lambda_{\mathrm{Ca}^{2+}}^{\infty}=\left(\mu_{\mathrm{Ca}^{2+}}\right) F ; \lambda_{\mathrm{SO}_{4}^{2-}}^{\infty}=\left(\mu_{\mathrm{SO}_{4}^{2-}}\right) F\)
\(\mu_{\mathrm{Ca}^{2+}}\) and \(\mu_{\mathrm{SO}_{4}^{2-}}\) - are ionic mobilities.
\(\Lambda_{\mathrm{CaSO}_{4}}^{\infty}=F(12.33+16.58) \times 10^{-4}\)
= 96500 x 10^{- 4}x 28.91 = 279
(ii) (c) : \(\alpha=\frac{\Lambda_{C}}{\Lambda^{\infty}} \Rightarrow 0.1=\frac{\Lambda_{C}}{279} \Rightarrow \Lambda_{C}=27.9\)
(iii) (a)
(iv) (c) : \(\Lambda_{m\left(\mathrm{CaCl}_{2}\right)}^{\circ}=\lambda_{\mathrm{Ca}^{2+}}^{\circ}+2 \lambda_{\mathrm{Cl}^{-}}^{\circ}\)
= (118.88 x 10^{- 4} ) + 2(77.33 x 10^{- 4} )
= 273.54 x 10^-4 m² mho mol^-1