CBSE Sample Papers for Class 11 Chemistry Set 8 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 11 Chemistry with Solutions Set 8 are designed as per the revised syllabus.

CBSE Sample Papers for Class 11 Chemistry Set 8 with Solutions

Time Allowed: 3 hours
Maximum Marks: 70

General Instructions:

• All questions are compulsory.
• This question paper contains 37 questions.
• Questions 1-20 in Section A are objective type-very short answer type questions carrying 1 mark each.
• Questions 21 – 27 in Section B are short answer type questions carrying 2 marks each.
• Questions 28 – 34 in Section C are long-answer 1 type questions carrying 3 marks each.
• Questions 35 – 37 in Section D are long-answer 11 type questions carrying 5 marks each.
• There is no overall choice. However, an internal choice has been provided in six questions of one mark, two questions of two marks, two questions of three marks and two questions of five marks. You must attempt only one of the choices in such questions.
• Use log tables, if necessary. Use of calculator is not allowed.

Section – A

Question 1.
Which of the following indicates that law of multiple proportion is being followed?
(A) Sample of carbon dioxide taken from any source will always have carbon and oxygen in the ratio 1: 2.
(Br Carbon forms two oxides namely C0₂ and CO, where masses of oxygen which combine with fixed mass of carbon are in the simple ratio 2 :1.
(C) When magnesium burns in oxygen, the amount of magnesium taken for the reaction is equal to the amount of magnesium in magnesium oxide formed.
(D) At constant temperature and pressure, 200 mL of hydrogen will combine with 100 mL oxygen to produce 200 mL of water vapour. [1]
Answer:
Option (B) is correct.

Explanation:
In CO₂, 12 parts by mass of carbon combine with 32 parts by mass of oxygen.
In CO, 12 parts by mass of carbon combine with 16 parts by mass of oxygen
So, simple ratio of masses of oxygen = 32 : 16
= 2 : 1
So, this follows law of multiple proportions.

Question 2.
For the electrons of oxygen atom, which of the following statements is correct? [1]
(A) Z_eff for an electron in a 2s orbital is the same as Z_eff for an electron in a 2p orbital.
(B) An electron in the 2s orbital has the same energy as an electron in the 2p orbital.
(C) Z_eff electron in Is orbital is the same as Z_eff for an electron in a 2s orbital.
(D) The two electrons present in the 2s orbital have spin quantum numbers m, but of opposite sign.
Answer:
Option (D) is correct.

Explanation:
According to Pauli’s Exclusion Principle, no two electrons in an atom can have the same set of four quantum numbers. It can also be stated as only two electrons may exist in the same orbital and they must be of opposite spin quantum numbers.

Question 3.
In a homologous series, a member differs from another member by the molecular mass of: [1]
(A) 10
(B) 12
(C) 14
(D) 28

OR

Which of the following is not correct ?
(A) ΔG is zero for a reversible reaction.
(B) ΔG is positive for a spontaneous reaction.
(C) ΔG is negative for a spontaneous reaction.
(D) ΔG is positive for a non-spontaneous reaction.
Answer:
Option (C) is correct.

Explanation:
In a homologous series, a member differs from another member by -CH₂ group, i.e., molecular mass of 12 + 2 = 14.
OR
Option (B) is correct.
Explanation:
If ΔG is positive, the process is non- spontaneous.

Question 4.
We know that the relationship between K_c and K_p is : [1]
K_p = K_c(RT)^Δn
What would be the value of An for the reaction :
NH₄Cl(S) ⇌ NH₃(g) + HCl(g) ?
(A) 1
(B) 0.5
(C) 1.5
(D) 2
Answer:
Option (D) is correct.

Explanation :
Δn = (Number of moles of gaseous products) – (Number of moles of gaseous reactants) = 2 – 0 = 2

Question 5.
Identify disproportionation reaction ? [1]
(A) CH₄ + 2O₂ → C0₂ + 2H₂O
(B) CH₄ + 4Cl₂ → CCl₄ + 4HCl
(C) 2F₂ + 2OH^– → 2F^– + OF₂ + H₂O
(D) 2NO₂ + 2OH^– → NO₂^– + NO₃^– + H₂O
Answer:
Option (D) is correct.

Explanation:
Reactions in which the same substance is oxidised as well as reduced are known as disproportionation reactions. In (d) N is both oxidized as well as reduced as the oxidation number increases from +4 in NO₂ to +5 in NO₃ and decreases from +4 in NO₂ to +3 in NO₂.

Question 6.
Choose the one which is not an example of arenes: [1]
(A) toluene
(B) biphenyl
(C) propanone
(D) naphthalene
Answer:
Option (C) is correct.

Explanation:
Propanone (CH₃COCH₃) is not an example of arenes as it does not contain benzene ring.

Question 7.
Catenation, i.e., linking of similar atoms depends on size and electronic configuration of atoms. The tendency of catenation in Group 14 elements follows the order : [1]
(A) C > Si > Ge > Sn
(B) C » Si > Ge = Sn
(C) Si > C > Sn > Ge
(D) Ge > Sn > Si > C
Answer:
Option (B) is correct.

Explanation:
Because the decrease in catenation property is linked with M – M bond energy which decreases from carbon to tin.

Question 8.
The addition of HCl to an alkene proceeds in two steps. The first step is the attack of H+ ion to >C =C< portion which can be shown as :

OR
Ionic species are stabilised by the dispersal of charge. Which of the following carboxylate ions is the most stable? [1]

Answer:
Option (B) is correct.

Explanation:
Since double bond is a source of electrons and the charge flows from source of more electron density. Therefore, electrons of the double bond attack the proton.
OR
Option (D) is correct.
Explanation :
In all the given carboxylate ions, the negative charge is dispersed which stabilises these ions. Here, the negative charge is dispersed by two factors, i.e., +R-effect of the carboxylate ion (conjugation) and effect of the halogens. It is evident from the above structures that +R-effect is common in all the four ions. Therefore, overall dispersal of negative charge depends upon the number of halogen atoms and electronegativity. Since F has the highest electronegativity and two F-atoms are present, thus, dispersal of negative charge is maximum.

Question 9.
Which statement is correct: [1]
(A) Aniline is purified by steam distillation.
(B) Glycerol is purified by distillation under reduced pressure.
(C) Naphthalene may be purified by sublimation.
(D) All of these
Answer:
Option (D) is correct.

Question 10.
Which of the following is not correct about pi bond? [1]
(A) p-bond is formed by sideways or lateral overlap of parallel π-orbitals.
(B) It is weaker bond than sigma bond as there is lower extent of overlapping.
(C) In π-bond formation, p-orbitals should be parallel and perpendicular to the plane of molecule.
(D) In π-bond rotation is not restricted.
OR
In which of the following, functional group isomerism is not possible?
(A)’ Alcohols
(B) Aldehydes
(B) Alkyl halides
(D) Cyanides
Answer:
Option (D) is correct.
OR
Option (C) is correct.

Explanation:
Alkyl halides do not show functional isomerism. Alcohols and ethers, aldehydes and ketones, cyanides and isocyanides are functional isomers.

Question 11.
Why Bohr’s orbits are called energy levels ?
Answer:
The Bohr’s orbits are called energy levels because they are associated with fixed amount of energy.

Question 12.
What will be the value of (pk_a + pk_b) at 25°C. [1]
OR
Why is ammonia termed as a base though it does not contain OH^– ions?
Answer:
pk_a + pk_b = 14
OR
When ammonia (NH₃) dissolves in water it accepts proton (H⁺) from the water to form NH₄⁺ ion due to presence of lone pair of electron. So, it acts as base known as Bronsted base.

Read the following text and answer the following questions on the basis of the same:
The VSEPR Theory is able to predict geometry of a large number of molecules, especially the compounds of p-block elements accurately. It is also quite successful in determining the geometry quite-accurately even when the energy difference between possible structures is very small.

Question 13.
According to VSEPR theory, pair of electrons ____________ each other : [1]
(A) attract
(B) repel
(C) stay as they are
(D) get positively charged
Answer:
Option (B) is correct.

Explanation:
The electron clouds are negatively charged.

Question 14.
Choose the correct order: [1]
(A) Lone pair (lp) – Lone pair (Ip) > Lone pair (Ip) – Bond pair (bp)
(B) Lone pair (Ip) – Bond pair (bp) > Lone pair (Ip) – Lone pair (Ip)
(C) Lone pair (lp) – Bond pair (bp) < Bond pair (bp) – Bond pair (bp)
(D) Lone pair (lp) – lone pair (Ip) < Bond pair (bp) – Bond pair (bp) 1 Answer: Option (A) is correct. Explanation: Lone pair (Ip) – Lone pair (Ip) > Lone pair (Ip) – Bond pair (bp)

Question 15.
Methane shows: [1]
(A) trigonal planar geometry
(B) tetrahedral geometry
(C) trigonal bipyramidal geometry
(D) Octahedral
Answer:
Option (B) is correct.

Explanation:
Methane shows tetrahedral geometry.

Question 16.
Which of the following angle corresponds to sp2 hybridisation? [1]
(A) 90°
(B) 120°
(C) 180°
(D) 109°
Answer:
Option (B) is correct.

Explanation:
In sp² hybridisation, geometry is generally trigonal planar making an angle of 120°.

Question 17.
Assertion (A): Simple distillation can help in separating a mixture of propan-l-ol (boiling point 97°C) and propanone (boiling point 56°C).
Reason (R): Liquids with a difference of more than 20°C in their boiling points can be separated by simple distillation. [1]
(A) Both A and R are correct and R is the correct explanation of A.
(B) Both A and R are correct but R is not the correct explanation of A.
(C) Both A and R are not correct.
(D) A is not correct but R is correct.
Answer:
Option (A) is correct.

Question 18.
Assertion (A): Significant figures for 0.200 is 3 whereas for 200 it is 1. [1]
Reason (R): Zero at the end or right of a number are significant provided they are not on the right side of the decimal point.
(A) Both A and R are correct and R is the correct explanation of A.
(B) Both A and R are correct but R is not the correct explanation of A.
(C) Both A and R are not correct.
(D) A is not correct but R is correct.
Answer:
Option (C) is correct.

Explanation:
Zero at the end or right of a number are significant provided they are on the right side of the decimal point.

Question 19.
Write IUPAC name of the compound

Give the IUPAC name of the following compound : [1]
CH₃CH(CH₃)CH(CH₂CH₃)CH = CHCH₃
Answer:
1 – Chlorocyclobutane
OR

Question 20.
What is the relation between wavelength and energy of radiation? [1]
OR
An atom having atomic mass number 13 has 7 neutrons. What is the atomic number of the atom?
Answer:
Greater the wavelength of radiation, lesser will be its energy.
Energy (E) = \(\frac{h c}{\lambda}\)
E = \(\frac{1}{\lambda}\)

Commonly Made Error
Students often write this answer as elaborated answer.

Answering Tip
Students should directly write the relation and the formula as per the marks for the question.

OR

Mass number = 13
Number of neutrons = 7
∴ Mass number = No. of protons + No. of neutrons
13 = No. of protons + 7
∴ No. of protons = 13 – 7 = 6
Atomic number = No. of protons = 6

Section – B

Question 21.
Ionisation energy of nitrogen is more then ‘O’ and ‘C’ both why ? [2]
Answer:
As we know that ionization energy increases from left to right in a period so, IE of N > IE of C this is due to decrease in atomic radius.
Electronic configuration of ₇N = 2, 5
= 1s², 2s², 2p³
Electronic configuration of ₈O = 2, 6
= 1s², 2s², 2p⁴
Since, nitrogen has stable electronic configuration due to presence of half-filled 2p-orbital. So, more energy is required to remove an electron in case of nitrogen than oxygen. So, I.E. of N > I.E. of O
Hence, I.E. of nitrogen is more than ‘O’ and ‘C’ both.

Question 22.
State application of intermolecular bonding.
OR
Group the following as linear and non-linear molecules: [2]
H₂O, HOCl, BeCl₂, Cl₂0
Answer:
Melting point and boiling point of water: Water has the lowest molecular weight among the hydrides of group 16 elements yet it has the highest melting and boiling points. It is due to intermolecular H-bonding in H2O. (2)
OR

Question 23.
Explain the following:
(a) Electronegativity of elements increase on moving from left to right in the periodic table.
(b) Ionisation enthalpy decreases in a group for top to bottom? [2]
Answer:
(a) Electronegativity of elements increases on moving from left to right in the periodic table because nuclear charge increases and atomic size decreases on moving from left to right in a period.
(b) Ionization enthalpy decreases in a group from top to bottom because atomic size increases down the group which results in lesser force of attraction between the nucleus and the outermost shell.

Question 24.
Calculate the mass of the photon with wavelength of 3.6 A.
OR
A 100 watt bulb emits monochromatic light of wavelength 400 nm. Calculate the number of photons emitted per second by the bulb. [2]
Answer:
Given, λ = 3.6 A = 3.6 × 10^-10 m
Velocity of photon= Velocity of light
= 3.0 × 10⁸ m/s
λ = \(\)
or m = \(\frac{h c}{\lambda}\)
= 6.135 × 10^-33 kg

OR

E = \(\frac{h}{m v}\)
= \(\frac{6.626 \times 10^{-34} \mathrm{Js} \times 3 \times 10^8 \mathrm{~ms}^{-1}}{400 \times 10^{-9}}\)
= 4.969 × 10^-19 J
Power of the bulb = 100 watt = 100 Js^-1
Number of photons emitted
= \(\frac{100 \mathrm{Js}^{-1}}{4.969 \times 10^{-19} \mathrm{~J}}\)
= 2.012 × 10²⁰ s^-1

Question 25.
How will you convert benzene into: [2]
(i) p – nitrotoluene
(ii) acetophenone? [2]
Answer:
(i) Benzene can be converted into p-nitrotoluene as:

(ii) Benzene can be converted into acetophenone as:

Question 26.
Describe the method, which can be used to separate two compounds with different solubilities in a solvent. [2]
Answer:
Fractional crystallisation is used for this purpose. A hot saturated solution of these two compounds is allowed to cool. The less soluble compound crystallises out while the more soluble remains in the solution. The crystals are separated from the mother liquor and the mother liquor is again concentrated and the hot solution is again allowed cool when the crystals of the second compound are obtained. These are again filtered and dried.

Question 27.
Complete the following reactions: [2]

Answer:

Section – C

Question 28.
The reaction 2C + O₂ → 2CO is carried oui by taking 240 g of carbon and 96.0 g of O. [3]
Find out:
(i) Which reactant is left in excess ?
(ii) How much of it is left?
(iii) How many grams of the other reactant should be taken so that nothing is left at the end of the reaction ?
OR
25 kg of N₂ and 6 kg of H₂ are mixed to produce NH₃.
(i) Identify the limiting reagent.
(ii) Calculate the amount of ammonia formed in this reaction.
Answer:

24 g of carbon will react with 32 g of O₂ to from 56 g of CO.
So, oxygen is left in excess.
(ii) Amount of oxygen left = 96 – 32 = 64 g
(iii) Mass of other reactant = 96 – 24 = 72 g

Commonly Made Error
Students wrongly identify the excess reagent and do calculations wrongly.

Answering Tip
Always understand the problem, given data and what needs to be calculated.
OR
N₂ + 3H₂ → 2NH₃
(i) 1 mole N₂ requires 3 moles of H₂
∵ 28 kg N₂ require 6 kg H₂
∴ 25 kg N₂ will require = \(\frac{6}{28}\) × 25 = 5.36 kg H₂
Hence, N₂ is the limiting reagent.

(ii) 1 mole N₂ produces 2 moles NH₃
∵ 28 kg N₂ produce 34 moles NH₃
∴ 25 kg N₂ will produce = \(\frac{34}{28}\) × 25
= 30.36 kg NH₃

Commonly Made Error
Students wrongly identify the limiting agent.

Answering Tip
Always understand the problem, given data and what needs to be calculated. Be thorough with the concepts like limiting agent, which helps to solve problems without confusion.

Question 29.
(a) How many unpaired electrons are there in Cr, when it is in its ground state? (Atomic number of Cr = 24)
(b) On the basis of Pauli’s exclusion principle show that the maximum number of electrons in the M-Shell (n=3) of any individual atom is 18. [3]
Answer:
(a) ₂₄Cr = 2, 8, 13, 1
= 1s², 2s²2p⁶, 3s²3p⁶ 3d⁵, 4s¹
∴ Number of unpaired electrons in Cr in ground state = 6

(b) According to Pauli’s exclusion principle, “Only two electrons may exist in the same orbital and these electrons must have opposite spin.”
For M-shell n = 3
Maximum number of electrons in the shell = 2n²
= 2(3)²
= 18

Question 30.
Electronic configuration of the elements are given below : [3]
(i) [Ar]4s²
(ii) [Ar] 3d¹⁰4s²
(iii) [Ar] 3d¹⁰4s²4p⁶5s²
(iv) [Ar] 3d¹⁰4s²4p⁶5s¹
Arrange these elements in increasing order of their metallic character. Give reasons for your answer.
Answer:
(i) [Ar]4s² is Calcium metal with At. no. = 20.
(ii) [Ar]3d¹⁰4s² is Zinc metal with At. no. = 30.
(iii) [Ar]3d¹⁰4s²4s¹5p⁶ is Strontium metal with At. no. = 38.
(iv) [Ar]3d¹⁰4s²4p⁶5s² is Rubidium metal with At. no. = 37.

Alkali metals are most metallic, followed by alkaline earth metals and transition metals. Among alkali metals – Rubidium (37) is most metallic. Among alkaline earth metals (Ca, Sr), Sr (Strontium) is more metallic than Calcium (Ca) as metallic character increases from top to bottom in a group. Zinc, the transition metal is least metallic.
Thus, metallic character increases from
Zn < Ca < Sr < Rb or (ii) < (i) < (iii) < (iv) (1)

Commonly Made Error
Students wrongly identify the elements and the order of metallic character.

Answering Tip
Learning elements and their electronic configurations and to which group it belongs, its special characters and other periodic properties is very important.

Question 31.
Explain why? [3]
(i) Reaction FeS0₄(aq) + Cu(s) → CuS0₄(aq) + Fe(s) does not occur.
(ii) Zinc can displace copper from aqueous CuS0₄ solution but Ag can not.
(iii) Solution of AgN0₃ turns blue when copper rod is immersed in it.
Answer:
(i) E°_Fe²⁺/Fe = – 0.44 V
E°_Cu²⁺/Cu = + 0.34 V
We know that a metal having more negative value of E° can displace the metal having less negative or positive value of E° from their salt solution. So, Fe can displace Cu from CuSO₄ solution.
But, in the reaction, it is showing that Cu displaces Fe from FeSO₄ solution which is not feasible.
So, the given reaction does not occur.

(ii) E°_Zn²⁺/Zn = -0.76 V
E°_Cu²⁺/Cu = +0.34 V
E°_Ag²⁺/Ag = + 0.80 V
Similarly, Zn has negative E° and Cu has positive E°. So, zinc can displace copper from aqueous CuSO₄ solution.
But silver has higher positive value of E° than Cu, so, Ag cannot displace Cu from CuSO₄ solution.

(iii) Cu has lesser E° than Ag. So, copper can displace Ag from AgNO₃ solution due to which the solution turns blue.

Question 32.
Represent the potential energy/enthalpy change in the following processes graphically. [3]
(a) Throwing a stone from the ground to roof.
(b) \(\frac{1}{2}\)H₂(g) + \(\frac{1}{2}\)Cl₂(g) → HCl(g) Δ_rH° – 92.32 kJ mol^-1
In which of the processes, potential energy/ enthalpy change is contributing factor to the spontaneity?
Answer:
(a) Throwing a stone from the ground to roof

(b) \(\frac{1}{2}\)H₂(g) + \(\frac{1}{2}\)Cl₂(g) ⇌ HCl(g)
Δ_rH° – 92.32 kJ

Since in process (a) energy increases and in process (b), enthalpy decreases. So, in the process (b), enthalpy change is contributing factor to the spontaneity.

Commonly Made Error
Students wrongly write the graphs and forget labelling it.

Answering Tip
Understanding the question is very important. Drawing and labelling correctly fetches full marks.

Question 33.
(i) Complete the following reaction : [3]

(ii) Complete the following and name the products:

OR
(i) Define and give an example of ring-chain isomerism.
(ii) What do you mean by cracking?
(iii) What are the main components of LPG?
Answer:

OR

(i) In ring-chain isomerism, one isomer is an open chain molecule and the other a cyclic molecule. For example, propene and cyclopropane.

(ii) The thermal decomposition of higher hydrocarbons into lower hydrocarbons in the presence or absence of a catalyst is called cracking.
(iii) Butane and isobutane.

Question 34.
(i) Write the resonance structure for CO₂.
(ii) Draw Lewis structure of H₂SO₄.
(iii) Why KHF₂ exist but KHCl₂ does not? [3]
Answer:

(iii) KF forms H-bonds with HF whereas KCl cannot form H-bond with HCl.
Hence, KHF₂ can dissociate to give HF₂ ion and therefore, KHF₂ exists. However, KHCl₂ cannot dissociate to HCl₂^– ion and therefore, KHCl₂ does not exist.

Section – D

Question 35.
Calculate the pH of a buffer which is 0.1 M in acetic acid and 0.15 M in sodium acetate. Given that the ionisation constant of acetic acid is 1.75 × 10^-5. Also calculate the change in pH of the buffer if to 1 L of the buffer : [5]
(i) 1 c.c. of 1 M NaOH are added
(ii) 1 c.c. of 1 M HCl are added.
Assume that the change in volume is negligible.
(iii) What will be the buffer index of the above buffer ?
OR
(i) Calculate the pH of 0.08 M solution of hypochlorous acid, HOCl. The ionisation constant of the acid is 2.5 × 10^-3. Determine the percent dissociation of HOCl.
(ii) Calculate the pH of a 0.10 M ammonia solution. Calculate the pH after 50.0 mL of this solution is treated with 25.0 mL of 0.10 M HCl. The dissociation constant of ammonia, K_b = 1.77 × 10^-5.
Answer:
Given,
[CH₃COOH] = 0.1 M
[CH₃COONa] = 0.15 M
pK_a = 1.75 × 10^-5

For an acidic buffer
pH = PK_a + log \(\frac{[\text { Salt }]}{[\text { acid }]}\) ………….(i)
pK_a = – log [K_a]
= – log [1.75 × 10^-5]
= – log 1.75 + 5 log 10
= – 0.2430 + 5 = 4.757
From equation (i)
pH = 4.757 + log \(\frac{(0.15)}{(0.10)}\)
= 4.757 + 0.1761
= 4.933

(i) 1 cc of 1 M NaOH contains NaOH = 10^-3 mol.
This will convert 10^-3 mol of acetic acid into the salt so the salt formed = 10^-3 mol
[Acid] = 0.10 – 0.001 = 0.099 M
[Salt] = 0.15 + 0.001 = 0.151 M
pH = pK_a + log \(\frac{\text { [salt }]}{\text { [acid] }]}\)
= 4.757 + log \(\frac{0.151}{0.099}\)
= 4.757 + 0.183
= 4.940
Increase in pH = 4.940 – 4.933
= 0.007 which is negligible

(ii) 1 cc of 1 M HCl contains HCl = 10^-3 M
This will convert 10^-3 mol CH₃COONa into CH₃COOH.
∴ [Acid] = 0.10 + 0.001 = 0.101 M
[Salt] = 0.15 – 0.001 = 0.149 M
pH = pK_a + log\(\frac{\text { [salt }]}{\text { [acid] }]}\)
= 4.757 + log\(\frac{0.149}{0.101}\)
= 4.757 + 0.169
= 4.925

∴ Decrease in pH = 4.933 – 4.925
= 0.008 which is negligible

(iii) No. of moles of HCl or NaOH added in buffer = 0.001 mol
Change in pH = 0.007

∴ Buffer index = \(=\frac{\text { No. of moles of acids or base added }}{\text { Change in } \mathrm{pH}}\)
= \(\frac{0.001}{0.007}\)
= \(\frac{1}{7}\)
= 0.143

OR

Given, C = 0.08 M
K_a = 2.5 × 10^-5

Ionization constant of acid,
K_a = \(=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{ClO}^{-}\right]}{[\mathrm{HOCl}]}\)
= \(\frac{x \times x}{(0.08-x)}\)
As x < < 0.08
So, 0.08 – x = 0.08
K_a = \(\frac{x^2}{0.08}\)
x² = 2.0 × 10^-6
x = 1.41 × 10^-3
[H⁺] = 1.41 × 10^-3 M
So, % dissociation of HOCl
= \(\frac{[\mathrm{HOCl}]_{\text {dissociated }}}{[\mathrm{HOCl}]_{\text {initial }}}\) × 100
= \(\frac{1.41 \times 10^{-3}}{0.08}\) × 100
= 1.76%
pH = – log [H⁺]
= – log [1.41 × 10^-3]
= – log 1.41 + 3 log 10
= – 0.1492 + 3
= 2.8508

(ii) Given, C = 0.10 M
K_b = 1.77 x 10^-5

Ionization constant of ammonia,
K_b = \(\frac{\left[\mathrm{NH}_4^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{NH}_3\right]}\)
1.77 × 10^-5 = \(\frac{x \times x}{(0.10-x)}\)
As x < < 0.10
So, 0.10 – x = 0.10
1.77 × 10^-5 = \(\frac{x^2}{0.10}/latex]
x² = 1.77 × 10^-6
x = 1.77 × 10^-3
[OH^–] = 1.33 × 10^-3 M

No, K_w = [H⁺][OH^–]
[H⁺] = [latex]\frac{\mathrm{K}_{\mathrm{w}}}{\left[\mathrm{OH}^{-}\right]}\)
= \(\frac{10^{-14}}{1.33 \times 10^{-3}}\)
= 7.51 × 10^-12
pH = – log [H⁺]
= – log [7.51 × 10^-12]
= – log 7.51 × 12 log 10
= – log 0.8756 + 12
= 11.12
On addition of 25 mL of 0.1 M HCl solution (i.e. 2.5 m mol of HCl) to 50 mL of 0.1 M NH₃ solution (i.e. 5 m mol of NH₃), 2.5 m mol of NH₃ molecules are neutralized.

The resulting 75 mL solution contains the remaining unneutralized 2.5 m mol of NH₃ molecules and 2.5 m mol of NH₄⁺.

The resulting 75 mL of solution contains 2.5 m mol of NH4+ ions (i.e., 0.033 M) and 2.5 m mol (i.e., 0.033 M) of unneutralized NH₃ exists in the following equilibrium:

where y = [OH^–] = [NH⁺]
The final 75 mL solution after neutralisation already contains 2.5 m mol NH4+ ions (i.e., 0.033 M), thus total concentration of NH₄⁺ ions is given as:
[NH₄⁺] = 0.033 + y
As y is small,
[NH₄OH] = 0.033 M
and [NH₄⁺] = 0.033 M
Now, K_b = \(\frac{\left[\mathrm{NH}_4^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{NH}_4 \mathrm{OH}\right]}\)
= 1.77 × 10^-5 = \(\frac{y \times 0.033}{0.033}\)
y = 1.77 × 10^-5 = [OH^–]
[H⁺][OH^–] = K_W
[H⁺] = \(\frac{K_b}{\left[O H^{-}\right]}\)
= \(\frac{10^{-14}}{1.77 \times 10^{-5}}\)
= 0.56 × 10^-9 = 5.6 × 10^-10
pH = – log [H⁺]
= – log [5.6 × 10^-10]
= – log 5.6 + 10 log 10
= – 0.7482 + 10
= 9.2518

Question 36.
(a) What is reversible process in thermodynamics ? [5]
(b) Name the thermodynamic process for which
(i) q = 0
(ii) ΔU = 0
(iii) ΔV = 0
(iv) Δp = 0
(c) Water decomposes by absorbing 286.2 kj of electrical energy per mole. When H₂ and O₂ combine to form one mole of H₂O, 286.2 kj of heat is produced, Which thermodynamic law is proved? Write its statement,
Answer:
(a) Reversible Process: It is defined as a system where change takes place infinitesimally slow and the direction of which at any point can be reversed by infinitesimal change in the state of the system.

(b) (i) For q = 0, heat is constant, so, the process is adiabatic process.
(ii) For DU = 0, internal energy is constant which takes place at constant temperature, so, the process is isothermal process.
(iii) For DV = 0, volume is constant, so, the process is isochoric process.
(iv) For Dp = 0, pressure is constant, so, the process is isobaric process.

(c) H₂O(g) → H₂(g) + 1/2O₂(g); Δ_rH° = 286.2 kJ mol^-1
H₂(g) + 1/2O₂(g) → H₂O(s); Δ_rH° = – 286.2 kJ mol^-1
It is referred to as first law of thermodynamics. According to this law, “Energy can neither be created nor be destroyed but can be converted from one form to another”.

Commonly Made Error
Students often make mistake while naming the thermodynamic process.

Answering Tip
Revise all the definition and thermodynamic process.

Question 37.
(a) Give condensed and bond line structural formulas and identify the functional group(s) present, if any, for : [5]
(i) 2,2,4-Trimethylpentane
(ii) 2-Hydroxy-l, 2,3-propanetricarboxylic acid ;
(iii) Hexanedial
(b) What is the relationship between the members of following pairs of structures? Are they structural or geometrical isomers or resonance contributors?

OR
(a) Identify the functional groups in the following compounds:

(b) Which of the following selected chains is correct to name the given compound according to IUPAC system?

Answer:
(a) (i) 2, 2 ,4-Trimethylpentane
Condensed formula-(CH₃)₂CHCH₂C(CH₃)₃

Bond line formula :

No functional group.

(ii) 2-hydroxy-1, 2, 3 – propanetricarboxylic acid Condensed Formula-(COOH)
CH₂C(OH) (COOH)CH₂(COOH)

Bond line formula:

Functional groups present in the given compound are -COOH and – OH.

(ii) Hexanedial
Condensed Formula-(CHO) (CH₂)₄(CHO)

Bond line formula:

Functional group present in the given compound is -CHO.

Commonly Made Error
Students miss carbons while writing the bond line formula.

Answering Tip
Always mention the functional group clearly. w Remember that while writing the bond line formula, the starting, end lines and the intersecting lines contain ‘C’ atoms.

(b) (i) The given compounds have the same molecular formula but they differ in the position of the functional group, so they are structural isomers. (1)
(ii) The given compounds having the same molecular formula, the same constitution, and the sequence of covalent bonds, but with different relative position of their atoms in space, so they are called geometrical isomers.

OR

(a) (i) Aldehyde (-CHO),
Hydroxyl (-OH),
Methoxy (-OMe),

(ii) Amino (-NH₂),

Diethylamine (n(C₂H₅)₂)
(iii) Nitro (-NO₂), and C = C double bond

Commonly Made Error
Some students get confused between methyl and methoxy group.
Some students get confused in -C = O (ketone) and -CHO (aldehyde) group.

Answering Tip
Learn all the types of functional groups. While writing the functional groups, do not forget to write the double bonds or triple bonds if present.

(b) The 4 carbon chain is correct according to the IUPAC system since it contains both the functional groups. The other three carbon chains are incorrect since none of them contains both the functional groups.