OLD is Gold [Asmita Publication] Chemistry SET 3 Complete Solution

Group A: MCQs

Tick the correct answer:
1. 0.1 M acetic acid ionise to an extent of 1.34% Ionization constant of acetic acid is:

1. 0.00134
2. 1.182
3. 1.82 x 10^-5
4. 2.8 x 10^-5

Solution:2.8 x 10^-5
2. Half-life of a 1st order and zero order reaction are same Then the ratio of the initial rates of 1 order reaction to the of the Zero order reaction is

1. 1/0.693
2. 2 x 0.693
3. 0.693
4. 2/0.693

Solution:2 x 0.693
3. What is the molarity of the solution of barium hydroxide, if 35 mL of 0.1 M HCI is used in the titration of 25 ml of the barium hydroxide solution?

1. 0.35
2. 0.07
3. 0.28
4. 0.14

Solution:0.07
4.The reaction, 3CIO^- (aq)→ CIO₃^- (aq) + 2Cl^-(aq) is an example of

1. Oxidation Reaction
2. Reduction Reaction
3. Disproportionation Reaction
4. Decomposition Reaction

Solution:Disproportionation Reaction
5.Different ions will split up by different compounds to give

1. same coloured complex
2. different coloured complex
3. same density complex
4. same density complex

Solution:different coloured complex
6.Which og the following are the correct matching of metals with the most commonly employed ores for their extraction?

1. Fe: Chalcorite; Al: Bauxite
2. Fe: Siderite; Al: Clay
3. Fe: Hameatite; Al: Corundum
4. Fe: Haematite; Al: Bauxite

Solution:Fe: Haematite; Al: Bauxite
7.In the nitration of benzene using a mixture of conc. H₂SO₄ and conc. HNO₃, the species which initiates the reaction is...

1. NO₂
2. NO₂⁺
3. NO-
4. NO₂^-

Solution:NO₂⁺
8.Which of the following compounds gives a secondary alcohol upon reaction with methylmagnesium bromide?

1. Butyl formate
2. 3-pentanone
3. Pentanal
4. Methyl butanoate

Solution:Pentanal
9. The specific gravity of cement is......

1. 2.5
2. 1.44
3. 3.15
4. 3.0

Solution:3.15
10. ...... is utilized for applying the pulp slurry to the screen.

1. Draining
2. Pressuring
3. Drying
4. Forming

Solution:Forming
11.In Nuclear reactor the control rod are made of

1. Graphite rod
2. Cadmium rod
3. Au
4. None of these

Solution:Cadmium rod

OLD is Gold [Asmita Publication] Chemistry SET 3 Complete Solution
Checkout✅: https://t.co/aJr8CVfPAF

— NepaliEducate (@nepalieducate) November 26, 2023

Group B: Short Answer Questions

Attempts all the Questions:

1. 20cm³ of a solution containing 7g/dm³ of metal hydroxide, XOH, were exactly neutralized with 25cm³ of 0.10M hydrochloric acid.

i) Write balanced chemical equation for the neutralization of the metal hydroxide, XOH, with hydrochloric acid.

ii)Calculate the concentration of the metal hydroxide in moles per dm³?

iii) Calculate molar mass of XOH.

iv) Identify element X.

Solution:

i) The balanced chemical equation for the neutralization of metal hydroxide, XOH, with hydrochloric acid, HCl, is:

XOH + HCl → XCl + H₂O

ii)

Number of moles of HCl used = 0.10 M x 0.025 dm³ = 0.0025 moles

From the balanced chemical equation,

1 mole of XOH reacts with 1 mole of HCl.

Therefore, number of moles of XOH = 0.0025 moles

Volume of solution used = 20/1000 dm³ = 0.02 dm³

Concentration of XOH in moles per dm³ = 0.0025 moles / 0.02 dm³ = 0.125 mol/dm³

Therefore, the concentration of the metal hydroxide XOH in the solution is 0.125 mol/dm³.

OR

Consider the exothermic reaction between reactants A and B?

A + B → E (fast)

E + B → C + D (slow)

a. What is the order with respect to reactants A and B?

b. What is the rate law for the reaction?

c. Sketch a potential energy diagram for the overall forward reaction. Identify the activation energy for the overall forward reaction. Identify the location of reactants, intermediate(s), activated complex(es), and products.

Solution:

(a) The reaction is first order with respect to reactant A and first order with respect to reactant B.

2. This question related to thermodynamics.

a. How is free energy change of a reaction related to enthalpy change and entropy change?

b. Calculate the enthalpy of formation of ethane at 298K, if the enthalpies of the combustion of C, H and C₂H₅ are -94.14, -68.47 and -373.3Kcal, respectively.

Solution:

(a) The free energy change of a reaction (ΔG) is related to the enthalpy change (ΔH) and entropy change (ΔS) by the equation:

ΔG = ΔH - TΔS

where T is the temperature in Kelvin. This equation is known as the Gibbs-Helmholtz equation.

If ΔG is negative, the reaction is spontaneous and can proceed in the forward direction. If ΔG is positive, the reaction is non-spontaneous and will proceed in the reverse direction. If ΔG is zero, the reaction is at equilibrium.

 

3. A metal M can be extracted from hematite ore. Steel is an alloy of metal M.

a. Write the principle involved in the manufacture of steel by open hearth process.

b. How is metal M rust by exposing in the moist air?

c. What is the function of limestone in the smelting of metal ‘M’.

Solution:

(a) Reaction involved in open Hearth furnace are:

Fe₂O₃ + S → Fe + SO₂

Fe₂O₃ + P → Fe + P₂O₅

Fe₂O₃ + Si → Fe + SiO₂

CaO + SiO₂→CaSiO₃

CaO + P₂O₅→Ca₃(PO₄)₂

(b)

When an iron object is left in damp air for a considerable time, it gets covered with a reddish-brown flaky substance called rust. This is called rusting of iron.

It occurs in the presence of oxygen and water.

Iron(Fe)+Oxygen(O₂​)+Water(H₂​O)→Rust(iron oxide Fe₂​O₃​) 

(c)

During the extraction of iron from haematite ore, limestone acts as a flux. Limestone decomposes froming CaO which reacts with silica forming slag.
 CaCO₃​→CaO+CO₂​
 CaO+SiO₂​→CaSiO₃​(slag)

4. (a) A haloalkane P reacts with aq. KOH to give Q. The compound Q on oxidation with K₂Cr₂O₇ + H⁺ gives R and R undergoes Clemmenson reduction to produce S. The compound P react with Sodium in presence of dry ether to form 2,3-dimethylbutane, write chemical reaction involved and identify P, Q, R, and S.

(b) What product would you expect when compound R is treated with hydrocyanide?

Solution: 

5. (a) Write down the isomeric alcohols of C₃H₆O and IUPAC name. Explain Victor- Meyer’s method to distinguish them.

(b) What happens when the product obtained by dehydrogenation of ethanol is allowed to react with Tollen’s reagents.

Solution:

(a) The Isomeric alcohols of C₃H₆O are:

(i) CH₃-CH₂-CH₂-OH : Propan-1-ol

(ii) $C{H_3} - \mathop {CH}\limits^{\mathop |\limits^{OH} }  - C{H_3}$: Propan-2-ol

(iii) CH₂=CH-CH₂OH : Prop-2-en-1-ol

 

(b) When the product obtained by dehydrogenation of ethanol (which is acetaldehyde) is allowed to react with Tollen's reagent (ammoniacal silver nitrate), it forms a silver mirror on the inner surface of the test tube or reaction vessel.

Reaction:

CH₃-CHO + [Ag(NH₃​)₂​]⁺→CH3​COO⁻+Ag

6. (a) An aromatic compound P on treating with aqueous ammonia and heating from Compound Q which on heating with Br₂ and KOH forms compound R of the molecular formula C₆H₇N

(b) How can you prepare p-hydroxy-azobenzene from compound ‘R’?

Solution:

(a) The aromatic compound A is benzoic acid C₆​H₅​COOH.

On treatment with aqueous ammonia and heating forms compound B, which is benzamide C₆​H₅​CONH₂​.

Benzamide on heating with bromine and KOH forms a compound C of molecular formula C₆​H₇​N, which is aniline C₆​H5​NH₂​. The reaction is called Hoffmann bromamide degradation.

(b)

7. (a) Write the name of one drug which relief pain and also draw structure.

(b) How can you distinguish addition and condensation polymer?

(c) What is the function of CaO in the manufacture of cement?

Solution:

(a) The name of pain reliving drug is paracetamol.

Structure:

(b)

One way to distinguish addition and condensation polymers is by looking at their reaction mechanisms and the byproducts produced during polymerization. Here's a simple comparison table:

Property	Addition Polymerization	Condensation Polymerization
Reaction mechanism	Monomers react to form a polymer with no byproducts	Monomers react to form a polymer with small molecules, such as water or alcohol, as byproducts
Polymerization conditions	Typically occurs at high temperatures and pressures	Can occur at lower temperatures and pressures
Examples	Polyethylene, polypropylene, PVC	Nylon, polyester, polyurethane

(c) The amount of lime in cement plays an important role in the formation of the silicates and aluminates of calcium, which are essential components for the strength and durability of the final product.

8. (a) A monohydroxyl substituted benzene (A) is prepared from hydrolysis of diazonium salt. Compound (A) is heated with zinc dust gives (B). The compound (B) on Friedel-Craft Alkylation with methyl chloride to give (C) which on oxidation with CeO₂ yield compound (D). Write the reaction involved and IUPAC name of A, B, C, D.

(b) Convert Compound A into m-nitrobenzoic acid.

Solution:

(a):

Since the compound A is prepared by hydrolysis of diazonium salt, which means it is Phenol.

Phenol(A) On Reduction with Zinc Dust Gives Benzene(B)

Benzene (B) on Friedel-Craft Alkylation with methyl chloride to give Toluene(C)

Toluene(C) on oxidation with CeO₂ yield Benzaldehyde(D)

A: Phenol

B: Benzene

C: Toluene

D: Benzaldehyde

(b):

OR

(a)Name the catalyst in the Haber process for the manufacture of ammonia.

(b) Name the catalyst used in the hydrogenation of carbon-carbon double bonds.

(c) Name the catalyst in the contact process for the manufacture of Sulphuric acid.

(d)Draw the structure of [Cu(H₂O)₅]²⁺ and [CuCl₄]^2- and write the shape of ion.

Solution:

(a) The catalyst used in the Haber process for the manufacture of ammonia is iron.

(b) The catalyst used in the hydrogenation of carbon-carbon double bonds is Nickel.

(c) The catalyst used in the contact process for the manufacture of Sulphuric acid is Vanadium Pentoxide.

(d)

Structure of [Cu(H₂O)₅]²⁺:

Structure of [CuCl₄]^2-:

Group C: Long Question

9. (a) Consider the reaction,

2Ag⁺ + Cd → 2Ag + Cd²⁺

The standard electrodes potentials for Ag⁺ → Ag and Cd⁺²→Cd couples are 0.80V and -0.40V, respectively.

i. What is the standard potential E⁰ for this reaction?

ii. For the electrochemical cell in which this reaction takes place which electrode is negative electrode?

(b) How is single electrode potential originated?

(c) Heat of combustion of compound are given as:

CH4	-210Kcl
C	-94Kcal
H2	-68Kcal

Calculate the heat of formation of CH₄.

Solution:

(a) i. The standard potential for the given reaction can be calculated using the formula:

E⁰ = E⁰(reduction) - E⁰(oxidation)

E⁰(reduction) is the standard reduction potential for the reduction half-reaction and E⁰(oxidation) is the standard oxidation potential for the oxidation half-reaction.

The balanced oxidation half-reaction is:

Cd → Cd²⁺ + 2e^-

The balanced reduction half-reaction is:

Ag+ + e^- → Ag

The standard potential for the reduction half-reaction is given as 0.80V (E⁰(Ag+ → Ag)) and for the oxidation half-reaction is -0.40V (E⁰(Cd → Cd²⁺ + 2e^-)).

Therefore, the standard potential for the given reaction is:

E⁰ = 0.80V - (-0.40V) = 1.20V

ii. The electrode with a more negative standard reduction potential will act as the negative electrode.

From the given standard electrode potentials, we can see that the standard reduction potential for Cd²⁺ → Cd is more negative than that of Ag⁺ → Ag. Therefore, Cd electrode will act as the negative electrode.

(b)

We Know:

Heat of Combustion of Carbon = -94Kcal

Heat of Combustion of Hydrogen = -68Kcal

Reaction for formation of methane:  C+2H₂​⟶CH₄

To calculate the heat of formation of CH₄, we need to use the following equation:

ΔH_reaction=∑ΔH_{Combustion of Product}​−∑ΔH_{Cobustion of Reactant}

=-94+ 2(-68)-(-210)

=-20Kcal

Thus, Heat of formation of Methane = -20Kcal.

OR

(a) Equal volumes of 0.02 M AgNO₃ and 0.02M HCN were mixed. Calculate [Ag⁺] at equilibrium given: K_sp=2.2×10^-15 K_a (HCN) = 6.2×10^-10

(b) A solution contains a mixture of Ag⁺ (0.1M) and Hg₂⁺² (0.1M) which are to be separated by selective precipitation. Calculate the maximum concentration of iodide ion at which one of them gets precipitated almost completely. What percentage of that metal ion is precipitated?

K_sp(AgI) = 8.5 ×10^-17, K_sp (Hg₂I₂) = 2.5 ×10^-26

Solution:

(a) 

(b) [I⁻] required to precipitate Ag+ and Hg₂²⁺​ are derived as:
For AgI:

[Ag⁺][I⁻]=K_spAgI​​;[0.1][I⁻]=8.5×10⁻¹⁷;[I⁻]=8.5×10⁻¹⁶M.......(i)

For Hg₂​I₂​:

[Hg2²⁺​][I⁻]2=K_spHg₂​I₂​​​;[0.1][I−]2=2.5×10⁻²⁶;[I⁻]=5×10⁻¹³M........(ii)

[I⁻] required to precipitate AgI are lesser than that required to precipitate Hg₂​I₂​ and thus, precipitation of AgI will take place first. It will continue till the [I⁻] becomes 5×10⁻¹³ when Hg₂​I₂​ begins to precipitate and thus,

Maximum [I⁻] for AgI preicipitation =5×10⁻¹³M. Also [Ag⁺]_left​ at this concentration of [I⁻] can be evaluated as:

[Ag⁺]_left ​[I⁻]=${{K_{s{p_{AgI}}}}}$

[Ag⁺]=$\frac{{{K_{s{p_{AgI}}}}}}{{\left[ {{I^ - }} \right]}} = \frac{{8.5 \times {{10}^{ - 17}}}}{{5 \times {{10}^{ - 13}}}}$​=1.7×10⁻⁴M

∵ 0.1M Ag+ will left =1.7×10⁻⁴M Ag+ in solution
∴ % of Ag⁺ left =$\frac{{1.7 \times {{10}^{ - 4}} \times 100}}{{0.1}}$=0.17%M Ag⁺
∴ % of Ag precipitated =100−0.17=99.83%

10. (a) Arrange the compound in the complete reaction sequence with the suitable reagent:
Aniline, Benzene diazonium Chloride, Benzonitrile, Benzamide, Benzoic Acid.
(b) Write the name of aldehyde which gives tollens test and shows aldol condensation.
(c) How is 2- Hydroxy propanoic acid obtained from Ethanal.

Solution:

(a)

(b)

(c)

Compound A is Aniline

Compound B is benzene diazonium chloride salt.

Compound C is iodobenzene.

11. (a) An organic compound a which characteristic order, on treatment with NaOH Forms two compounds B & C which on oxidation with CrO₃ gives back compound A. Compound C is the sodium salt of acid. Compound C when heated with soda lime yields an aromatic hydrocarbon D. Deduce the structure of A, B, C and D. Write down the chemical equation for all the reaction taking place.

(b) Why NH₂ group of aniline is protected before nitration?

(c)Write a product which is obtained by the reduction of acetic anhydride.

Solution:

(a)

(b)

To prevent unwanted side reactions and to ensure selective nitration at the ortho and/or para positions. Aniline contains a reactive amino group (-NH2) that can react with nitric acid during the nitration reaction, leading to the formation of undesirable by-products. For example, the amino group can be nitrated, leading to the formation of di- and tri-nitroanilines instead of mono-nitroanilines. These by-products are difficult to separate from the desired product and can reduce the overall yield of the reaction.

(c) 

CH₃​CO−O−COCH₃​ (acetic anhydride) +LiAlH₄​→C₂​H₅​−OH (Ethanol)

OR

(a) Write the structure A, B and C in the following:

\[{C_6}{H_5} - CON{H_2}\mathop  \to \limits_\Delta ^{B{r_2}/KOH} A\mathop  \to \limits_{0 - {5^0}C}^{NaN{O_2}/HCl} B\mathop  \to \limits^{KI} C\]

(b) What happens when compound C is heated with sodium metal in the presence of dry ether?

(c) What product would you get when compound A and B are heated?

Solution:

(a) Compound A is Aniline

Compound B is Benzene Diazonium Chloride Salt.

Compound C is Iodobenzene

The Structures are:

Aniline (A)	Benzene Diazonium Chloride Salt (B)	Iodobenzene (C)

(b) When Iodobenzene reacts with shoddy metal in presence of dry ether to give di-phenyl or bi-phenyl.

(c) When Compound A (Aniline) and B (Benzene Diazonium Chloride Salt) are heated p-amino azobenzene is formed.