C1545203513visc Chemistry_self Assessment Paper_6.pdf

Time : 3.00 Hours Maximum Marks : 70

CHEMISTRY ISC

Solutions

Self Assessment Paper

6

1. (a) (i) dehydration, alkene (ii) pale yellow, aldehyde (iii) chemical, electrical 

[ISC Marking Scheme, 2014]

Examiner’s Comment Some candidates wrote the answers in the reverse order. The correct answer was ‘chemical’ and ‘electrical’ respectively, but some candidates wrote ‘electrical’ and ‘chemical’ instead.

(iv) ferromagnetism (b) (i) (1)  Formic acid (ii) (3) Phenyl isocyanide (iii) (2) Mg (iv) (1) K2SO4 (c) (i) Nylon 6—Polycaprolactam (ii) Vitamine B1 — Beri-Beri (iii) Liquid dispersed — Aerosol in gas (iv) One step reaction — Elementary (d) (i) Acetamide can be converted to methylamine by the reaction with bromine in the presence of alkali solution. CH3CONH2 + Br2 + 4KOH → CH3NH2 + 2KBr + K2CO3 + 2H2O The reaction is Hoffmann’s bromamide reaction.  (ii) d2sp3 hybridization and paramagnetic. [ISC Marking Scheme, 2015]

Examiner’s Comment Many candidates reported sp3d2 and diamagnetic whereas the correct answer was d2sp3 hybridization and paramagnetic.

(iii) In transition elements, the oxidation state can vary from +1 to the highest oxidation state by removing all its valence electrons. Also, in transition elements, the oxidation states differ by 1 (Fe2+ and Fe3+; Cu+ and Cu2+). In non-transition elements, the oxidation states differ by 2, for example, +2 and +4 or +3 and +5, etc.  (iv) The effect of pressure on the solubility of a gas in a liquid is governed by Henry’s Law. It states that the solubility of a gas in a liquid at a given temperature is directly proportional to the partial pressure of the gas. Mathematically, P = KHX where P is the partial pressure of the gas, X is the mole fraction of the gas in the solution and KH is Henry’s Law constant.

OSWAAL ISC Sample Question Papers, CHEMISTRY, Class - XII

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Applications of Henry’s law : (i) In the production of carbonated beverages (as solubility of CO2 increases at high pressure) (ii) In the deep sea diving. (iii) For climbers or people living at high altitudes, where low blood O2 causes climbers to become weak and make them unable to think clearly. 2. (a) Half-life, t1/2 = 0.693 / k = 0.693 / 200s−1 = 3.47s  (Approximately) (b) t1/2 = 0.693 / k = 0.6932 min−1 = 0.35 min (Approximately)  OR (b) Here, k=

0.693 t1/ 2

  = 0.693 /5730 years−1 It is known that,  

t=

2.303 R0 × log k R

t=

2.303 100 × log 0.693 / 5730 80

  = 1845 years  (approximately) Hence, the age of the sample is 1845 years.  3. They are sodium or potassium salts of sulphonated long chain hydrocarbons. e.g., sodium alkyl benzene sulphonate, i.e., sodium dodecyl benzene sulphonate. –

CH3(CH2)11

SO3 Na+

4. Phenol is formed. CHO



OH + NaOH (aq)

673 K 3000 atm

+NaCl



5. (i) PAN (Polyacrylonitrile). (ii) Triethylaluminium and titanium tetrachloride (Ziegler-Natta catalyst). 6. (i) Peptide linkage is the linkage that exists between the monomeric amino acids in a polypeptide chain by —COOH group of one amino acids and the —NH2 group of next amino acid by condensation. — R

R

O

—

— —

H

— —

O

—

H

NH2 — C — C — OH + H — N — C — C — H  H

—

O

—

— —

H

— —

O

—

H

H

R

NH2 — C — C — N — C — C —OH R

 

Peptide linkage

Solutions

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(ii) Denaturation is the disruption of native conformation of protein when subjected to change in conditions like pH, temperature, etc. resulting in change of secondary and tertiary structure but not in primary structure. e.g., Change of milk to curd. 7. (a) (i) C2H5-O-C2H5 + PCl5 → 2C2H5Cl + POCl3  [ISC Marking Scheme, 2016]

Examiner’s Comment A number of candidates wrote wrong products such as C2H5COCl or C2Cl5 — O — C2Cl5 although correct answer was C2H5Cl and POCl3.

(ii) C2H5-O-C2H5+ HI → C2H5I +C2H5OH 

OR

(b) (i) CH3—CH2—CH2—OH Propan-1-ol (ii) OH

— CH3 — CH — CH3

     Propan–2– ol  (i) and (ii) are the position isomers. 8. Let the concentration of the reactant be [A] = a Rate of reaction, R= k [A]2 = ka2 (a) If the concentration of the reactant is doubled, i.e [A] = 2a, then the rate if the reaction would be R′ = k(A)2 = 4ka2 = 4R Therefore, the rate of the reaction now will be 4 times the original rate. (b) If the concentration of the reactant is reduced to half, i.e [A]=1 / 2a, then the rate of the reaction would be R”= k(1/2a)2  = 1 / 4ka  = 1 / 4R Therefore, the rate of the reaction will be reduced to 1/4th. 9. Molar mass of ethane (C2H6) = 2 × 12 + 6 × 1 = 30 g mol−1 Therefore, number of moles present in 6.56 × 10−2g of ethane = 6.56 × 10−2 / 30 = 2.187 × 10−3mol Let ‘x’ be the number of moles of the solvent, According to Henry’s law, p = KHX 1 bar = KH.2.187 × 10−3 / 2.187×10−3 + x 1 bar = KH.2.187 × 10−3 / x KH = x / 2.187 × 10−3 bar         (Since x >> 2.187 × 10−3) Number of moles present in 5 × 10−2 g of ethane = 5 × 10−2 / 30 mol = 1.67 × 10−3mol According to Henry’s law, p=KHX = x / 2.187 × 10−3 × 1.67 × 10−3 / (1.67 × 10−3) + x = x / 2.187 × 10−3 × 1.67 × 10−3x           (Since, x>> 1.67 × 10−3) = 0.764 bar Hence, partial pressure of the gas shall be 0.764 bar.

OSWAAL ISC Sample Question Papers, CHEMISTRY, Class - XII

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OR (b) Vapour pressure of heptanes, p1 = 105.2 kPa, Vapour pressure of octane, p2 = 46.8 kPa, We know that, Molar mass of heptanes (C7H16) = 7 × 12 + 16 × 1 = 100 g mol−1 Therefore, number of moles of heptane = 26 / 100 = 0.26 mol Molar mass of octane (C8H18) = 8 × 12 + 18 × 1 = 114 g mol−1 Therefore, number of moles of octane = 35 / 114 = 0.31 mol Mole fraction of heptane, x1=0.26 / 0.26+0.31 = 0.456 And, mole fraction of octane, x2=1–0.456 = 0.544 Now, partial pressure of heptane, p1=x1p1 = 0.456 × 105.2 = 47.97 kPa Partial pressure of octane, p2 = x2p2 = 0.544 × 46.8 = 25.46 kPa Hence, vapour pressure of solution, ptotal = p1 + p2 = 47.97 + 25.46 = 73.43 kPa 10. Given that, edge length, a = 3.61 × 10−8 cm, As the lattice is fcc type, the number of atoms per unit cell, z = 4, Atomic mass, M = 63.5 g mol−1 We also know that, NA = 6.022 × 1023 mol−1 By applying the relation: d = =

z× M 3

a × NA 4 × 63.5

(3.61 × 10−8 )3 × 6.022 × 10 23

= 8.97 g cm-3 The measured value of density is given as 8.92 g cm−3 hence, the calculated density 8.97 g cm−3 is in agreement with its measured value.  11. Multimolecular Colloids Macromolecular Colloids Associated Colloids In these type of colloids, colloidal particles are aggregates of atoms of small molecules with molecular size less than 1 nm. e.g., Gold sol.

 12. (a) (i) [Cr(NH3)3(H2O)3]Cl3 (ii) K3[Fe(CN)6]

In these type of colloids, colloidal particles are themselves large molecules having colloidal dimensions. e.g., Proteins.

There are certain substances which behave as normal, strong electrolyte at low concentration but at higher concentration behave as colloidal solution due to aggregation. e.g., Soap and detergents

Examiner’s Comments Many candidates gave incorrect formula (i) In place of [Cr(NH3)3(H2O)3]Cl3 some candidates wrote [Cr(NH2)3(H2O)3]Cl3. (ii) In place of K3[Fe(CN)6] some candidates wrote .3K4[Fe(CN)6].

[ISC Marking Scheme, 2014]

Solutions

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(b) EDTA (Ethylene diamine tetraacetate) is used in lead poisoning which is a hexadentate i.e., the denticity of EDTA is 6. It binds with metal in octahedral manner by two N-atoms and four acetate oxygen atoms. 13. (a) Potassium permanganate (KMnO4) can be prepared from pyrolusite (MnO2). The ore is fused with KOH in the presence of either atmospheric oxygen or an oxidising agent, such as KNO3 or KClO4, to give K2MnO4. heat      → 2K2MnO4+2H2O 2MnO2+4KOH+O2  The green mass can be extracted with water and then oxidized either electrolytically or by passing chlorine/ozone into the solution. Electrolytic oxidation K2MnO4 ↔ 2K+ + MnO42−  H2O ↔ H+ + OH− At anode, manganate ions are oxidized to permanganate ions. MnO42− ↔ MnO4− + e− Oxidation by chlorine 2K2MnO4 + Cl2 → 2KMnO4 + 2KCl 2MnO42− + Cl2 → 2MnO4− + 2Cl− Oxidation by ozone 2K2MnO4 + O2 + H2O → 2KMnO4 + 2KOH + O2 2MnO42− + O2 + H2O → 2MnO42− + 2OH− + O2 OR (b) (i) Most of the complexes of transition metals are coloured. This is because of the absorption of radiation from visible light region to promote an electron from one of the d−orbitals to another. In the presence of ligands, the d-orbitals split up into two sets of orbitals having different energies. Therefore, the transition of electrons can take place from one set to another. The energy required for these transitions is quite small and falls in the visible region of radiation. The ions of transition metals absorb the radiation of a particular wavelength and the rest is reflected, imparting colour to the solution. (ii) The catalytic activity of the transition elements can be explained by two basic facts. (a) Owing to their ability to show variable oxidation states and form complexes, transition metals form unstable intermediate compounds. Thus, they provide a new path with lower activation energy, Ea, for the reaction. (b) Transition metals also provide a suitable surface for the reactions to occur. 14. [A] – Benzene (C6H6) [B]–Nitrobenzene (C6H5NO2) [C]– Aniline (C6H5NH2) [D]–Phenyl isocyanide (C6H5NC) [E]– Methyl phenylamine (C6H5NHCH3) [F]–Acetyl Chloride (CH3COCl) 15. (a) Cryolite performs two functions in the electrolysis of alumina. (i) It lowers the melting point of the mixture of about 1250 K. (ii) It improves the electrical conductivity of the melt. (b) The choice of a reducing agent in a particular case depends on thermodynamic factor. For a reaction to be feasible, the reaction of metal oxide with the reducing agent should have negative ∆G°. Therefore, the reducing agent is suitable for which ∆G° for the reduction is negative. Thus ZnO + C → Zn + CO ∆rG° = negative (Feasible) Cr2O3 + 3C → 2Cr + 3CO ∆rG° = positive (Not feasible)

OSWAAL ISC Sample Question Papers, CHEMISTRY, Class - XII

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16. Λ°m(HCOOH) = λ°(H+) + λ° (HCOO)− = 349.6 + 54.6 = 404.2 S cm2 mol−1 ΛCm = 46.1 S cm2 mol−1, Now degree of dissociation: ∴ α = Λcm/ Λ0m = 46.1/ 404.2 = 0.114 HCOOH HCOO− + H+ Thus, dissociation constant: ∴ Ka = cα2 / (1- α) = (0.025) × (0.114)2 / (1 - 0.114) = 3.67 × 10-4 mol L-1 OR −5 −1 (b) Given, κ = 7.896 × 10  S m c = 0.00241 mol L−1 Then, molar conductivity, Λm= k × 1000/c = 7.896 × 10−5 S cm−1 / 0.00241 mol L−1 × 1000 cm3 / L = 32.76 S cm2 mol−1 Λ0m = 390.5 S cm2 mol−1 Again, α = Λm / Λ0m = 32.76 S cm2 mol−1 / 390.5 S cm2 mol−1 = 0.084 Dissociation constant, Ka = cα2(1−α) = (0.00241 mol L−1)(0.084)2(1–0.084) = 1.86 × 10−5 mol L−1 17. (a) (i) Almost all halogens are coloured. This is because halogens absorb radiations in the visible region. This results in the excitation of valence electrons to a higher energy region. Since the amount of energy required for excitation differs for each halogen, each halogen displays a different colour. (ii) The steps which are required in the production of sulphuric acid by the contact process Step (1)- Sulphide ores or sulphur are burnt in air to form SO2.  Step (2)- By reaction with oxygen, SO2 is converted into SO3 in the presence of V2O5 as a catalyst. V O

2 5     2SO2 (g) +O2 (g)      → 2SO3 (g)



Step (3)-SO3 produced is absorbed on H2SO4 to give H2S2O7 (oleum).

  SO3 + H2SO4 → H2S2O7 This oleum is then diluted to obtain H2SO4 of the desired concentration. In practice, the plant is operated at 2 bar (pressure) and 720 K (temperature). The sulphuric acid thus obtained is 96-98% pure. OR (b) (i) In ammonia there is a lone pair of electron on nitrogen atom which makes ammonia a Lewis base. (ii) Fluorine does not have d-orbital. Chlorine has vacant 3d orbital. Electrons jump from 3s and 3p orbitals to 3d orbitals. [ISC Marking Scheme, 2014]

Examiner’s Comment Most of the candidates wrote that fluorine gives only one oxide due to high electronegativity but did not mention that fluorine does not have d orbital whereas chlorine forms series of oxides due to vacant d orbital.

Solutions

7

ANSWERING TIP... Chlorine shows variable covalency due to the presence of empty d-sub shell whereas fluorine has no d-orbital

(iii) Noble gases being monoatomic have no interatomic forces except weak dispersion forces and therefore, they are liquefied at very low temperatures. Hence, they have low boiling points. (iv) Halogens have the smallest size in their respective periods and therefore high effective nuclear charge. As a consequence, they readily accept one electron to acquire noble gas electronic configuration [5] (v) Because of the compact nature of oxygen atom, it has less negative electron gain enthalpy than sulphur. 18. (a) (A) (i)  A – acetylene    B – acetaldehyde    C – acetic acid    D – acetyl chloride    E – ethyl acetate (ii)  Isomer of acetaldehyde H H

— —

— —

C—C

H OH



(iii)  CH = CH+H2O

H2SO4+HgSO4

CH3CHO B

A

(B) Clemmensen’s reduction: Zn / Hg CH3CHO + 4H           → CH3-CH3 + H2O Conc. HCl OR

Zn / Hg

CH3COCH3+ 4H           → CH3CH2CH3+H2O  Conc. HCl

[ISC Marking Scheme, 2014]

Examiner’s Comment For Clemmensen’s reduction, the correct condition for the reaction was not given by many candidates.

OR Conc. HNO3 /

NaOH +CaO Sn / HCl             → C6H5NO2  (b) (i) C6H5COOH             → C 6H 5          → C6H5NH2 Conc.H SO ∆ 2

4

[ISC Marking Scheme, 2016]

Examiner’s Comment The conversion of benzoic acid to aniline was answered correctly. However, some candidates failed to write the conditions for the reaction.

(ii) CH3Cl

2Na + CH3 Cl

C2 H6 +2NaCl +Cl2 hv

C2H5Cl + HCl

+aq. KOH C2 H5OH +KCl +O2 K2Cr2O7 + H2 SO4

CH3COOH + H2O Acetic acid

CH3

OSWAAL ISC Sample Question Papers, CHEMISTRY, Class - XII

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+ CH3Cl

(iii)

Benzene

+ HCl

Touene

CH3

CH3 + CH3Cl 2 Benzene

+ 3O2 Touene

+Coacetate, HCl Mn acetate 443 k and pressure COOH

CH3 2

+ 3O2

Coacetate, Mn acetate 443 k and pressure COOH

+ 2H2 O

Benzoic acid

2 + 2H2 O CH3COOH NaOH CH3COONa + H2 O Benzoic acid

C —O H— Acetaldehyde

CH3MgI H3C—

OMgI C H— CH3

—

—

H C—

3 (v) C —O H— Acetaldehyde

CH3MgI H3C—

OMgI C H— CH3

— + OH H , H2O H3C C — H CH3



—

H3C—

—

CaO

—

CH4 + Na2 CO3

Mathane Sodium carbonate

—

(iv)

2



Propan-2-ol

— + OH H , H2O H3C C H— CH3 —

—



Propan-2-ol





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