Oswal 36 Sample Question Papers ISC Class 12 Chemistry Solutions
Section-A
Answer 1.
- A. (i) Donor, acceptor
- (ii) dehydration, alkenes
- (iii) rochelle salt
- (iv) isoelectric point
- B. (i) (c) Nearly same atomic size
- (ii) (d) Tetrahedral and diamagnetic
- (iii) (b) Unpaired electrons
- (iv) (a) Carboxylic acid
- C. (i) (c)
- (ii) (a)
- (iii) (b)
- (iv) (d)
The unit of rate constant for second order reaction is mol–1 L s–1 and the rate of second order reaction depends on initial concentration of reactant. Thus, both assertion and reason are true but reason is not the correct explanation for assertion.
(ii) (a) Both assertion and reason are true and reason is the correct explanation of assertion.
Liquids and solids exhibit practically no change of solubility with changes in pressure. Gases as might be expected, increase in solubility with an increase in pressure. Thus both assertion and reason are true and reason is the correct explanation of assertion.
Section-B
Answer 2.
- Given:
- Vapour pressure of pure CCl4 at 25°C (PCCl4 ° ) = 115.0 torr
- Vapour pressure of pure SnCl4 at 25°C (PSnCl4 ° ) = 238.0 torr
- Molar mass of CCl4 [M(CCl4)] = 154 g mol–1
- Molar mass of SnCl4 [M(SnCl4)] = 170 g mol–1
- Mass of CCl4 in solution (wCCl4) = 10 g
- Mass of SnCl4 in solution (wSnCl4) = 15 g
- From Dalton's law of partial pressure:
$$\text{And}\space \text{P}_{total}=X_{CCl_4}-P\degree_{CCl_4}+X_{SnCl_4}-P\degree_{SnCl_4}\\\qquad\text{And}\space X=\frac{N_{multiple}}{n_{total}}=\frac{w}{M×n_{total}}\\\qquad\qquad\text{Now},\space\space n=\bigg[\frac{w}{M}\bigg]_{CCl_4}=\frac{10}{154}=0.065\\\text{and}\space \bigg[\frac{w}{m}\bigg]_{SnCl_4}=\frac{15}{170}=0.09$$
Answer 3.
The general electronic configuration of d-block elements can be written as (n – 1)d1–10 ns1–2. Where n is the
valence shell and (n – 1) is penultimate shell.
Transition elements show variable state oxidation in their compounds because their valence electrons are
in two different sets of orbitals, i.e., (n – 1)d and ns and there is a very small energy difference in between
(n – 1)d and ns orbitals. As a result, electrons of (n – 1)d orbitals as well as ns-orbitals take part in bond
formation. Thus, transition elements have variable oxidation states.
Answer 4.
Answer 5.
- (i) Potassium cyanide (KCN).
- (ii) 4-Nitroaniline < Aniline < 4-Methyl aniline.
Answer 6.
$$\text{we know that }:\\\Delta G\degree=\Delta G\degree_f(\text{Products})-\Delta G_f(\text{Reactants})\\=(-154.0 \text{kJ} + 0)-(0 + 64.4 \text{kJ})=-218.4 \text{kJ}\\\Delta G\degree=nFE\degree\\\text{Since, the reaction involves a 2-electron change, n = 2}\\\text{E}\degree=-\frac{\Delta G \degree}{nF}=-218.4\text{kJ}\space\text{mol}^{-1}/2×96485 \text{C\space mol}^{-1}\\= 1.13 V\\\text{Hence, the standard e.m.f. of the cell is 1.13 V.}$$
Answer 7.
- Since there are no unpaired electrons,
- complex is diamagnetic.
- The complex has two unpaired electrons, therefore, it will be paramagnetic.
Answer 8.
- (i) (a) When sodium phenoxide is heated with CO2 at 400 K and at 4-7 atm pressure, sodium salicylate is formed as the major product, which on acidification yields salicylic acid. This reaction is known as Kolbe’s reaction.
- (b) The preparation of acetophenone from phenol is a two-step process. In the first step, phenol is converted into benzene in the presence of zinc under heating. In the second step, the benzene formed in the first step reacts with acetic anhydride in the presence of aluminium chloride and forms the desired product acetophenone.
OR
- (ii) Alcohols undergo dehydration (removal of a molecule of water) on treating them with protonic acids(conc. H2SO4 or H3PO4) or catalysts such as anhydrous zinc chloride or alumina to form alkenes. For example, tert-butanol undergoes dehydration when it is treated with 20% H3PO4 at 358 K :
- The relative case of dehydration of alcohols follows the order: Tertiary alcohol > Secondary alcohol > Primary alcohol.
Answer 9.
- For the first order reaction,
$$\text{t =}\frac{2.303}{k}\space\text{log}\frac{[\text{A}]_0}{[\text{A}]}$$
$$\text{Let initial concentration,}[A_0]=a\\\text{For}\space 99\%\space\text{completion of reaction, [A]}\\a-\frac{a×99}{100}=0.01\space a\\\qquad t_{(99\%)}=\frac{2.303}{k}\text{log}\frac{a}{0.01a}\\=\frac{2.303}{k}\text{log}100=\frac{2.303×2}{k}\space ….(i)\\\text{For} 90\% \text{completion of reaction,}\\\begin{bmatrix}\text{A}\end{bmatrix}=a-\frac{a×90}{100}=0.1a\\t_{(90\%)}=\frac{2.303}{k}\text{log}\frac{a}{0.1a}\\\frac{2.303}{k}×1\space ….(ii)\\\text{Dividing equation (i) by (ii),}\\\frac{t_{(99\%)}}{t_{(90\%)}}=2\\\text{or}\space t_{(99\%)}=2×t_{(90\%)}\\\text{Hence, it is proved that time required for}\space99\%\space \text{completion is twice the time required for the completion of 90} \%\space\text{reaction}.$$
Answer 10.
- (i) Ethanal and propanal undergo cross aldol condensation when heated in presence of sodium hydroxide :
- (ii) FCH_{2}COOH < CHCl_{2}COOH < NC–CH_{2}COOH < CF_{3}COOH
Answer 11.
- With the successive filling of the inner orbitals, 4f, there is a gradual decrease in the atomic and the ionic sizes of inner transition elements along the series. This is known as ‘lanthanoid contraction’. This phenomenon can be attributed to the imperfect shielding of one electron by another in the same sub-shell.
- Causes of lanthanide contraction: The lanthanide contraction is due to the imperfect shielding of one 4f electron by another in the same sub shell. As we move along the lanthanide series, the nuclear charge and the number of 4f electrons increase by one unit at each step. However, due to imperfect shielding, the effective nuclear charge increases causing a contraction in electron cloud of 4f-subshell.
Section-C
Answer 12.
$$\text{For 0.01 M KCl:}\\\text{Cell constant = Conductivity × Resistance}\\= 1.29 × 100 = 129m^{\normalsize–1}\\\text{For 0.02 M KCl:}\\\text{Cell constant = Conductivity × Resistance}\\129 = \text{Conductivity} × 520\\\therefore\space \text{Conductivity =}0.248 \text{S m}^{\normalsize–1}\\=2.48 × 10^{\normalsize–3} \text{S cm}^{\normalsize–1}\\\text{Also Molar conductivity = Conductivity}×\frac{1000}{\text{M}_{(mole/1)}}\\=\frac{2.48×10^{-3}×1000}{0.02}=124 \text{S cm}^2 \text{mol}^{\normalsize–1}\\\text{or}\space\text{Molar conductivity = Conductivity (S m}^{\normalsize–1})×\frac{1}{\text{M}_{(mole/m^{3})}}\\=\frac{0 248 × 1}{0 02 × 10^{3}}=12.4 × 10^{\normalsize–3}\\= 124 × 10^{\normalsize–4}\text{S m}^2 \text{mole}^{\normalsize–1}\\\text{M}=0.02\frac{\text{mole}}{\text{litre}}=0.02 × \frac{\text{mole}}{m^3}\\\therefore\space =\frac{k×1000}{\text{N}}=\frac{1.11×10^{−2} ×1000}{0.1}\\= 111 \text{S cm}^2 eq^{\normalsize–1} = 111 × 10^{\normalsize –4} \text{S m}^2 \text{eq}^{\normalsize –1}\\\lambda=\frac{k_{(sm^{-1})}}{N_{(eq/m^3)}}=\frac{1.11}{0.1×10^3}\\= 111 × 10^{\normalsize–4} \text{S m}^2 \text{eq}^{\normalsize–1}$$
Answer 13.
The products will be as follows:
Answer 14.
- (i) (a) D-glucose reacts with hydroxylamine to form oxime.
- (b) D-glucose reacts with acetic anhydride to give penta-acetate.
- (ii)
Globular Protein | Fibrous Protein |
1. Globular proteins have almost spheroidal shape due to folding of the polypeptide chain. | Polypeptide chains of fibrous proteins consist of thread like molecules which tend to lie side by side to form fibres. |
2. Globular proteins are soluble in water. | Fibrous proteins are insoluble in water. |
3. Globular proteins are sensitive to small changes of temperature and pH. Therefore, they undergo denaturation on heating or on treatment with acids/bases. | Fibrous proteins are stable to moderate changes of temperature and pH. |
4. They possess biological activity thats why they act as enzymes. Example: Maltase, invertase, etc., hormones (insulin) antibodies, transport agents. | They do not have any biological activity but serve as chief structural material of animal tissues. Example: Keratin in skin, hair, nails and wool. |
OR
- The following reactions cannot be explained by the open chain structure of glucose.
- 1. Despite of having aldehyde group, glucose does not gives 2, 4-DNP test, Schiff’s test and it does not forms the hydrogen sulphite addition product with NaHSO_{3}.
- 2. The pentacetate of glucose does not reacts with hydroxyl amine indicating the absence of free – CHO group.
- 3. D-glucose on treatment with methyl alcohol in the presence of dry HCl gas gives two isomers, methyl α–D glucoside and methyl β-D glucoside. These glucosides do not reduces Fehling’s solution and also do not react with hydrogen cyanide indicating the absence of free-CHO group. A ring structure called pyranose structure (α- and β) is proposed for glucose molecule.
Answer 15.
- The plot of ln k vs 1/T gives a straight line according to the equation given below :
- ln k = – E_{a}/RT + ln A
- Activation energy can be calculated by the value of slope (= – E_{a}/R) of the graph.
- Given, Slope = – 5841 K and R = 8.314 JK^{– 1} mol^{–1}
- Putting the values :
- Slope = – E_{a}/R
- – E_{a} = – 5841 K × 8.314 JK^{– 1} mol^{–1}
- = 48562 J mol^{–1}
- Hence, the energy of activation for the reaction is 48562 J mol^{–1.}
Answer 16.
- (i)
- (ii) The resonating structures of phenol can be shown as :
- 1. The carboxyl group (—COOH) imparts an acidic character to carboxylic acids.
- 2. A carboxyl group is made of —OH group bonded to a carbonyl group.
- 3. In aqueous solution, the H atom in OH of carboxyl group dissociates as proton and carboxdylate ion is formed as the conjugate base.
- 4. Carboxylate ion is resonance stablised by two equivalent resonance structures as shown below:
- 5. Carboxylate ion has two resonance structures (i) and (ii) and both of them are equivalent to each other.
- 6. This gives good resonance stabilisation to carboxylate ion, which in turn gives an acidic character to carboxylic acids.
Answer 17.
Answer 18.
$$\text{(i) Rate of reaction = Rate of disappearance of A = –}\frac{1}{2}\frac{\Delta[\text{A}]}{\Delta t}\\\text{Rate}=-\frac{1}{2}\frac{[A_2]-[A_1]}{t}=-\frac{1}{2}\frac{0.6-0.7}{10}\\\therefore\space\text{Rate}=-\frac{1}{2}×\frac{-(0.1)}{10}=\frac{(0.1)}{20}=\frac{1}{200}=0.005 = 5 × 10^{-3} \text{mol lit}^{\normalsize–1} \text{m}^{\normalsize–1}\\\text{(ii)}\space \text{K}=\frac{0.693}{t_{12}}\\k=\frac{0.693}{6.93s}\\k=0.1s^{-1}$$
- (iii) Since, the reaction is first order reaction and the rate of first order reaction is directly proportional to the concentration of reactants the rate of reaction will also double.
Section-D
Answer 19.
- (i) (a) Freons are chlorofluorocarbon compounds of methane and ethane.
- (b) Freon 12 (CCl_{2}F_{2}).
- c) Freon’s chemical components degrade the ozone layer as it passes through the air. Depletion of the ozone layer increases the amount of ultraviolet radiation reaching the earth’s surface, providing a serious health risk to people.
- (ii) (a) Glucose is also known as dextrose and is an aldohexose.
- (b) The secondary structure of protein refers to the shape in which a long polypeptide chain can exist. They are found to exist in two different types of structures namely, a-helix and b-pleated sheet structure.
Answer 20.
- (i) (a) Both shows coordination isomerism because both cationic and anionic entities and isomers differ in the distribution of ligands in the coordination entity of cationic and anionic part.
- (b) In [Ni(H_{2}O)_{6}]^{2+} Ni is in +2 oxidation state with electronic configuration 3d^{8}. In the presence of weak ligand H_{2}O the two unpaired electrons do not pair up and hence the complex has two unpaired electrons. Therefore, it is coloured and shows d-d transitions which absorbs red light and emits green complimentary light. In case of [Ni)CN)_{4}]^{2–} Ni also shows +2 oxidation state but CN ligand is strong ligand and two unpaired electrons undergo pairing, as a result, no d-d occurs. So, it is colourless.
- (c) Pentaamminecarbonatocobalt (III) chloride.
- (ii) (a) Hexaamminecobalt(III) ion
- (b) In [Co(NH_{3})_{6}]_{3+} Co is in +3 state and has configuration 3d^{6}. In the presence of NH3, 3d electrons pair up leaving two d-orbitals empty. Therefore, the hybridisation of the complex is d^{2}sp^{3}. Hence, it is inner orbital complex.
- (c) Primary valency is zero because there is no ion outside coordination sphere and secondary valency is 6 because there are six monodentate ligand attached with cobalt metal ion.
- (d) Octahedral geometry.
Answer 21.
- (i) The cell can be represented as :
$$\text{Mg}|\text{Mg}^{2+}(0.001 \text{M})||\text{Cu}^{2+}(0.01\text{M})|\text{Cu}\\E_{\text{cell}}=E\degree_{\text{cell}}-\frac{0.059}{n}\text{logkc}\\\text{Putting the given values,}\\\text{E}_{\text{cell}}=\text{E}\degree_{\text{cell}}-\frac{0.059}{2}\text{log}\frac{10^{-3}}{10^{-2}}\\= 2.71 + 0.0295\\\text{E}_{\text{cell}}=2.7395V$$
- (a) When an external opposite potential is applied less than 2.71 V the direction of flow of current would remain same.
- (b) When an external opposite potential is applied more than 2.71 V the direction of flow of current would be reversed.
- (ii) Reactions which will occur in the solution are:
$$\text{CuCl}_2(s) \xrightarrow{} \text{Cu}^{2+} + 2\text{Cl}^ {\normalsize {–}}\\H_2O \xrightarrow{} H^{+} + OH^{\normalsize–}\\\text{Reaction at cathode :}\\Cu^{2+} + 2e^{\normalsize–} → Cu(s)\\\text{Because}\space\space E° Cu^{2+}/Cu \text{\textgreater}E° H^+/H_2\\\text{Reaction at anode :}\\2Cl^{\normalsize–}\xrightarrow{} Cl_2 + 2e^{\normalsize–}$$
- This reaction occurs at the anode due to over-potential of O_{2} oxidation of Cl° is preferred. Hence, Cu will deposit at cathode and Cl_{2} gas will be liberated at anode.
OR
- (i) (a) Molar conductivity is defined as the conducting power of all the ions produced by dissolving one mole of an electrolyte in solution.
- Λ_{m} = K/C
- (b) Secondary batteries are those batteries which can be recharged by passing electric current through them and hence can be used over again e.g., Lead storage battery.
- (ii) (a) The cell can be represented as:
- Hence, the emf of the cell is 1.42 V.
$$\text{Al}\bigg|\text{Al}^{3+}(0.01 \text{M})\bigg|\bigg| \text{Ni}^{2+}(0.1)\bigg|\text{Ni}\\\text{Cell potential is given by the following equation in this case:}\\\text{E}_{cell}=\text{E}\degree_{cell}=\frac{0.059}{6}\text{log}\frac{\begin{bmatrix}Al^{+3}\end{bmatrix}^2}{\begin{bmatrix}Ni^{+2}\end{bmatrix}^3}\\\text{Given:}\\\text{E\degree cell = 1.41 V, concentration of Al}^{3+} \text{ions is 0.01 M and Ni}^{2+} \text{ions is 0.1 M}\\\text{Putting the values in equation}\\E_{\text{cell}}=1.41V-\frac{0.059}{6}\text{log}\frac{[0.01]^2}{[0.1]^3}\\= 1.4198 \text{V}\\= 1.42 \text{V}$$
$$\Lambda\degree_m=\text{v}_{\normalsize+}\lambda\degree_{\normalsize +}+\text{v}_{\normalsize -}\lambda\degree_{\normalsize -}$$
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