Oswal Practice Papers CBSE Class 12 Chemistry Solutions (Practice Paper - 8)

Section-A

1. (c) sec^–1

Explanation :

For first order,
n = 1
K = (mol)^1–n L^{n – 1} s^–1
So, K = (mol)^1–1 L^{1 – 1} s^–1
K = s^–1

2.(d) Alkaline KMnO₄

Explanation :

Baeyer’s reagent is an alkaline KMnO₄ solution. This shows a redox reaction because Baeyer’s reagent is an alkaline solution of cold potassium permanganate, which is a potent oxidant. The colour of an organic material fades from purplish-pink to brown as it reacts with double or triple bonds.

3. (a) 1-(r), 2-(p), 3-(q)

Explanation :

The electrolyte used in dry cell, nickel-cadmium cell and leads storage cell is zinc chloride, potassium hydroxide and aqueous H2SO₄ respectively.

4. (a) I, II, III

Explanation :

D-configuration of the compounds is always compared to the structure of (+) glycerides.
Among the given options I, II, and III shows D-configuration with -OH group on the right side of asymmetric carbon.

5. (d)
$$\frac{-E_a}{2.303R}$$

Explanation :

Rate constant of any reaction is given by the formula,

$$k=Ae^{-E}a/^{RT}\\\text{Taking log on both side:}\\log k=log A–\frac{E_a}{2.303R}\\\text{Therefore, slope of log k vs 1/T graph is given by}\frac{-E_a}{2.303R}$$

6. (c) Phenyl isocyanide

Explanation :

When aniline reacts with chloroform and alcoholic KOH it gives an offensive smelling liquid i.e., phenyl isocyanide as product. It is called as isocyanide test. It is given by aliphatic and aromatic primary amines.

7. (c) Gluconic acid

Explanation :

On oxidation with a mild oxidising agent like Br₂/H₂O, the glucose is oxidised to six carbon carboxylic acid (gluconic acid). This indicates that the carbonyl group is present as an aldehydic group.

8. (c) CHCl₂

Explanation :

In the reaction of phenol with CHCl₃ and aqueous NaOH at 70°C, the electrophile attacking the ring is CCl₂. This is known as Reimer Tiemann Reaction.
Reimer Tiemann Reaction makes ortho substituted phenol:

9. (b) Ethylene, C₂H₄

Explanation :

When ethanol is heated with conc. H₂SO₄ at 443K, the product obtained is ethylene.
$$\underset{Ethyl alcohol}{CH_3CH_2OH}\xrightarrow[443K]{Conc.H_2So_4}\underset{Ethene}{CH_2=CH_2}+\underset{Water}{H_2O}$$

10. (b) 32 times

Explanation :

For every 10°C rise of temperature, the rate is doubled. Thus, the temperature coefficient of the reaction = 2
When temperature is increased by 50°C, rate becomes = 2^(50/10) = 2⁵ times = 32 times

11. (a) 1.0 × 10¹⁰

Explanation :

For a cell reaction in equilibrium at 298K.
$$E_{cell}°=\frac{0.0591}{n}\space log\space K_c\\⇒ log K_c=\frac{E_{cell}×n}{0.0591}\\\text{Substitute the given values of}E_{cell}°\text{and n in the above equation, we get}\\log K_c=\frac{0.295×n}{0.0591}\\log K_c=10\\⇒log K_c=1×10^{10}.$$

12. (b) 1° > 2° > 3°

Explanation :

Esterification is the process in which an organic acid (RCOOH) combines with an alcohol (ROH) to form a fruity smell product ester (RCOOR) and water. The relative order of esterification of alcohols is 1° > 2° > 3°. This is because as the steric hinderance (or bulkiness) increases from primary to secondary to tertiary alcohol, the removal of H+ will be difficult, hence, the order of esterification decreases.
Question No. 13 to 16 consist of two statements – Assertion (A) and Reason (R). Answer these questions selecting the appropriate option given below:
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.

13. (c) A is true, but R is false.

Explanation :

KCl undergoes dissociation in solution, hence, observed molar mass will be lower. Experimentally, determined molar mass can be higher or lower depending upon whether solute undergoes dissociation or association. Hence, assertion is true but reason is false

14. (c) A is true, but R is false.

Explanation :

Rate of reaction does not remain constant during the complete reaction because rate depends upon the concentration of reactants which decreases with time.

15. (a) Both A and R are true and R is the correct explanation of A.

Explanation :

Many trivalent lanthanoid ions are coloured both in the solid state and in aqueous solutions and colour of these ions is due to the presence of f-electrons. Thus, both assertion and reason are correct statements and reason is the correct explanation of assertion.

16. (b) Both A and R are true and R is not the correct explanation of A.

Explanation :

Alcohol which forms the more stable carbocation undergoes dehydration more readily. Since tert-butyl alcohol forms more stable tert-butyl cation, therefore, it undergoes dehydration most readily than propanol. Primary (1°) and secondary (2°) alcohols give Victor-Meyer’s test but not tertiary (3°) alcohol. Hence, both assertion and reason are correct but reason is not the correct explanation for assertion.

Section-B

17. (a) A solution of water and methanol is minimum boiling azeotropic mixture.
(b) De-icing is a process of removal of snow, ice and frost from surfaces such as roads using de-icing agents. De-icing agents such as common salt decreases the freezing point of ice and eventually melts the ice deposited and clears the surface.

W_B = 30 g M_B = 60 g mol^–1
W_A = 846 g M_A = 18 g mol^–1
P^° = 23.8 mmHg
P_s = x
Relative lowering of vapour pressure

$$\frac{p°-P_s}{p°}=\frac{W_B×M_A}{M_B×W_A}\\\frac{23.8-x}{23.8}=\frac{30×18}{60×846}\\23.8–x=0.253\\x=23.8–0.253=23.547\\\text{So, vapour pressure of water for this solution = 23.547 mmHg}$$

18. (a)

will undergo S_N ¹ reaction faster due to stable carbocation (3°carbocation).
(b) Boiling point decreases with increasing branching.

19. Glucose does not have open chain structure and hence it does not have a free —CHO group. Actually —CHO group combines with C₅ – OH to form a hemiacetal.
Glucose largely exists in the cyclic hemiacetal form along with a very small amount (0.5%) of the open chain form. Since the concentration of the open chain form is low and its reaction with 2, 4-DNP is reversible, therefore, formation of 2, 4-DNP derivative cannot disturb the equilibrium to generate more of the open chain form from the cyclic hemiacetal form and hence, it does not react with 2, 4-DNP.
The capital letter D in the name D-(+)- glucopyranose indicates that the C₅—OH group is oriented towards right while the sign (+) indicates that glucopyranose is dextrorotatory.

20. (a)

21. (a) In aqueous solution 2° amine is more basic than 3° amine due to the combined effect of the inductive effect, solvation effect and steric factor.
(b) As aniline is an aromatic primary amine on with HNO₂ at Oº–5ºC followed by treatment with an alkaline salution of β–naphthol gives an orange coloured azo dye. But ethylamine does not give this test.

Section-C

22. (a) Non-essential amino acids: The amino acids which can be synthesised in the body, are known as non-essential amino acids. For example, Glycine, Alanine, etc. Essential amino acids: The amino acids which cannot be synthesised in the body and must be obtained through diet are known as essential amino acids. For example, Valine, Leucine, etc.

(b) Nucleoside: A nucleoside contains only two basic components of nucleic acids i.e., a pentose sugar and a nitrogenous base. During their formation 1-position of the pyrimidine or 9-position of the purine moitey is linked to C₁ of the sugar (ribose or deoxyribose) by a β-linkage.

Nucleotides: A nucleotide contains all the three basic components of nucleic acids. i.e., a phosphoric acid group, a pentose sugar and nitrogenous base. These are formed by esterification of C₅′—OH of the sugar of the nucleoside with phosphoric acid.

23. C₄H₈O (A) gives precipitate with 2, 4-DNP, it indicates that compound A contains a carbonyl group. It does not reduce Tollen’s reagent so it is a ketone compound. It gives Iodoform test as it contains a methyl ketone group (–COCH₃). Let us assume tentative structure of A as X—COCH₃. Now, A gives compound B on NaBH4 reduction, so the structure of compound B becomes X-CHOH-CH₃. Compound B further is subjected to dehydration which means the new compound C will contain a double bond. Compound C gives two molecules of ethanal on Ozonolysis, this tells us that compound C is symmetrical and the double bond divides the molecule in two equal parts. Collecting all the above given information, the reaction sequence can be formulated as:

24. (a) Acetamide (CH₃CONH₂) undergoes Hoffmann Bromamide degradation in presence of bromine and NaOH to give Methanamine.
CH₃CONH₂ + Br₂ + 4NaOH → CH₃NH₂ + Na₂CO₃ + 2NaBr + 2H₂O

(b) Aniline reacts with bromine water at room temperature to give 2, 4, 6-tribromoaniline and hydrobromic acid.

(c) Ethylamine being a primary amine forms foul-smelling methyl isocyanide on heating with chloroform and alcoholic potassium hydroxide :
$$\underset{Ethylamine}{CH_3CH_2NH_2}\xrightarrow[CHCl_3]{Alc KOH}\underset{methyl isocyanide}{CH_3NC}$$

25. The given information is :
P₁ (vapour pressure of solvent) = 54.2 mm Hg; Molecular weight of glucose (C₆H₁₂O₆), M₂ = 180 g mol–1; Lowering of vapour pressure (i.e., P₁ – p₁) = 0.23 mm Hg; M₁ = 18 g; w₁ = 100 g.
Relative lowering of vapour pressure can be given by the below formula :
P₁ – p₁ /P₁ = w₂ × M₁/M₂ × w₁
Here, P₁ and p₁ are vapour pressure of solvent and solution respectively, w1 and w₂ are masses and M₁and M₂ are molar masses of the solute and solvent respectively.
Now, putting the values in equation :
0.23 mm Hg/54.2 mm Hg = w₂ × 18 g mol^–1/180 g mol^–1 × 100 g
w₂ = 4.24 g
Hence, weight of glucose is 4.24 g.

26. (a) Unit of specific conductance is S m^–1.
(b) Electrochemical equivalent (ECE) of a substance is the amount of a substance in grams produced or consumed by the passage of one coulomb of electricity in an electrochemical reaction.
(c) A voltmeter is used to measure the electrochemical equivalent of an element.

27. The rate of law may be expressed as
Rate = k[A]^p[B]^q
Comparing experiments 2 and 3
(Rate)₂ = k[0.2]^p [1.0]^q = 8.4 × 10^–3 …(i)
(Rate)₃ = k[0.2]^p [2.0]^q = 8.4 × 10^–3 …(ii)
$$\text{Dividing equation by (ii) by (i)}\\\frac{(Rate)_3}{(Rate)_2}=\frac{k[0.2]^p[2.0]^q}{k[0.2]^p[1.0]^q}=\frac{8.4×10^{-3}}{8.4×10^{-3}}\\{[2]^q=[2]^°orq=0}\\\text{Comparing experiments (i) and (ii)}\\(Rate)_2=k[0.2]^p[1.0]^q=8.4×10^{–3} …(iii)\\(Rate)_1=k[0.10]^p [1.0]^q=2.1×10^{–3}…(iv)\\\text{Dividing equation, (iii) by (iv)}\\\frac{(Rate)_2}{(Rate)_1}=\frac{k[0.20]^p[1.0]^q}{k[0.10]^p[1.0]^q}=\frac{8.4×10^{-3}}{2.7×10^{-3}}= 4\\{[2]^q = [2]^2 or q=2}\\\text{order with respect to A=2}\\\text{order with respect to B = 0}$$

28. (a) For ns⁰ (n – 1)d⁴ configuration i.e., the crystal field splitting energy is less than pairing energy, shows that the ligand must be a weak field ligand. Hence the 4th electron will go to eg level rather than pairing in t_2g level. So configuration will be t_2g ³ e_g ¹.

(b) In [Ni(CO)₄] here Ni has oxidation number 0, i.e., 3d⁸4s² state. Also CO is a strong field ligand.

Either for sp³ or dsp² hybridization for 4 ligands the s-orbital should be empty. The ligand CO forces the electrons to pair and the 4s electrons are transformed to 3d-orbitals.

Inner orbital complex, diamagnetic and tetrahedral shape.

Section-D

29. (a) As during S_N² reaction back-side approach of the nucleophile takes place, thus inversion of configuration occurs.
(b) Neopentyl halide having three alkyl groups at the b-carbon and thus, due to steric hindrance it does not undergo S_N ² reaction.

is a tertiary substrate and also the back side attack on a-carbon is not at all possible because of its cage like structure. So, it does not undergo S_N² reaction.

CH₃CH₂Cl is a primary substrate and back side approach of the nucleophile is more feasible and easy. Hence SN₂ reaction occur here.

30. (a) The different colours are due to the different mode of d-d electron transition
(b) [Fe(CN)₆]^4– is a spin paired complex as there is no unpaired electron in the t_2g level because of the strong field CN ligand..
(c)

In [Fe(CN)₆]^4–, CN is a strong field ligand and Δ₀ is large and thus, it exhibits pale yellow colour.

In [Fe(H₂O)₆]²⁺, H₂O is weak field ligand and Δ₀ is small and thus, it exhibits light green colour.

Section-E

31.

Transition metals forms large number of complexes due to:
- Small size of atoms and ions of transition metals.
- High nuclear charge.
- Presence of incompletely filled d-orbitals.
As the oxidation state increases, the size of ion goes on decreasing thus the covalent character increases as a result of this amphoteric and acidic strength increases. While in case of lower oxides of transition metals ionic size increases and thus basic character increases.
This is because Mn²⁺ has 3d⁵ as a stable oxidation state which is half filled and stable. Mn has very high third ionisation energy for change from d⁵ to d⁴ but in case of Cr³⁺, 3d³ is more stable due to t2g
(crystal field splitting theory) that is why Mn³⁺/Mn²⁺ is highly positive as compared to Cr³⁺/Cr²⁺.
Transition metals show paramagnetic behaviour due to the presence of unpaired electrons in the d-orbitals. Each electron having a magnetic moment associated with its spin angular momentum. However, in the first transition series, the orbital angular momentum is quenched. Therefore, the resulting paramagnetism is only because of the unpaired electron.
Transition elements have large number of d-subshell valence electrons. Therefore, they form very strong metallic bonds. As a result, the enthalpy of atomisation of transition metals is high.
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 that is d-d transition. 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. Hence reflected radiation is visible in the form of coloured solutions.
The catalytic activity of the transition elements can be explained by two basic facts:
- Because of 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 to proceed.
- Transition metals also provide a suitable surface for the reactions to occur, i.e., act as a solid adsorption medium.

32. (a) Chemical equation for the given cell can be represented as :

Zn(s)+2Ag⁺(aq) → Zn²⁺+2Ag(s)

And the Nernst equation can be written as :
$$E_{cell}=E°_{cell}–\frac{RT}{nF}In\frac{[Zn^{+2}]}{[Ag^+]^2}$$

With increase in Zn²⁺ concentration, cell potential drops from its initial value.
(b) Electrolysis of aqueous solution of NaCl yields NaOH, Cl₂ and H₂.
The net reaction can be shown as :
$$NaCl(aq)+H2_O(l)→Na+(aq)+OH–(aq)+½H_2(g)+½Cl_2(g)$$

(c) The half cell reactions are :
$$Ni^{2+}(aq)+2e^–→Ni(s);E^°_{cell}=–0.25V+\xrightarrow[2]{0.0591}log[Ni^{+2}]\\=–0.25V+\xrightarrow[2]{0.0591}log0.01\\=–0.25V+0.03(–2)=–0.31V\\Cu^{2+}(aq)+2e^–→Cu(s);E°_{cell}=+0.34V+\xrightarrow[2]{0.0591}log[Cu^{+2}]\\=+0.34V+\xrightarrow[2]{0.0591}log[0.1]\\=+0.34V+0.03(–1)=+0.33V\\\text{So, the overall cell reaction is :}\\Ni(s)+Cu^{2+}(aq) Ni2+(aq)+Cu(s);\\\text{Now emf of the cell could be given by:}\\E°=0.33V–(0.31 V)=0.64 V$$

(a) Kohlrausch law of independent migration of ions states that limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte. Thus if ^λ° Ca2+ and ^λ° Cl- are limiting molar conductivity of the calcium and chloride
ions respectively, then the limiting molar conductivity for calcium chloride is given by the equation:

^λ° m Ca Cl (CaCl2 ) = ^λ° Ca ²⁺ + ^2λ° Cl⁻

(b) Conductivity always decreases with the dilution, so as the case with given NaCl solution here, This can be explained by the fact that the number of ions per unit volume that carry the current in a solution decreases on dilution.

Molar conductivity of a solution at a given concentration is the conductance of the volume V of the solution containing one mole of electrolyte in a standard volume increases with decrease in concentration, same is the case here with NaCl solution. This is because the total volume, V, of solution containing one mole of electrolyte also increases.

(c) Molar conductivity = L_m = k/c
Where k is conductivity of the solution.
Now, Cell constant = G* = Conductivity × resistance
or Conductivity = Cell constant/resistance
k = 1.29 cm^–1/100 Ω
= 1.29 × 10^–2 S cm^–1
Molar conductivity = Lm = 1.29 × 10^–2 S cm^–1 × 1000 cm³ L^–1 molarity^–1
= 1.29 × 10^–2 S cm^–1 × 1000 cm³ L^–1/0.1 mol L^–1
= 1.29 S cm² mol^–1.

33. From the formula C₈H₇O₂Br seems like a carboxylic acid. Formation of two isomers [B] and [C] of formula C₈H₆O₂Br₂ must have given an ortho and para isomer.

Compound [A] can be any of these I, II or III structures.
As vigorous oxidation of [A] alongwith acid [B] and [C] produces strongest acid [D] from structure [A] hence compound [A] must be structure [I].

[D] is strongest acid because of ortho-effect whereas presence of Br molecule at the meta position weakens the acidic strength.

(a) (i) Fehling’s solution is alkaline solution of CuSO₄ along with some Rochelle salt.
(ii) Ethanol converts Cu(II) of Fehling’s solution to Cu(I) i.e., + 1 state.