Std. XII (CBSE)− Chemistry : Solution to Prelim Test - 1 (Full Length Test)
SECTION – A 1. Reaction taking place at cathode when the battery is in use:
PbO2(s) + SO4(aq)2− + 4H(aq)+ + 2e− PbSO4(s) + 2H O2 ( )
2. 2F 3. Molarity = 38 1.294 1000 5.02 M 98 100 = 4. Ascorbic acid. 5. At anode : O2(g) At cathode : H2(g) 6. Negative 7. Nucleotide 8. Amino acids 9. Iodoform test. 10. Benzaldehyde 11. (D)
−NO2 group is a deactivating and m-directing group so it directs the incoming Br− ion to the m-position
thereby leading to the formation of m-bromonitrobenzene. Upon reduction with Sn/HCl, the −NO2 group
reduces to −NH2 i.e., the final product is m-bromoaniline.
12. (A)
[Co(NH3)3Cl3] on ionisation does not give Cl− ion, hence it does not give a precipitate of AgCl with
AgNO3 solution.
13. (A)
Mn2+ with electronic configuration 3d54s0 has 5 unpaired electrons.
Spin-only magnetic moment () = n(n+2) For Mn2+ : 3d5 4s0
= 5(5+2) = 5.916 = 5.92 B.M.
This is the calculated value of s. Experimental for Mn2+ is 5.96. While the calculated for
Fe2+ = 4.90 B.M. ; V2+ = 3.87 B.M. ; Cu2+ = 1.73 B.M.
14. (A) 15. (B) 16. (C) 17. (A) 18. (D) 19. (D) 20. (D) SECTION – B
21. Lone pairs : 2 [1M]
(2) Vidyalankar : Std. XII (CBSE) − Chemistry
22. Since Rate = k [A] [B]2
(i) When the concentration of B is increased to 3 times, then rate would be Rate 1 = k [A] [3B]2 = 9 k [A] [B]2 = 9 Rate
Rate is increased 9 times. [1M]
(ii) When the concentration of A as well as B are doubled, then rate would be Rate 2 = k [2A] [2B]2 = 8 k [A] [B]2 = 8 Rate
Rate is increased 8 times [1M]
23. When there is dissociation of solute into ions, in dilute solutions (ignoring interionic attractions) the
number of particles increases. As the value of colligative properties depends on the number of particles of the solute, the experimentally observed value of colligative property will be higher than the true value, therefore the experimentally determined (observed) molar mass is always lower than the true value. [1M] For KCl(electrolyte) the experimentally determined molar mass is always lower than the true value when water is used as solvent. Glucose (non-electrolyte) does not show a large variation from the true value.
[1M]
24. (a) CH3−CH(Cl)−COOH [1M]
(b) C6H5CHO [1M]
25. Complexes in which metal is bound to only one kind of donor groups are known as homoleptic
complexes.
e.g. [Co(NH3)6]3+. [1M]
27. In S 1 reaction, carbocation intermediate is formed. In allylic and benzylic halides those carbocations N are stabilised by resonance. So, they show high reactivity towards S 1 reaction. N [1M]
28. Molar mass of C6H12O6 = 12 6 + 1 12 + 16 6 = 72 + 12 + 96 = 180 g mol−1
Molar mass of C12H22O11 = 12 12 + 1 22 + 16 11 = 144 + 22 + 176 = 342 g mol−1
Total number of moles = 18 68.4
180+ 342 = 0.1 + 0.2 = 0.3 (nB) [1M] Tf = Kf m [½M] Tf = B f A n 1000 K W Tf = 1.86 0.3 10000 200 [½M] = 1.86 0.3 5 = 2.79 K [½M]
Hence, freezing point of aqueous solution = 273 − 2.79 = 270.21 K [½M]
29. When reaction is 99.9 % completed
[R]t = [R]0 − 0.999 [R]0 = 0.001 [R0] [½M] k = 0 t [R] 2.303 log t [R] k = 0 0 [R] 2.303 log t 0.001[R] [½M] or, t = 2.303log 103 k t = 2.303 3 k … (i) [½M] For half-life, t1/2 = 0.693 k … (ii) [½M]
Dividing (i) by (ii) 1/2 t t = k 0.693 k 1/2 t t 10 or t 10.t1/2 [1M] OR
(a) The formula of rate constant for first order reaction is
k = 0 t [A] 2.303 log t [A] [½M] k1 = 2 1 2 1 2.303 (1.6 10 ) mol L log 300 s (0.8 10 ) mol L − − − − = 2.3 10−3 s−1 [½M] Similarly, k2 = 2 1 2 1 2.303 (1.6 10 ) mol L log 600 s (0.4 10 ) mol L − − − − = 2.3 10 −3 s−1 [½M]
Unit and magnitude of rate constant shows the given reaction is of first order. [½M]
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30. Formula : M WRT V =
Here, W = ?, M = 342 g mol−1, = 4.82 atm : T = 293 K, V = 1 L, R = 0.0821 L atm mol−1 K−1 So, W = 342 4.82 1 1648.44 68.52 g
0.0821 293 24.055
= =
[1M]
Mass the solution = 1015 g
So, Mass of the solvent = 1015 g − 68.52 g = 946.48 g [1/2M]
Molality of the solution =
1 68.52 g 1000 946.48 g 342 g mol− [1/2M] = 68520 0.211 molal. 323696.16 = [1M]
31. (a) The stability of the highest oxidation state (+ 5) decreases as we move down in group 15. The
oxidation state (+ 5) in Bi is less stable due to inert pair effect. So, bismuth in pentavalent state is a
strong oxidising agent. [1M]
(b) X–X’ bond in inter-halogens is weaker than X–X bond in halogens except F–F bond. [1M] (c) Due to the ease with which it liberates atoms of nascent oxygen. [1M]
32. Mechanism : The formation of ether is nucleophilic bimolecular reaction.
Step I : [1M] Step II : [1M] Step III : [1M] 33. (a) [1M] (b) [1M] (c) [1M] OR
(a) Benzoic acid to benzamide
(b) Methanol to ethanol [1M] (c) Ethanol to acetone [1M] 34. (a) (i) … [1 M] (ii) [1 M]
(b) In substituted phenols, the presence of electron releasing group decreases the acidic strength of phenol. This effect is more pronounced when such a group is present at ortho and para-positions. It is due to the increased negative charge in phenoxide ion. Thus m-aminophenol is stronger acid than
o-aminophenol. [1 M]
35. (a) Kohlrauschs law states that at infinite dilution, when the dissociation of electrolyte is complete each
ion makes a definite contribution towards the molar conductivity of electrolyte, irrespective of the nature of other ion with which it is associated. [1M]
= + m + + − −
(b) The given cell reaction is
3Sn4+
(aq) + 6e− → 3Sn2+(aq)
2Cr(s) → 2Cr3+ + 6e−
3Sn4+ + 2Cr → 3Sn2+ + 2Cr3+
Number of electrons transferred = 6, E0 = 0.89 V
Using formula rG0 = −nFE0
= − 6 96500 C 0.89 V
= − 515310 J = − 515.31 kJ [1M]
(c) When the solution of a weak electrolyte is diluted, its degree of ionisation increases. There is an increase in the number of ions in the solution. Hence the molar conductivity of a weak electrolyte increases on dilution. Electrolytic conductivity, = 3.72 10−2 S cm−1
(6) Vidyalankar : Std. XII (CBSE) − Chemistry = 349.6 + 40.9 = 390.5 S cm2 mol−1 [½M] = m 1000 c [½M] = m 5 1 3 1 1 8.0 10 S cm 1000 cm L 0.0024 mol L − − − − = 33.33 S cm2 mol−1 [½M] = mo m [½M] = 2 1 2 1 33.33 S cm mol 390.5 S cm mol − − = 0.085 [½M]
(b) Electrolyte B is a strong electrolyte.
Limiting molar conductivity increases only to a smaller extent for a strong electrolyte, as on dilution the interionic interactions are overcome. Limiting molar conductivity increases to a larger extent for a weak electrolyte, as on dilution the degree of dissociation increases, therefore the number of ions in
total volume of solution increases. [2M]
36. OR (a) (i) [1M] (ii) [1M] (iii) C6H5N2Cl + CH3CH2OH → + N2 + HCl + CH3CHO [1M] A B C D
CONH2
Benzamide
NH2
Aniline
(b) Aniline undergoes isocyanide test (carbylamine reaction) whereas N, N-dimethylaniline does not.
[1M]
(c) Stronger the base, lower will be the pKb value.
b
2 5 2 6 5 3 6 5 2
pK 3.29 4.63 9.38
C H NH C H NHCH C H NH [1M]
37. (a) (i) Transition metals form complex compounds because of comparatively smaller size of their metal
ions, high ionic charge and availability of vacant d-orbitals. [1M] (ii) As oxidation number of transition metals in their oxides increases, their acidic character increases due to lack of d-electrons. In the lowest oxide, oxidation state is lower and hence, it becomes basic and in highest oxide, oxidation state is higher and therefore, it becomes acidic or
amphoteric. [1M]
(iii) Much larger third ionization energy of Mn (where the required change is d5 to d4) is mainly
responsible for this. [1M]
(b) (a) (i) Carbylamine Reaction : Aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide form isocyanides or carbylamines which are foul smelling substances. Secondary and tertiary amines do not show this reaction. This reaction is carbylamine reaction or isocyanide test and is used as a test for primary amines.
2 3 Heat 2
Carbylamine (foul smell)
R−NH +CHCl +3KOH ⎯⎯⎯→ R−NC + 3KCl + 3H O [1M]
(ii) Hofmanns bromamide reaction: When a primary amide is heated with an aqueous or ethanolic solution of NaOH or KOH and bromine (i.e., NaOBr or KOBr), it gives a primary amine with one carbon atom less.
2 2 2 2 3 2
Acid amide 1 amine
R CONH Br 4NaOH R NH Na CO 2NaBr 2H O
− + + ⎯⎯→ − + + +
e.g. + Br2 + 4KOH ⎯⎯→ + K2CO3 + 2KBr + 2H2O [1M]
OR
(a) (i) Ti4+ has highest oxidation state among the given ions. Ti4+ has stable inert gas configuration and
hence, most stable in aqueous solution.
On the other hand, V2+, Mn3+, Cr3+ have unstable electronic configuration and hence, are less
stable. [1M]
(ii) Due to presence of highest oxidation state of Ti, it acts as the strongest oxidising agent among
the given ions. [1M]
(iii) Due to absence of unpaired electron in Ti4+, it is a colourless ion.
Electronic configuration of Ti4+ = [Ar]3d04s0 [1M]
(b) (i) A : CH3CH2CONH2 [1/2M]
B : CH3CH2NH2 [1/2M]
(ii) A : CH3CH2CH2NH2 [1/2M]
B : CH3CH2CH2OH [1/2M]
