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FIITJEE NSEC MOCK TEST-1

Time Allotted: 2 Hours CODE: Maximum Marks: 216

 Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.
 You are not allowed to leave the Examination Hall before the end of the test.

INSTRUCTIONS
Write the question paper code (mentioned above) on YOUR OMR Answer Sheet (in the space provided), otherwise
your Answer Sheet will NOT be evaluated. Note that the same Question Paper Code appears on each page of the
question paper.
Instructions to Candidates:
1. Use of mobile phone, smart watch, and ipad during examination is STRICTLYPROHIBITED.
2. In addition to this question paper, you are given OMR Answer Sheet along with Candidate's copy.
3. On the OMR sheet, make all the entries carefully in the space provided ONLY in BLOCK CAPITALS as well
as by properly darkening the appropriate bubbles.
Incomplete/ incorrect/ carelessly filled information may disqualify your candidature.
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5. Your Enrollment number entered on the OMR Answer Sheet shall remain your login credentials means login
id and password respectively for accessing your performance / result in NSE.
6. Question paper has two parts. In part A1 (Q. No.1 to 48) each question has four alternatives, out of which only
one is correct. Choose the correct alternative (s) and fill the appropriate bubble (s), as shown.
Q. No. 22 a b c d
In part A2 (Q. No. 49 to 60) each question has four alternatives out of which any number of alternative (s) (1, 2,
3, or 4) may be correct. You have to choose all correct alternative(s) and fill the appropriate bubble(s), as shown
Q. No. 54 a b c d
7. For Part A1, each correct answer carries 3 marks whereas 1 mark will be deducted for each wrong answer. In
Part A2, you get 6 marks if all the correct alternatives are marked. No negative marks in this part.
8. Rough work may be done in the space provided. There are 15 printed pages in this paper
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NSEC-MOCK TEST-1-CHEMISTRY-(2023)-2

INDIAN ASSOCIATION OF PHYSICS TEACHERS

NATIONAL STANDARD EXAMINATION IN CHEMISTRY
(NSEC-2023)
Time: 120 minutes Max. Marks: 216
Attempt All Sixty Questions
A–1
ONLY ONE OUT OF FOUR OPTIONS IS CORRECT BUBBLE THE CORRECTION OPTION

1. We have a solution of NH4Cl. What effect will addition of NaCl have on the pH of the solution?
(A) Increase pH (B) Decrease pH
(C) No effect (D) Cannot tell from the given information

2. A sample of 100 ml of 0.10 M acid HA (Ka = 1 × 10–5) is titrated with standard 0.10 M KOH. How many
ml of KOH is required to be added when the pH in the titration flask will be 5.00?
(A) 0 ml (B) 10 ml
(C) 100 ml (D) 50 ml
–12 2
3. Ksp of CoS is 4 × 10 M . Which of the following statements are correct?
(I) Solubility of CoS is 2 × 10–6 m/L.
(II) Solubility of CoS is greater than 2 × 10–6 m/L because of the hydrolysis of S2  .
(III) Solubility of CoS is greater in an acidic solution than that in pure water.
(IV) Solubility of CoS is decreased when it is simultaneously dissolved along with MnS.
(A) I and III (B) I, III and IV
(C) II, III and IV (D) I and IV

4. A weak acid HX has the dissociation constant 1 × 10–5 M. It forms a salt NaX on reaction with alkali.
The degree of hydrolysis of 0.1 M solution of NaX is
(A) 0.0001% (B) 0.01%
(C) 0.1% (D) 0.15%
o
5. Following reaction occurs at 25 C,


2NO(g)(1 105 atm)  Cl2 (g)(1 102 atm)  2 o
 2NOCl(1 10 atm); the value of G is
(A) –45.65 kJ (B) –28.53 kJ
(C) –22.82 kJ (D) –57.06 kJ

6. 

Reaction, A  B 
 C  D, is reversible. If for the reaction, A  B  C  D, G  ve then G for
C  D  A  B, is
(A) +ve
(B) –ve
(C) +ve and –ve both


(D) –ve for those reactions where Kc for the reaction, A  B 
 C  D, is smaller than 1.

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7. One mole of an anhydrous salt AB dissolves in water with the evolution of 21 Jmol–1 of heat. If the
heat of hydration of AB is –29.4 Jmol–1, then the heat involved in the conversion of A+(g) and B–(g)
into AB(s) is
–1 –1
(A) 50.4 Jmol (B) 8.4 Jmol
–1 –1
(C) –50.4 Jmol (D) –8.4 Jmol

8. The heat of combustion of yellow phosphorus and red phosphorus are –9.91 kJ/mol and –8.78 kJ/mol
respectively. The heat of transition from yellow phosphorus to red phosphorus is
(A) –1.13 kJ (B) –18.69 kJ
(C) +18.69 kJ (D) +1.13 kJ

9. A 3 mol sample of a triatomic ideal gas at 300 K is allowed to expand under isobaric adiabatic
condition from 5 L to 40 L. The value of H is
(A) -12.46 KJ (B) -14.965 KJ
(C) - 24.62 KJ (D) - 10.24 KJ

10. The ratio of van’t Hoff factor, if PBr5 and PCl5 are assumed to be soluble in water and remains same
as in solid state, is
(A) 2 (B) 1
(C) 1.5 (D) 2.5

11. K 4 [Fe(CN)6 ] may have the same freezing point at equal concentration of which of the following
solution.
(A) K 3Fe(CN)6 (B) MgSO 4
(C) Ba3 (PO4 )2 (D) NaCl

12. For the given electrolyte AxBy. The degree of dissociation ‘’ can be given as
i 1
(A)  = (B) i  (1   )  x  y
 x  y  1
1 i
(C)   (D) all of these
1  x  y 
13. For the cell reaction,

Cu2  (aq)  Zn(s)  2
 Zn (aq)  Cu(s)
C1 C2
of an electrochemical cell, the change in free energy, G , at a given temperature is a function of
C 
(A) n C1 (B) n  2 
 C1 
(C) n C2 (D) n (C1  C2 )

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NSEC-MOCK TEST-1-CHEMISTRY-(2023)-4

14. Consider following reaction,

Zn  2H  Zn2   H2
half-life period is independent of concentration of Zn at constant pH. At constant Zn concentration,
half-life is 10 minutes at pH = 2 and half-life is 100 minutes at pH = 3. Hence, rate law is
(A) k [Zn] [H+] (B) k [Zn] [H+]2
0 +
(C) k [Zn] [H ] (D) k [Zn]0 [H+]2
o
15. What will be the solubility of molar solution of AgCl at 25 C if specific conductivity of saturated AgCl
solution is 2.3 × 10–6 S.cm–1 at 25oC and molar conductivities of Ag+ and Cl– at infinite dilution are
61.9 and 76.3 Scm2mol–1 respectively?
–3 –5
(A) 2.382 × 10 M (B) 1.66 × 10 M
–8 –6
(C) 5 × 10 M (D) 7.2 × 10 M

16. The Keq. value in HCN addition to the following aldehydes are in the order:
O
O C H CHO

H3CO
(I) (II)
CHO
(III)
(A) I > II > III (B) III > II > I
(C) III > I > II (D) II > I > III

17. In Cannizaro reaction which species is best hydride donor (in conc. Alkaline medium)
(A) O (B) O
H C O H C H
OH
(C) O (D) O
H C H H5C6 C H
O O

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18. CH 2 NH 2
HNO

2
 A, A is
OH
(A) CH 2OH (B) CH2NH2

OH NO
(C) NH2 (D) O

19. O
H2 C C NH2 IV  O
HC C6 H4 NH C CH3
III
H2 C NH CH CH2 NH2 I
II CH3
Arrange the above mentioned nitrogen atoms of the compound in decreasing order of basic nature?
(A) I > II > IV > III (B) II > I > IV > III
(C) II > I > III > IV (D) I > IV > III > II

20. Ag2S  NaCN  Zn  Ag

This method of extraction of Ag by complex formation and then its displacement is called:
(A) Parke’s method (B) Mc – Arthur Forset method
(C) Serpeck method (D) Hall’s method

21. In the extraction of copper the reaction which takes place in Bessemer converter is
(A) 2CuFeS2 + O2  Cu2S + FeS + SO2
(B) 2Cu2O + Cu2S  6Cu+SO2
(C) 2Cu2S+3O2  2Cu2O + 2SO2
(D) 2FeS + 3O2  2FeO + 2SO2

22. pH values of some of the solutions are compared below. Which of the following statements is correct?
(A) NaF < NaCN ; Na2SO3 < Na2TeO3 ; NaNO2 < NaAsO2
(B) NaF < NaCN ; Na2SO3 < Na2TeO3 ; NaNO2 > NaAsO2
(C) NaF > NaCN ; Na2SO3 > Na2TeO3 ; NaNO2 > NaAsO2
(D) NaF > NaCN ; Na2SO3 < Na2TeO3 ; NaNO2 > NaAsO2

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NSEC-MOCK TEST-1-CHEMISTRY-(2023)-6

23. The incorrect statement regarding ClO2 molecule is that

(A) it contains odd number of electrons and hence paramagnetic.
(B) ClO2 dimerises.
(C) the Cl – O bond length is shorter than the single Cl – O bond length.
(D) it is a highly reactive species.

24. Which pair is amphoteric?

(i) Zn(NH2)2 (ii) Al(OH)3
(iii) Zn(OH)2 (iv) Be(OH)2
(A) (ii), (iii), (iv) (B) (i), (ii), (iii)
(C) (i), (ii), (iii), (iv) (D) (ii), (iv)

25. The hydroxide which has lowest value of Ksp at 25oC is

(A) Be(OH)2 (B) Ba(OH)2
(C) Ca(OH)2 (D) Mg(OH)2

26. A compound which can be used in space vehicles both to absorb CO2 and liberate O2 is
(A) NaOH (B) NaO
(C) Na2O2 (D) CaO & NaOH

27. Select the incorrect statement about the compound NO[BF4]

(A) it is a diamagnetic species
(B) it has 5 and 2 bond
(C) B – F bond length in compound is larger than BF3
(D) nitrogen – oxygen bond is longer than in nitric oxide (NO)

28. Which of the following is tetrahedral and paramagnetic?

(I) MnO 4 (II) CrO42
(III) Cr2 O72 (IV) MnO 42
(V) ClO 4 (VI) ClO3
(VII) NO2 (VIII) N2O4
(A) (I), (II) and (III) (B) (IV), (V) and (VI)
(C) (IV) only (D) (V) only

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29. Which of the following order are correct?

I. Acidity order : o-nitrobenzoic acid > p-nitrobenzoic acid > m-nitrobenzoic acid
II. Basicity order : NH2  EtO  OH  RCOO  Cl
III. Heat of hydrogenation : cis-2-butene > trans-2-butene
IV. Ease of decarboxylation :
C6H5COCOOH> Ph C CH2 COOH > Ph CH COOH > Ph CH COOH
O OH NH2
(A) I and II (B) I and III
(C) I and IV (D) I, II and III

30. For the given reaction,

R1 R1
HOH
R C X   R C OH
R2 R2
Which substance will give maximum racemisation?
(A) CH3 (B) CH3

H 5 C6 C Br H2C CH C Br

CH3 CH3
(C) Br (D) Br
H 5C 6 C OCH3 H 5C 6 C NO 2

CH3 NH3

31. A mixture of gas contains N2 and C2H2. 20 ml of this mixture is added to 70 ml of O2 and combustion
is allowed to take place over mercury. After cooling the volume of the gas is found to be 68.5 ml.
when the resultant gas mixture is passed through KOH solution. The volume of the residual gas is
38.5 ml. The percentage composition of N2 in the mixture will be:
(A) 20 (B) 25
(C) 40 (D) 75

32. N2Cl

O O
K L M
  

The reagents/conditions K, L and M are respectively

(A) Ba(OH)2, KCN and CHCl3/NaOH heat, H3O+
(B) H2O (boil), CO2/KOH and Ac2O – AcONa, heat
(C) Steam ; CHCl3/NaOH, heat, H3O+ and Ac2O – AcONa, heat
(D) Cu2Cl2/HCl, NaOH/high P and CO2|NaOH, heat

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NSEC-MOCK TEST-1-CHEMISTRY-(2023)-8

33. A cinema hall has equidistant rows 1m apart. The length of cinema hall is 287m and it has
287 rows. From one side of cinema hall, laughing gas N2O is released and from the other side,
weeping gas (C6H5COCH2Cl) is released. In which rows, spectators will be laughing and weeping
simultaneously?
th th
(A) 100 row from weeping gas side (B) 100 row from laughing gas side
th th
(C) 187 row from the weeping gas side (D) 185 row from the laughing gas side

34. An ionic crystal of AB2 type crystallizes in fluorite style with density 6.45 g ml-1. Edge length of unit cell
o
is 400 A . If all the ions lying on a body diagonal are removed from all the unit cells then what will be
the new density? (B = 19).
-1 -1
(A) 6.45 g ml (B) 5.3 g ml
-1 -1
(C) 3.3 g ml (D) 4.5 g ml

35. Which one of the following statements is incorrect?

(A) Carbon is unaffected by alkalis
(B) Diamond is unaffected by concentrated acids
(C) -Graphite is in the form of ABCABCABC……..
-Graphite is in the form of ABABAB………..
o
(D) Graphite reacts with F2 at 500 C, forming intercalation compounds or graphite fluoride (CF)n

36. Argon crystallizes in a structure in which the atoms are located at the following coordinates:
1 1 1 1 1 1 
 0,0,0  ;  0, ,  ;  ,0, 
 , ,0 
 2 2   2 2  2 2 
The unit cell is:
(A) simple cubic (B) b. c. c.
(C) f. c. c. (D) h. c. p

37. Which of the following has different colour than others?

(A) AgI (B) Ag2CO3
(C) Ag3PO4 (D) Ag2S

38. At 20oC, the osmotic pressure of a solution of 100 ml of which contains 6.33 g of the colouring
substance of the blood haematin is 243.4 kPa. What is the molecular formula of the haematin; if the
elementary composition of haematin is C = 64.4%, H = 5.2%, N = 3.8%, O = 12.6% and Fe 8.8%?
(A) C34H33N4O5Fe (B) C33H32N4O5Fe
(C) C34H30N3O5Fe (D) C32H28N4O5Fe
O3
39. X  C9H14  
CH2Cl2
1 mole ethanal  2 mole glycerol  1 molePropanal

 one mole 
No. of stereo isomers possible for compound X are:
(A) 8 (B) 6
(C) 4 (D) 10

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40. In the following reactions:

Br2 2NaNH2 Na/Liq.NH3
Me CH CH Me  CCl4
 A   B  C
The product (C) is:
(A) Me H (B) Me Me
C C C C
H Me H H
(C) Me – CH2 – CH2 – Me (D) Me  C  C  Me

41. H2 O,H
 X


 i Hg OAc 
 ii NaBH4  Y
 2

 CH3 3 C  CH  CH2
 i BH  THF
 ii H2 O2  Z
 3

The correct statement from the above sequence is:

(A) X and Y are optically active where as Z is inactive
(B) Y and Z are optically active where as X is inactive
(C) X and Z are optically inactive where as Y is active
(D) Y and Z are optically inactive where as X is active

CH3
CH2 CH3
(i) NaNH ,NH
42. 
2 3
,
(ii) CH3I

N
(A) CH3 (B) CH3
CH2CH3 CH2CH2 CH3

H3C N N
(C) CH2CH3 (D) CH3
CH2CH3 CH2CH3

+
N N I—
CH3

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NSEC-MOCK TEST-1-CHEMISTRY-(2023)-10

43. Concentrated NaOH

X

(S)-2-bromo propanoic acid

Dil. NaOH
Y
Ag2O
In the above sequence of reactions X and Y are respectively:
(A) (S) – Lactic acid and (R) – Lactic acid (B) (R) – Lactic acid and (S) – Lactic acid
(C) (S) – Lactic acid and (S) – Lactic acid (D) (R) – Lactic acid and (R) – Lactic acid

44. Consider the following reaction sequence:

NaCN  alc 
 A  
S 1
N

B  C
  i H3 O
Isomeric products
 ii red P / HI
 i H3O 
 ii red P / HI 2,5  dimethyl pyrrole

2  ethyl pyrrole
Hence organic compound (A) may be:
(A) (B)
CH2 Cl
N CH2 Cl
N
H
(C) (D)
Cl CH2 Cl
N H3C N
H H

45. (i) CF CO H Conc. H SO

 3
(ii) H O
3
 A 
 
2 4
B
2

A and B are respectively.

(A) OH OH (B) OH

and and

OH
(C) O (D) O
OH OH OH
and and

OH

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46. An ether (A) C6H14O when heated with excess of hot conc. HI produced two alkyl halides which on
hydrolysis form compounds (B) and (C). Oxidation of (B) gave on acid (D) where as oxidation of (C)
gave a ketone (E). What are structural formula of (A), (B) and (C) respectively?
(A) H5C2 O CH CH2 CH3 , CH3CH2OH H3C CH C2H5
CH3 OH
(B) C2H5OCH2C3H7, C2H5OH, OHC – CH2CH2CH3
(C) H3C CH2 CH2 O CH CH3, HO CH CH3, CH3CH2CH2OH
CH3 CH3
(D) H3C CH O CH2 CH CH3, H3C CH OH, CH3CH2CH2OH
CH3 CH3 CH3

47. Which of the following compounds on oxidation with (CH3COO)4Pb would most readily give the
product OHC (CH2)4CHO?
(A) (B)
H H OH H

OH OH H OH
(C) H H (D) H OH

OH OH OH H

48. O

NH OH

2
HCl
?

The major product of the following reaction is.....?

(A) Cl (B) N
OH

(C) O (D) OH
N
N

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NSEC-MOCK TEST-1-CHEMISTRY-(2023)-12

ANY NUMBER OF OPTIONS 4, 3, 2 or 1 MAY BE CORRECT

MARKS WILL BE AWARDED ONLY IF ALL THE CORRECT OPTIONS ARE BUBBLED

49. Select the correct statement(s) regarding 3Py orbital:

(A) total number of nodes are 2
(B) number of maxima in the curve 4r 2R2 r  Vs r is two
(C) quantum number n, l and m for orbital may be 3, 1 and 1 respectively
(D) the magnetic quantum number may have a positive value

50. In which of the following pairs of species the central atom has same hybridization?
+ –
(A) ClF3O, ClF3O2 (B) (ClF2O) , (ClF4O)
–
(C) ClF3, ClF3O (D) (ClF4O) , (XeOF4)

51. Which of the following reaction will not be feasible?

1
(A) R  C  CH     Na  s   R  C  CNa  H2  g
2
(B) OH O Na

    Na2CO3  aq.   aq.  H2 O     CO2  g

(C) R  C  CH     NaOH    
 R  C  CNa   aq.  H2O   
(D) R  C  CH     NaNH2  s  
 R  C  CNa   aq.  NH3  g 

52. Which of the following statements is/are correct?

(A) In Serpeck’s process for the purification of bauxite, ammonia is obtained as a by product
(B) In the Hall’s process for the purification of bauxite, the ore is fused with NaOH
(C) In Hall-Heroult’s process for the reduction of alumina, a mixture of alumina, cryolite and fluorspar
is used
(D) In the Bayer’s process for the purification of bauxite, the ore is digested with Na2CO3

53. Electrolysis of molten MgCl2 is the final production step in the isolation of magnesium from sea water
by Dow’s process. Assuming that 35.6 g of Mg metal forms. Which of the following statement (s)
is/are correct?
(A) 2.96 moles of electrons are required
(B) 3.83 × 105 C charge is required
(C) 31.8 A of current is required for 2.5 hours
(D) 41.5 A of current is required for approximately 115 mins

54. Which of the following statement(s) is/are correct?

(A) Pot.tetracyanocuperate(I) is tetrahedral complex
(B) Sod.tetracyanonickelate(II) is a square planar complex
(C) Pot.tetrachloridoplatinate(II) is a tetrahedral complex
(D) Tetraamminecopper(II)sulphate is a square planar complex

Space for rough work

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55. Which of the following statement(s) is/are correct?

(A) At Boyle’s temperature a real gas behaves as ideal gas for a wide range of pressure.
(B) At Boyle’s temperature van der Waal’s constant ‘a’ and ‘b’ compensate each other.
(C) The area under the Maxwell distribution molecular speed curve remains same irrespective of
temperature of gas.
(D) The fraction of molecules with most probable speed increases with increase of temperature.

56. A saturated solution of AgCl was treated with 1.5 M Na2CO3 solution giving Ag2CO3. The remaining
solution contained 0.0026 gm of Cl¯ per litre. [Ksp of Ag2CO3 = 8.2 x 10-12 M3]. Which of the following
statements is/are correct.
-5
(A) The concentration of NaCl at equilibrium is 7.32 x 10 M.
+ -6
(B) The concentration of Ag at equilibrium is 2.33 x 10 M.
(C) Ksp of AgCl from the above data is 1.71 x 10-10 M2
(D) The equilibrium concentration of CO3-2 is 8.2 x 10-12 M

57. Which of the following pairs can be distinguished by Fehling’s solution?

(A) Glucose and fructose
(B) Glucose and sucrose
(C) Benzaldehyde and ethanal
(D) Hydroxy propanone and benzaldehyde

58. Which of the following dicarboxylic acids on heating form anhydride?

(A) malonic acid (B) succinic acid
(C) glutaric acid (D) adipic acid

NOCl/H LiAlH4 NaNO2  HCl
59. CH3  C  CH2  Ph   ' X'  ' Y ' 
0  5o C
'Z'
||
O
(A) Compound ‘X’ is CH3  C  C H  Ph (B) ‘Y’ is optically active
|| |
O NH2

(C) Compound ‘Z’ is CH3 C  CH2  Ph (D) Compound ‘Y’ is CH3CH2 CH  Ph

|| |
O NH2

60. Which of the following compounds are formed by the oxidation of

CH3
H3C C CH CH2
OH OH OH
with HIO4
(A) CH3 – CHO (B) CH3COCH3
(C) H – COOH (D) HCHO

Space for rough work

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NSEC-MOCK TEST-1-CHEMISTRY-(2023)-14

ANSWERS

1. (C) 2. (D) 3. (B) 4. (B)

5. (A) 6. (B) 7. (D) 8. (A)

9. (B) 10. (A) 11. (C) 12. (D)

13. (B) 14. (B) 15. (B) 16. (B)

17. (C) 18. (D) 19. (B) 20. (B)

21. (B) 22. (A) 23. (B) 24. (C)

25. (A) 26. (C) 27. (D) 28. (C)

29. (D) 30. (C) 31. (B) 32. (C)

33. (A) 34. (B) 35. (C) 36. (C)

37. (D) 38. (A) 39. (A) 40. (A)

41. (C) 42. (C) 43. (B) 44. (A)

45. (C) 46. (A) 47. (A) 48. (D)

49. (A, B, C, D) 50. (A, C, D) 51. B, C 52. (A, C)

53. (A, C, D) 54. (A, B, D) 55. (A, B, C) 56. (A, B, C)

57. (C, D) 58. (B, C) 59. (B, C) 60. (B, C, D)

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HINTS AND SOLUTIONS

1. (C)

2. (D)
[salt]
pH  pK a  log
[acid]
 N2 V2 
5  5  log  
N V
 1 1  N V
2 2 

0.1 V2
1
0.1 100  0.1 V2
10 – 0.1V2 = 0.1 V2
V2 = 50 ml

3. (B)
Ksp values are experimentally determined. So, they take into account extra solubility due to hydrolysis.

4. (B)
NaX  Na  X
X  H2 O 

 HX  OH


t0 1.0 excess 0 0

1   
t  eq. 0.1(1   ) 0.1 0.1
[HX] [OH ]
Kh 
[X  ]
[HX] [H ][OH ]
 
[H ] [X  ] 1
K w 0.1  0.1
   0.1  2 1    1 as 1   for weak acid
Ka 0.1(1   )
1014
  0.1  2
10 5
   1 104  0.01%

5. (A)
2
PNOCl 8
K eq  2
= 10
PNO  PCl2
 G  RT n K eq  2.303  8.314  298  8  45.65 kJ

6. (B)

7. (D)

8. (A)
Pyellow  Pred
HR = –9.91 – [–8.78]
= –9.91 + 8.78 = –1.13 kJ

9. (B)
For triatomic gas
 = 1.33, Cp = 4 R
1
 V1 
T2 = T1  
 V2 

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NSEC-MOCK TEST-1-CHEMISTRY-(2023)-16

1.33 1 1/ 3
 5   1
= 300     300    =3001/2 =150
 40  8
H = n Cp(T2 – T1) = 348.314x(150 – 300) = - 14.965 KJ

10. (A)
PBr5 (s) [PBr4 ] (s)  Br  (s); [i  2]
 
2PCl5 (s) 
[PCl6 ]  [PCl4 ] ; [i  1]
2
ratio  2
1
11. (C)
12. (D)
AxBy  XA y+  yB x 
1  x y
I = (1-) + x + y 
Hence (D) is correct.
13. (B)
14. (B)
Half-life is independent of concentration of Zn, thus order w.r.t. [Zn] = 1
At pH = 2, [H+] = 10–2, t1/2 = 10 minutes
pH = 3, [H+] = 10–3, t1/2 = 100 minutes
n 1
[t1/ 2 ]1  a2 
 
[t1/ 2 ]2  a1 
n 1
10  103 
 
100  10 2 
 n = 2, thus w.r.t. [H+] order = 2
Rate law = k [H+]2 [Zn]
Hence, (B) is correct.
15. (B)
K  1000
S
 oAgCl
2.3  106  1000

(61.9  76.3)
 1.66  10 5 M

16. (B)
III > II > I
In I – OCH3 is electron donating group (+M) so will decrease the electron deficiency at carbonyl
carbon to the most.
In II there is no electron donating group.
In III electron withdrawing group is present which will increase by electron deficiency at carbonyl
carbon.
17. (C)
In very conc. base, the rate equation is:
rate = k[HCHO]2 [OH–]2

O H O
H C 

CH3 
 H
OH
C H 
 H
 OH
C H
O O O
The dianion will clearly be a much powerfull hydride donor.
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18. (D)
Mechanism
H
+
CH2NH2 CH2 N2 CH2 OH O
HNO2
    
OH OH OH

19. (B)

20. (B)

21. (B)

22. (A)
HF is stronger acid than HCN.
H2SO3 is stronger acid than H2TeO3.
HNO2 is stronger acid than HAsO2.

23. (B)
It does not dimerise due to delocalization of odd electrons.

24. (C)
Zn(NH2)2 is amphoteric in NH3.
Rest are amphoteric in H2O.

25. (A)
Solubility of hydroxides increases down the group, hence Be(OH)2 is least soluble. Hence, least Ksp.

26. (C)

27. (D)

28. (C)

29. (D)
NO 2 NO 2 NO 2
COOH
I: > >

COOH
Strong -I effect COOH
-R effect -I effect Only -I effect
-R effect
II: Higher the acidity of an acid lower the basicity of its conjugate base.
III: Higher the stability of an alkene lower the heat of hydrogenation.
IV: Ph - C- CH2 - COOH is the -keto acid so it would be having maximum ease for decarboxylation.
||
O

30. (C)
Since (C) produces the most stable carbocation, hence, this will undergo maximum racemisation.

31. (B)
2C2H2  g  5O2  g   4CO2  g  H2 O   
Volume of CO2 = 68.5 – 38.5 = 30 ml
30
Volume of C2H2 =  15ml
2
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NSEC-MOCK TEST-1-CHEMISTRY-(2023)-18

Volume of N2 = 20 – 15 = 5 ml
5
% of N2 =  100  25%
20

32. (C)
N 2 Cl OH OH
O O
CHO
steam (i)CHCl3 / NaOH /  (CH 3 CO ) 2 O
    
CH 3 COONa
(ii) H O 
3

Perkin's reaction

33. (A)

N2O C8H7OCl

287 m
Molar mass of C8H7OCl = 8  12 + 7  1 + 16 + 35.5 = 154.5 g
Molar mass of N2O = 2  14 + 16 = 44g
According to Graham’s Law of diffusion
rN2 O MC8H7OCl 154.5
   3.5  1.87 : 1
rC8H7 OCl MN2O 44
1.87 th
dN2O   287 = 187 row from N2O side
2.87
1.0 th
dC8H7OCl   287 = 100 row from weeping gas side
2.87
Therefore, the spectator from the side of N2O in the 187th row will be laughing and weeping
simultaneously Alternatively, the spectator from the side of weeping gas in 100th row will laugh and
weep.

34. (B)
ZM
d
a3  No
6.45  64  1024  6.02  1023
M
4
(Z = 4 as it is fluorite type) M = 62
Now if all the ions removed from one body diagonal then new mass of effective unit cell = M  Z  204
Z  M -
d = 5.30 g ml
a3  No

35. (C)
(C) -Graphite is in the form of ABABAB………..
-Graphite is in the form of ABCABCABC……..
36. (C)
37. (D)
All others are yellow in colour while Ag2S is black.
38. (A)
 293.4
  CRT,So,C    0.1 mol / L
RT 8.31 293
If 100 ml contains 6.33 g of haematin.
 1 L will contain 63.3 g of haematin.
Molecular mass of haematin = 633 u
Estimating the empirical formula of compound
64.6 5.2 8.8 12.6 8.8
C:H:N:O:F= : : : :
12 1 14 16 56
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 34 : 33 : 4 : 5 : 1
So, empirical formula is C34H33N4O5Fe, whose mass corresponds to the molecular mass.

39. (A)
(X)
No. of stereo site = 3
3
No. of stereo isomer = 2 = 8

40. (A)
Br Me H
Br2 2NaNH2 Na/Liq.NH3
Me  CH  CH  Me 
CCl4
 Me CH CH Me   Me  C  C  Me  
H Me
Br

41. (C)
(A) OH
CH3 CH3
H OH
CH3  H3C C CH CH3
 1,2  CH3 shift
H3C C CH CH2 H
 H3C C CH CH3   H3C C CH
CH3 CH3 CH3 CH3 CH3 CH3
Optically inactive
(X)
(B) On oxymercuration demercuration the product is alcohol without rearrangement
CH3 OH
H3C C CH CH3 Optically active

 Y is CH3
(C) On hydroboration – oxidation the addition is against to Markonikov’s rule

CH3
H3C C CH2 CH2OH Optically inactive

 Z is CH3

42. (C)
43. (B)
With conc. NaOH the reaction is SN2, hence inversion in configuration. With Ag2O, NaOH,
neighbouring group participation. Retention in configuration.
44. (A)


CN
CH2 Cl   CN 

N
SN 1
CH2  
N CH2 CN
H N
H
H
A
B 
i  H3O 
H
ii  red P / HI

CN
N C CH2   CH2
N N CH2 CH3
N
H H
H
 2  ethyl pyrrole 
i H O
 3
N C CH3 
ii red P /HI 
H3C CH3
N N
H H
 2,5  Dimethyl pyrrole 
Hence, (A) is the correct answer.
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NSEC-MOCK TEST-1-CHEMISTRY-(2023)-20

45. (C)
OH OH2
(i) CF3CO3H Conc. H2SO4

(ii) H2O

 
, Pinacol Pinacolone reaction


OH OH

-H2O


H
  

H O
O H O

46. (A)
CH3 CH2OCH CH2CH3 
 CH3 CH2Cl 
 CH3CH2OH
Cl CH2 CH3 HO
CH3 CH CH2 CH3
CH
CH3 CH3

47. (A). A is cis 1, 2-diol.

48. (D)
HO
O N OH
O NH2 OH HO NH
H2 N OH
H2O




IMPT

 
 
 


N OH 2 OH

N N
 H2O
H

 
 
 
HO
 
2
H

49. (A, B, C, D)
For 3Py orbital
Total nodes = n – 1 = 2
50. (A, C, D)
In all ClF3O, ClF3O2, ClF3 hybridisation of Cl is sp3d and ClF4O – and XeOF4 hybrdisation of central
3 2
atom is sp d .

51. (B, C). Weak acid-base pair cannot react to form strong acid-base pair.
52. (A, C)
In the Hall’s process for the purification of bauxite the ore is fused with Na2CO3.
In the Bayer’s process for the purification of bauxite the ore is digested with NaOH.
53. (A, C, D)
35.6
 2.966 equivalents
12
For 1 equivalent 96500 C is required
So, for 2.966 equivalent 286283.33 C is needed so.
2.97 moles of electron carries same charge hence (A) is correct.
Q = It
Q = 31.8 × 2.5 × 60 × 60
Q = 286283.33
Hence, (C) is correct.
Q = 41.5 × 115 × 60
= 286283.33
Hence, (D) is correct.
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54. (A, B, D)
[Cu(CN)4]3- → tetrahedral (sp3)
2- 2
[Ni(CN)4] → square planar (dsp )
2- 2
[PtCl4] → square planar (dsp )
2+ 2
[Cu(NH3)4] → square planar (dsp )
Hence, (A), (B) and (D) are the correct answer.
55. (A, B, C)
56. (A, B, C)
[CO32-] = [Na2CO3] = 1.5 M
0.0026
At equilibrium [Cl-] = [NaCl] =  7.32  10 5 M
35.5
2AgCl(s) + Na2CO3 Ag2CO3(s) + 2NaCl
Initial 1.5 M 0
At equilibrium (1.5 – 7.32 x 10-5)M 7.32 x 10-5M
[Ag+]2 [CO32-] = Ksp [Ag2CO3]
8.2  10 12
[Ag+] =  2.33  10 6 M
1.5
+ - -6 -5 -10
Ksp of AgCl = [Ag ] [Cl ] = (2.338 x 10 ) x (7.32 x 10 ) = 1.71 x 10 .
57. (C, D)
Glucose, fructose, ethanal and sucrose gives Fehling’s test. But HCHO and C6H5CHO undergo
Cannizzaro reaction in alkaline medium and does not gives Fehling’s test.
58. (B, C)
According to Blanc’s rule only 1,4 & 1,5 dicarboxylic acids can give anhydride
59. (B, C)

CH 3  C  CH2  Ph CH3  C = CH  C6H5

O OH
NOCl

+
CH 3  C  CH  Ph + CH 3  C  CH  Ph
H
O NO OH NO
‘X’

OH
LiAlH4 NaNO2
CH3  C  C  Ph CH3  C H  CH  Ph
 HCl
o
O N  OH NH2 0 5 C
‘Y’

+
CH3  CH  CH  Ph CH 3  C  CH2Ph

OH O

60. (B, C, D)
CH3 O O O
2HIO4
H3C C CH CH2   H3C C CH3  H C OH  H C H
OH OH OH

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