1
CHEMISTRY
MULTIPLE CHOICE QUESTIONS (SINGLE OPTION CORRECT)
31. What will be the molar solubility of molar solution of AgCl at 25o
C if specific conductivity of saturated
AgCl solution is 2.3 × 10–6
S.cm–1
at 25o
C and molar conductivities of
Ag+
and Cl–
at infinite dilution are 61.9 and 76.3 Scm2
mol–1
respectively?
(A) 2.382 × 10–3
M (B) 1.66 × 10–5
M
(C) 5 × 10–8
M (D) 7.2 × 10–6
M
32. PA = (235 y – 125 xy) mm of Hg
where PA is partial pressure of A
x is mole fraction of B in liquid phase in the mixture of two liquids A and B and y is mole fraction of A in
vapour phase, then o
B
P in mm of Hg is
(A) 235(B) 0
(C) 125(D) 110
33. Which of the following plots represent the behaviour of an ideal binary liquid solution?
(A) Plot of PTotal vs YA is linear (B) Plot of
B
Total
1
vs Y
P
is linear
(C) Plot of PTotal vs YB is linear (D) None of the above
34. Following reaction occurs at 25o
C,
5 2 2 o
2
2NO(g)(1 10 atm) Cl (g)(1 10 atm) 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
35. The approximate heat necessary to raise the temperature of 18.4 gm of ethyl alcohol from 45o
C to
55o
C at 1 atm pressure (Cp = 26.8 cal/mole) is
(A) 53.6 cal (B) 107.2 cal
(C) 214.4 cal (D) 536 cal
36. Find H
 at 358 K for the reaction,
2 3 2 2
Fe O (s) 3H (g) 2Fe(s) 3H O( )
 
 
Given that, 298
H 33.29 kJ/mole
   and Cp for Fe2O3(s), Fe(s), 2 2
H O( ) and H (g) are 103.8, 25.1, 75.3 and
28.8 J/K mole.
(A) 85.9 kJ/mole (B) –28.136 kJ/mole
(C) +28.136 kJ/mole (D) 85.9 J/mole
37. List – 1 (complex ion) List – 2 (number of unpaired electrons)
(a) [CrF6]-4
(1) one
(b) [MnF6]-4
(2) zero
(c) [Cr(CN)6]-4
(3) three
(d) [Mn(CN)6]-4
(4) four
(5) five
a b c d
(A) 4 1 2 5
(B) 2 1 3 5
(C) 4 5 2 1
(D) 2 5 3 1

2
38. Which of the following cations can be identified by using Na2CrO4 solution?
(I) Ba2+
(II) Pb2+
(III) Ag+
(IV) Sr2+
(A) (I), (II) (B) (I), (II), (III)
(C) (II), (III) (D) (I), (II), (III), (IV)
39. A cupric salt on being heated with metallic Cu and conc. HCl, the colour of the solution is
discharged due the formation of a
(A) CuCl2 (B) Cu2Cl2
(C) H[CuCl2] (D) H[CuCl3]
40. A compound (A) forms an unstable pale blue colour solution in water. The solution decolourizes
Br2 water and an acidified solution of KMnO4. The compound (A) is
(A) HNO2 (B) HNO3
(C) N2O5 (D) None of the above
41. The increasing order of affinity of P4O10, Cl2O7 and I2O5 for water is
(A) P4O10 < I2O5 < Cl2O7 (B) Cl2O7 < I2O5 < P4O10
(C) P4O10 < Cl2O7 < I2O5 (D) I2O5 < Cl2O7 < P4O10
42. Borazene, B3N3H6, is isoelectronic and isostructural with benzene. Which of the following
statement(s) is/are true about borazene?
(i) Borazene is aromatic.
(ii) There are four isotopic disubstituted borazene molecules, B3N3H4X2.
(iii) Borazene is more reactive towards addition reactions than benzene.
(A) Only (i) (B) (i) and (ii)
(C) (i) and (iii) (D) (i), (ii) and (iii)
43. Which of the following statement regarding pyrophosphoric acid is not correct?
(A) It is obtained by heating equimolar mixture of ortho and metaphosphoric acid at 100o
C.
(B) On boiling with water, it produces orthophosphoric acid.
(C) On strong heating, it produces metaphosphoric acid.
(D) It forms four series of salts, viz., NaH3P2O7, Na2H2P2O7, Na3HP4O7 and Na4P2O7.
44. In Castner’s process some sodium is lost because
(A) It reacts with dissolved O2 in molten NaOH
(B) Due to its high reactivity it attacks other substance present in the system
(C) At high temperature some of it vapourises
(D) All of the above
45. When dilute H2SO4 and H2O2 are added to a solution of chromate ions, an intense blue colur is
produced which is stable in ether. This is due to the formation of
(A) 2
2 7
Cr O  (B)
2 3
Cr O
(C)
5
CrO (D)
3
CrO
46. BCl3 does not exist as dimer but BH3 exist as dimer (B2H6) because
(A) chlorine is more electronegative than hydrogen
(B) there is p - p back bonding in BCl3 but BH3 does not contain such multiple bonding.
(C) Large sized chlorine atoms do not fit in between the small boron atoms whereas small sized
hydrogen atoms get fitted in between boron atoms
(D) None of the above
47. Which of the following statements regarding sulphur trioxide is not correct?

3
(A) it exists in three polymorphic forms
(B) It is an anhydride of sulphuric acid
(C) SO3 is an acidic oxide
(D) The SO bond length is smaller than the expected value
48. For the equilibrium at 298 K
2 3
2H O H O OH
 

G is approximately:
(A) 100 KJ (B) 80 KJ
(C) 80 KJ (D) 100 KJ
49. The emf of a cell corresponding to the reaction:
        
2
2
Zn s 2H aq. Zn 0.1M H g 1atm
 
 
 
is 0.28 V at 15C. Calculate the pH of the solution at the hydrogen electrode:
2
2
0 0
Zn / Zn H /H
E 0.76V; E 0
 
  
(A) 6.68 (B) 8.62
(C) 9.45 (D) 3.25
50. NCl3 on hydrolysis will give NH3 and HOCl but PCl3 on hydrolysis gives H3PO3 & HCl because
(A) N is more electronegative than phosphorus
(B) size of P is larger than nitrogen
(C) P is having vacant d-orbital
(D) A & C are correct
51. Amongst 2 3 2
6 6 2 2 4
TiF , CoF ,Cu Cl & NiCl
  
     
     
. The colourless species are
(A) 2
6 2 2
TiF & Cu Cl

 
 
(B) 3 2
6 4
CoF & NiCl
 
   
   
(C) 2 3
6 6
TiF & CoF
 
   
   
(D) 2
2 2 4
Cu Cl & NiCl

 
 
52. The relative lowering of vapour pressure of an aqueous solution of a non volatile solute of
molecular weight 60 (which neither dissociates nor associates in the solution) is 0.018. If f
K of water is
1
1.86 m
 , the depression in freezing point will be
(A) 0.93°C (B) 3.72°C
(C) 1.86°C (D) 0.018°C
53. A 0.001 molal solution of Pt(NH3)4Cl4 in water has freezing point depression of 0.0054C. If Kf for
water is 1.8, the correct formula for the above molecule is
(A)  
3 2 2
4
Pt NH Cl Cl
 
 
(B)  
3 3
4
Pt NH Cl Cl
 
 
(C)  
3 4
4
Pt NH Cl
 
 
(D)    
3 3 3
2 2
Pt NH Cl NH Cl
 
 
54. It is found that elevation in boiling point of a given aq. NaCl solution is equal to depression in
freezing point of 0.25 molal aq. Na2CO3 solution. If Kf = 1.86 mol 1
kg & Kb = 0.52 mol 1
kg. Calculate
molality of NaCl solution; assume complete ionization of NaCl & Na2CO3 in aq. solution.
(A) 1.11 (B) 1.78
(C) 1.34 (D) 1.56
55. The density of metallic hydrides
(A) is lower than the parent metal (B) is higher than the parent metal

4
(C) is same (D) cannot be predicted
56. Calculate resonance energy of N2O from the following data:
 
0 1
f 2
H N O 82KJ mol
 
BE of N  N = 946 KJ mol 1
BE of N = N = 418 KJ mol 1
BE of O = O = 498 KJ mol 1
BE of N = O = 607 KJ mol 1
(A) 68 KJ mol 1
(B) 42 KJ mol 1
(C) 88 KJ mol 1
(D) 56 KJ mol 1
57. On the addition of mineral acid to an aqueous solution of borax, the following compound is formed?
(A) boron hydride (B) boric acid
(C) metaboric acid (D) pyroboric acid
58. The standard reduction potential values of three metallic cations x, y, z are 0.52, 3.03 and 1.18
V respectively. The order of reducing power of the corresponding metal is
(A) y > z > x (B) x > y > z
(C) z > y > x (D) z > x > y
59. Among the following which is not the -bonded organometallic compound?
(A)
 
5
5 5
2
Fe C H
  (B)
 
5
6 6
2
Cr C H
 
(C)  
3 4
CH Sn (D)
 
2
3 2 4
K PtCl C H
 
 
 
60. Consider the complex [Co(NH3)5CO3]ClO4. The co-ordination number, oxidation number, number
of d electrons and number of unpaired d electrons on the metal are, respectively,
(A) 6, 3, 6, 0 (B) 7, 2, 7, 1
(C) 7, 1, 6, 4 (D) 6, 2, 7, 3

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ANSWERS
31. B
32. D
33. B
34. A
35. B
36. B
37. C
38. B
39. C
40. A
41. B
42. D
43. A
44. A
45. C
46. C
47. D
48. C
49. B
50. D
51. A
52. C
53. A
54. C
55. A
56. C
57. B
58. A
59. C
60. A
HINTS & SOLUTIONS
31.
o
AgCl
K 1000
S



6
2.3 10 1000
(61.9 76.3)

 


5
1.66 10 M

 
32. PA = (235 – 125x) × y
PA = PT × yA
PT = 235 – 125x
o
B
P 110 mm of Hg

Hence, (D).
33. o
B B
B
Total
P X
Y
P


6
Graph
B
Total
1
Y vs
P
is linear.
34.
2
2
NOCl
eq 2
NO Cl
P
K
P P


= 108
eq
G RT n K 2.303 8.314 298 8 45.65 kJ
          
35. 18.4
n 0.4
46
 
p
Q nC T 0.4 26.8 10 107.2 cal/mole
     
36. p
C 2 25.1 3 75.3 (103.8 3 28.8) 85.9 J/mole
        
358 298
p
2 1
H H
C
T T
  
 

358
H 28.136 kJ/mole
  
38. BaCrO4 – yellow precipitate
PbCrO4 – yellow precipitate
Ag2CrO4 – red precipitate
40. HNO2 is pale blue colour due to dissociation of N2O5 in concentrated solution. It decomposes as
2 2 2
2HNO NO NO H O

  
and in dilute solution it decomposes as
2 3 2
3HNO HNO H O 2NO

  
Reaction with (Br2 water)
(a)
2 2 2 3
HNO H O Br HNO 2HBr
  
 
(b)
4 2 4 2 2 4 4 2 3
KMnO 3H SO 5HNO K SO 2MnSO 3H O 5HNO
  
   
43.
Pyrophosphoric acid H O P
O
O
OH
P OH
OH
O
H4P2O7
45. Due to the formation of CrO5, it show blue colour (colour is due to charge transfer spectra)
48. G = 2.303 RT log Keq
= 2.303 (8.314)  298 log 10 14
= 79881.87 J
= 79.9 KJ  80 KJ
49. Half cell reactions will be
2
Zn 2e Zn ...(i)
 

We know that

7
2 2
2
0
Zn/ Zn Zn/ Zn
Zn
RT
E E ln
nF Zn
 

 
 
 
 
 
2
Zn/ Zn
8.314 298 0.1
E 0.76 ln 0.79V
2 96500 1


  

Similarly,
2 2
2
0
2 2
H /H H /H
H 1
RT 8.314 298
E E ln 0 ln
nF 2 96500
H H
 
 
   

   
   

   
   
=
10
0.05915 log H 0.05915 pH

   
 
Since,
2
2
Zn/ Zn H /H
E E E
 
 
= 0.28 = 0.79 0.05915 pH
0.15
pH 8.62
0.05915
 
51. In case of Ti, it is in +4 oxidation state and has no d-electrons in case of Cu d-orbitals are filled so
in both cases d-d-transition will not take place and compound will be colourless.
52. 0.018 = mole fraction of solute
0.018 = x
x 55.5

where x is the no. of moles of solute in 1 litre water.
0.999
x
1 0.018
 

x = 0.99
or 1 = m
f f
T K m
  
f
T 1.86 1 1.86
   
53. f
f
T
m
K


m = 0.003
54. Given,    
b f 2 3
T NaCl T Na CO
  
0 1 0 1
b f
K 0.52 mol kg K 1.80 mol kg
 
 
m =? m = 0.25
NaCl = Na+
+ Cl 2
2 3 3
Na CO 2Na CO
 

i = [1 + (y – 1)x] i = 1 + 2x = 3
= 1 + x = 1 + 1 = 2
(x = 1 since compete ionization) f f i
T K m 1.86 0.25 3
    
Tb = Kbm = 0.52  m  2
Since Tb = Tf
m 0.52 2 1.86 0.25 3
    
1.86 0.25 3
m
0.52 2
 


= 1.34 molal

8
55. Metallic hydrides are made of d-block elements. In these elements void is very small thus when
hydrogen enter in to it then it expands due to that density decreases.
56. From the given BE values
       
0
f 2 N N O O N N N O
1
H N O Be BE BE BE
2
   
   
    
   
 
= 498
946 418 607 1195 1025
2
 
    
     
     
 
 
= 170 KJ
Resonance energy = observed Hf – calculated Hf
 = 82 – 170 = 88 KJ mol 1
57.
2 4 7 2 3 3
Na B O 2HCl 5H O H BO 2NaCl
  
 
58. Lower the reduction potential, stronger is the reducing agent.
60. C.N. = 6, O.N. = + 3, no. of d electrons = 6,
No. of unpaired d electrons = 0
Hence, (A) is the correct answer.