Chemistry-Question-Bank-píginas-59

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MOCK TEST 1 MT-3
31. The number of enantiomers of the compound
COOHCHBrCHBrCH3 is :
(a) 2 (b) 3
(c) 4 (d) 6
32. Equivalent weighs of KMnO4 acidic medium, neutral
medium and concentrated alkaline medium respectively are
M M M, , .
5 1 3
 Reduced products can be
(a) 2 2
2 4MnO , MnO , Mn
(b) 2 2
2 4MnO , Mn , MnO
(c) 2 2
4 2Mn , MnO , MnO
(d) 2 2
2 4Mn , MnO , MnO
33. Which of these have no unit?
(a) Electronegativity (b) Electron affinity
(c) Ionisation energy (d) Excitation potential
34. Which of the following statements is not correct for sigma
and pi-bonds formed between two carbon atoms?
(a) Sigma-bond determines the direction between carbon
atoms but a pi-bond has no primary effect in this
regard
(b) Sigma-bond is stronger than a pi-bond
(c) Bond energies of sigma- and pi-bonds are of the order
of 264 kJ/mol and 347 kJ/mol, respectively
(d) Free rotation of atoms about a sigma-bond is allowed
but not in case of a pi-bond
35. The reactivity of metals with water is in the order of
(a) Na > Mg > Zn >Fe > Cu
(b) Cu > Fe > Zn > Mg > Na
(c) Mg > Zn > Na > Fe > Cu
(d) Zn > Na > Mg > Fe > Cu
36. The emf of Daniell cell at 298 K is E1
Zn | ZnSO4 (0.01 M) | | CuSO4 (1.0 M) | Cu
When the concentration of ZnSO4 is 1.0 M and that of
CuSO4 is 0.01 M, the emf changed to E2 What is the relation
between E1 and E2?
(a) E1 = E2 (b) 2 20E E
(c) 1 2E E (d) 1 2E E
37.
2CH
2CH
O
O
O
2CH
The above shown polymer is obtained when a carbonyl
compound is allowed to stand. It is a white solid. The
polymer is
(a) Trioxane (b) Formose
(c) Paraformaldehyde (d) Metaldehyde.
38. The correct order of atomic/ionic sizes is
(a) N II > III.
13. (d)
14. (b) 4 2 2 2NH NO N 2H O
Volume of N2 formed in successive five minutes are
2.75 cc, 2.40 cc and 2.25 cc which is in decreasing
order. So rate of reaction is dependent on
concentration of NH4NO2. As decrease is not very
fast so it will be first order reaction.
15. (d) We know that
2 6B H
3 2
ether, 0°C1, Pr opene
6 CH CH CH
2 2H O
3 2 2 3
OH
2(CH CH CH )
3 2 2 3 3
Pr opanol
6CH CH CH OH 2H BO
MOCK TEST 1 MT-5
16. (b) totp 740mm
2 2tot O Hp p p
Number of moles of 2O and 2H are equal
2 2O Hp p
2 2O Hp p 740
2 2O Hp 370mm p
17. (d) Formation of only CH3COOH by ozonolysis indicates
that the compound Y should be CH3CH = CHCH3
which can be formed by all of the three given
compounds
21H / Pt
2 2
X
CH CH CH CH
3 3
Y
CH CH CH CH
21H / Pt
3 3 3 3
X Y
CH C CCH CH CH CHCH
21H / Pt
2 3
X
CH C CHCH
3O
3 3 3
Y
CH CH CHCH 2CH COOH
18. (b)
1pH [p log1]
2 aK 
p
2
aK
pH' (twice of pH) = pKa
1p [p log ]
2a aK K C
– log C = pKa = – log Ka
C = Ka = 1.8 × 10–5 M
dilution 1
C
 5.55 × 104 times
19. (d) The sulphide ore is roasted to oxide before reduction
because the o
fG of most of the sulphides are greater
than those of CS2 and H2S, therefore neither C nor H
can reduce metal sulphide to metal. Further, the
standard free energies of formation of oxide are much
less than those of SO2. Hence oxidation of metal
sulphides to metal oxide is thermodynamically
favourable.
20. (d) Buffer solution contains weak base + salt of weak
base with strong acid or weak acid + salt of weak acid
with strong base.
In option (d) the acid used is HClO4 which is strong
acid and KClO4 is salt of this acid with strong base.
Soit is not an example of buffer solution.
21. (d) The most abundant rare gas found in the atmosphere
is argon and not helium.
22. (c)
Co
en
en
Cl
Cl
Cis-d-isomer
+
Co
en
en
Cl
Cl
Cis- -isomer
+
Mirror
23. (d) Given,
2 2 3 1N 3H 2NH ; ....(i)K
2 2 2N O 2NO; ....(ii)K
2 2 2 3
1H O H O; ....(iii)
2
K
We have to calculate
3 2 24NH 5O 4NO 6H O; ?K
or 3 2 2
52NH O 2NO 3H O
2
For this equation, 
2 3
2
2 5/ 2
3 2
[NO] [H O]
[NH ] [O ]
K
but 
2 2
3
1 23
2 22 2
[NH ] [NO],
[N ] [O ][N ] [H ]
K K
2
3 ½
2 2
[H O]
&
[H ][O ]
K
3
2
3 3 3/ 2
2 2
[H O]or
[H ] [O ]
K
Now operate, 
3
2 3
1
.K K
K
3 32
2 2 2
3 3 / 2 2
2 2 2 2 3
[H O] [N ][H ][NO] .
[N ][O ] [H ] [O ] [NH ]
2 3
2
2 5 / 2
3 2
[NO] [H O]
[NH ] [O ]
K
3
2 3
1
.K KK
K
24. (c) For third ionization enthalpy last configuration of
V – 4s0 3d3
3d 4s
Cr – 4s0 3d4
Mn – 4s0 3d5
Fe – 4s0 3d6
For third Ionization enthalpy Mn has stable
configuration due to half filled d-orbital.
25. (d) 2
35 ]NH)CO(Co[ . In this complex. Co-atom attached
with NH3 through bonding with CO attached with
dative -bond.
26. (a) Given reaction is
3 2 2IO aI bH cH O dI
Ist half reaction
2I I ...(i)
– 1 0 (oxidation)
IInd half reaction
3 2IO I ...(ii)
+ 5 0 (reduction)
On balancing equation (ii) we have
3 2 210e 2IO 12H I 6H O ...(iii)
Now, balance equation (i)
MT-6 CHEMISTRY
22I I 2e ....(iv)
Multiply eqn (iv) by 5 and add it to eqn (iii), we get
3 2 22IO 10I 12H 6I 6H O
or, 3 2 2IO 5I 6H 3I 3H O
Hence a = 5, b = 6, c = 3, d = 3
27. (b) This is because of inter-electronic replusions between
lone pairs.
B.E. : F – F Cl – Cl
(kJ mol–1) : 158.8 242.6
..
..
..
..
:FF:
28. (b) Molecules having higher molecular weight and less
branching have higher boiling point.
29. (b) Always transition metals combines with more than
one carbonyl group.
30. (b) Hydration energy of Cl+ is very less than H+ hence it
doesn’t form Cl+ (aq) ion.
31. (c) No. of asymmetric carbon = 2
No. of enantiomers = 22 = 4.
32. (c) Change Equiv. mass
H 2
4
27
MnO Mn 5
M
5
2H O 2
4 4
7 6
MnO MnO 1 M
OH
4 2
7 4
MnO MnO 3
M
3
33. (a) Electronegativity is the tendency of the atom to attract
electrons to itself when combined in a compound as
defined by Pauling. Electronegativity, is a relative term
so it does not have any unit.
34. (c) As sigma bond is stronger than the (pi) bond, so it
must be having higher bond energy than (pi) bond.
35. (a) The reactivity may be attributed to size factor, larger
is the size, higher is tendency to lose electron (low
I.E.). Zn is the last element of 3d series and it has
larger size than Cu due to 10 2(3d 4s ) configurations.
Hence, the reactivity order is
Na > Mg > Zn > Fe > Cu.
36. (c) Using the relation
0
cell cell
0.0591 [anode]E E log
[cathode]n
2+
0
cell 2+
0.0591 [Zn ]E log
[Cu ]n
Substituting the given values in two cases.
0
1
0.0591 0.01E E log
2 1.0
 0 20.0591E log10
2
 0 00.0591
E 2 or (E 0.0591)V
2
0
2
0.0591 1E E log
2 0.01
 0 20.0591E – log10
2
 0 02 0.0591E – or (E – 0.0591)V
2
Thus, 1 2E E
37. (a) 3 HCHO 
On keeping
aq. solution
CH2
CH2
O
O
CH2
O
Trioxane 
(meta formaldehyde)
38. (b) Amongst the isoelectronic species, the anion having
more negative charge would have the larger size.
39. (c) Ksp = [Ag+] [Cl–]
1.8 × 10–10 = [Ag+] [0.1]
[Ag+] = 1.8 × 10–9 M
Ksp = [Pb+2] [Cl–]2
1.7 × 10–5 = [Pb+2] [0.1]2
[Pb+2] = 1.7 × 10–3 M
40. (a) Excess of HCl is used to convert free aniline to aniline
hydrochloride otherwise free aniline would undergo
coupling reaction with benzenediazonium chloride.
41. (b) In lake test of Al3+, there is formation of coloured
floating lake. It is due to the adsorption of litmus by
Al(OH)3.
42. (c) 2Cl(g) — Cl2(g)
Entropy is decreasing (–ve) in the reaction. Further
the reaction is exothermic since a bond is being
formed, i.e., H is also –ve.
43. (a) Due to H-bonding, the boiling point of ethanol is much
higher than that of the isomeric diethyl ether.
44. (d) Phenol has active (acidic) hydrogen so it reacts with
CH3MgI to give CH4, and not anisole
6 5 3C H OH CH MgI 4 6 5CH C H OMgI
45. (d) From given data, we have
C + O2 CO2 – x kJ … (i)
2 2 2
1H O H O y kJ
2
…(ii)
4 2 2 2CH 2O CO 2H O z kJ …(iii)
The required equation is
2 4C 2H CH Q
To get the required equation, operate
(i) + 2 × (ii) – (iii)
Thus, we get
2 4C 2H CH [( ) ( 2 ) ( )]x y z
Thus, heat of formation of methane is
 (–x – 2y + z) kJ
Time : 1 hr Max. Marks -180
Mock Test-2
1. The angular momentum of the electron in first excited
energy state of hydrogen atom is
(a)
h
(b) 2
h
(c)
2
)12(2 h (d) None of these
2. When NaCl is dopped with 1.0 × 10–3 mole of SrCl2, the
number of cation vacancy is
(a) 6.023 × 1018 (b) 6.023 × 1020
(c) 2 × 6.023 × 1020 (d) 3.011 × 1020
3. A 0.5 M NaOH solution offers a resistance of 31.6 ohm in a
conductivity cell at room temperature. What shall be the
approximate molar conductance of this NaOH solution if
cell constant of the cell is 0.367 cm–1.
(a) 23.4 S cm2 mole–1 (b) 23.2 S cm2 mole–1
(c) 46.45 S cm2 mole–1 (d) 54.64 S cm2 mole–1
4. Ammonium dichromate on heating gives
(a) chromic acid & ammonia
(b) chromium sesquioxide & nitrogen
(c) chromium sesquioxide & ammonia
(d) chromic acid and N2
5. Predict the relative acidic strength among the following
(a) H2O, H2S, H2Se, H2Te
(b) H2O Cs > Rb (b) Rb > Cs > Li
(c) Cs > Li > Rb (d) Li > Rb > Cs
12. Paramagnetism of Cr (Z = 24), Mn2+ (Z = 25) and Fe3+
(Z = 26) are x, y and z respectively. They are in the order
(a) x = y = z (b) x > y > z
(c) x = y > z (d) x > y = z
13. Formaldehyde reacts with ammonia to give urotropine is
(a) (CH2)6N4 (b) (CH2)4N3
(c) (CH2)6N6 (d) (CH2)3N3
14. Indicate the wrongly named compound
(a) CHO2CH2CHH
CH
|
C3CH
3
(4-methyl -1- pentanal)
(b) COOHCCH
CH
|
C3CH
3
(4- methyl -2- pentyn -1- oic acid)
(c) COOHH
CH
|
C2CH2CH3CH
3
(2- methyl -1- pentanoic acid)
(d) 323 CH
O
||
CCHCHCHCH
MT-8 CHEMISTRY
(3- hexen -5- one)
15. The favourable condition for a process to be spontaneous
is :
(a) , ive, iveT S H H S
(b) , ive, iveT S H H S
(c) , ive, iveT S H H S
(d) , ive, iveT S H H S
16. Vapour pressure (in torr) of an ideal solution of two liquids
A and B is given by : P = 52XA + 114
where XA is the mole fraction of A in the mixture. The vapour
pressure (in torr) of equimolar mixture of the two liquids
will be :
(a) 166 (b) 83
(c) 140 (d) 280
17. The decomposition of a substance follows first order
kinetics. Its concentration is reduced to 1/8th of its initial
value in 24 minutes. The rate constant of the decomposition
process is
(a) 1/24 min–1 (b) 1min
24
692.0
(c)
1min
8
1log
24
303.2
(d)
1min
1
8log
24
303.2
18. Fluorine does not show highest oxidation state opposite
to other halogens, because
(a) it is most electronegative
(b) it has no d-orbital
(c) its atomic radius is very small
(d) F– ion is stable and isoelectronic with neon
19. Glucose contains in addition toaldehyde group.
(a) one secondary –OH and four primary –OH groups
(b) one primary –OH and four secondary –OH groups
(c) two primary –OH and three secondary –OH groups
(d) three primary –OH and two secondary –OH groups
20. Which of the following product is obtained by treating 1-
butyne with HgSO4 and H2SO4?
(a) CH3CH2COCH3 (b) CH3CH2CH2CHO
(c) CH3CH2CH2COOH (d) CH3CH2CH = CH2
21. An example of Perkin’s reaction is
(a) 6 5 3 2C H CHO CH NO
KOH
6 5 2C H CHCHNO
(b) O)COCH(CHOHC 2356
CHCOOHCHHC 56
COONaCH3
(c) 6 5 3C H CHO CH CHO
NaOH
6 5C H CH= CHCHO
(d) 2256 )COOH(CHCHOHC
HCHCOCHHC 256
NH.Alc 3
22. The product obtained on reaction of C2H5Cl with hydrogen
over palladium carbon is :
(a) C3H8 (b) C4H10
(c) C2H6 (d) C2H4
23. Which of the following can be predicted from
electronegativity values of elements ?
(a) Dipole moment of a molecule
(b) Valency of elements
(c) Polarity of bonds
(d) Position in electrochemical series
24. The reaction )aq()aq(3)aq( Cl2ClOClO3 is an
example of
(a) oxidation reaction
(b) reduction reaction
(c) disproportionation reaction
(d) decomposition reaction
25. The concentration of a reactant X decreases from 0.1 M to
0.005 M in 40 min. If the reaction follows first order kinetics,
the rate of the reaction when the concentration of X is 0.01
M will be
(a) 1.73 × 10–4 M min–1 (b) 3.47 × 10–4 M min–1
(c) 3.47 × 10–5 M min–1 (d) 7.5 × 10–4 M min–1
26. Mark the false statement?
(a) A salt bridge is used to eliminate liquid junction
potential
(b) The Gibbs free energy change, G is related with
electromotive force E as G = –nFE
(c) Nernst equation for single electrode potential is
n
o
M
RTE E log a
nF
(d) The efficiency of a hydrogen-oxygen fuel cell is 23%
27. The paramagnetism of transition element compounds is
due to
(a) paired eletrons spining in opposite directions
(b) unpaired eletrons in d and f-orbitals
(c) shared valance electrons
(d) unpaired electrons in s or p-orbitals.
28. Aniline, chloroform and alcoholic KOH react to produce a
bad smelling substance which is
(a) phenyl isocyanide
(b) phenyl cyanide
(c) chlorobenzene
(d) benzyl alcohol.
29. The species with a radius less than that of Ne is
(a) Mg2+ (b) F–
(c) O2– (d) K+
30. Vapour density of the equilibrium mixture of the reaction
SO2Cl2(g) SO2(g) + Cl2(g) is 50.0. Percent dissociation
of SO2Cl2 is :
MOCK TEST 2 MT-9
(a) 33.33 (b) 35.0
(c) 30.0 (d) 66.67
31. What will be the emf for the given cell
Pt | H2 (P1) | H
+ (aq) | H2 (P2) | Pt
(a)
2
1
P
Pln
F
RT
(b)
2
1
P
Pln
F2
RT
(c)
1
2
P
Pln
F
RT
(d) None of these
32. H3PO3 is
(a) neutral (b) basic
(c) a tribasic acid (d) a dibasic acid
33. Of the following which is diamagnetic in nature?
(a) 3
6 ]FCo[ (b) 2
4 ]ClNi[
(c) 2
4 ]ClCu[ (d) 2
4[Ni(CN) ]
34. Which of the following products is formed when
benzaldehyde is treated with CH3MgBr and the addition
product so obtained is subjected to acid hydrolysis ?
(a) A secondary alcohol (b) A primary alcohol
(c) Phenol (d) tert-Butyl alcohol
35. Mole fraction of methanol in its aqueous solution is 0.5.
The concentration of solution in terms of percent by mass
of methanol is
(a) 36 (b) 50
(c) 64 (d) 72
36. The unit cell of an ionic compound is a cube in which
cations (A) occupy each of the corners and anions (B) are
at the centres of each face . The simplest formula of the
ionic compound is
(a) AB2 (b) A3B
(c) AB3 (d) A4B3
37. For orthorhombic system axial ratios are a b c and
the axial angles are
(a) 90
(b) 90
(c) 90,90
(d) 90
38. Electrolytic reduction of alumina to aluminium by Hall-
Heroult process is carried out
(a) in the presence of NaCl
(b) in the presence of fluorite
(c) in the presence of cryolite which forms a melt with
lower melting temperature
(d) in the presence of cryolite which forms a melt with
higher melting temperature
39. Consider the following complex
[Co(NH3)5CO3]ClO4.
The coordination number, oxidation number, number of d-
electrons and number of unpaired d-electrons on the meal
are respectively
(a) 6, 3, 6, 0 (b) 7, 2, 7, 1
(c) 7, 1, 6, 4 (d) 6, 2, 7, 3
40. Nylon is a :
(a) polysaccharide (b) polyester
(c) polyamide (d) all of the above
41.
Br2OH SO H3 H2O X,
X is identified as
(a) 2, 4, 6-tribromophenol
(b) 2-bromo-4-hydroxylbenzene sulphonic acid
(c) 3, 5-dibromo-4-hydroxybenzene sulphonic acid
(d) 2-bromophenol
42. The non-polar molecule is :
(a) NF3 (b) SO3
(c) CHCl3 (d) ClO2
43. The hybridization of P in 3
4PO is the same as of
(a) S in SO3 (b) N in NO–
3
(c) S in SO4
– – (d) I in ICl4
–
44. Solution of potash alum is acidic in nature. This is due to
hydrolysis of
(a) 2
4SO (b) K+
(c) Al2(SO4)3 (d) Al3+
45. When conc. HNO3 acts on our skin, the skin becomes
yellow, because
(a) HNO3 acts as an oxidising agent
(b) HNO3 acts as a dehydrating agent
(c) Nitro-cellulose is formed
(d) The proteins are converted into xanthoproteins
MT-10 CHEMISTRY
ANSWER KEY
1. (a) 2. (b) 3. (b) 4. (b) 5. (a) 6. (c) 7. (b) 8. (d) 9. (b) 10. (c)
11. (a) 12. (d) 13. (a) 14. (d) 15. (b) 16. (c) 17. (d) 18. (b) 19. (b) 20. (a)
21. (b) 22. (c) 23. (c) 24. (c) 25. (d) 26. (c) 27. (b) 28. (a) 29. (a) 30. (b)
31. (b) 32. (d) 33. (d) 34. (a) 35. (c) 36. (c) 37. (b) 38. (c) 39. (a) 40. (c)
41. (c) 42. (b) 43. (c) 44. (a) 45. (d)
HINTS & SOLUTIONS
1. (a) Angular momentum, 
2
hmvr n
(n = 2 for first excited state)
2. (b) Two Na+ ions are replaced by one Sr2+ion to maintain
electrical neutrality.
Hence, number of vacancies
= Number of Sr2+ ions doped
= 20233 1002.61002.6100.1
3. (b) Here, R = 31.6 ohm
C = 11 1
ohm
R 31.6
 = 0.0316 ohm–1
Specific conductance = conductance × cell
constant
= 0.0316 ohm–1 × 0.367 cm–1
= 0.0116 ohm–1 cm–1
Now, molar concentration = 0.5 M
(given)
= 0.5 × 10–3 mole cm–3
Molar conductance = 3
K 0.0116
C 0.5 10
 = 23.2 S cm2 mol–1
4. (b) heat
4 2 2 7(NH ) Cr O 2 3 2 2Cr O 4H O N
5. (a) Assume that each has lost a proton . So we get : HO–
, HS–, HSe–, HTe–
It can be easily seen that the volume available for the
negative charge is increasing from HO– to HTe–,
therefore
(i) volume available for the negative charge is
increasing from left to right
(ii) charge density is decreasing from left to right
(iii) basicity is decreasing from left to right
(iv) acidity of conjugate acids is increasing from left
to right
 H2Oin Cr, Mn2+ and Fe3+
are 6, 5 and 5 respectively.
13. (a) 36HCHO 4NH
 2 6 4 2
urotropine
hexamethylene
tetramine
(CH ) N 6H O
14. (d) 3
1234
2
5
3
6
HC
O
||
CHCHCHCHC
(hex 3-ene-5-one)
15. (b)
16. (c) Total V.P.,
)X(PXPXPXPP AºBAºABºBAºA 1
 = ºBAºBºA P)XP(P
Thus, torr114ºBP ; 52ºBºA PP
or torr166ºAP
Hence torr140
2
1114
2
1166P
17. (d)
2.303 ak log
t a x
 8log
24
303.2
8
1
1log
24
303.2
18. (b)
19. (b) Structural formula of glucose is
|
4
|
2
CHO
(CHOH)
CH OH
In addition to – CHO group it contains one primary
and four secondary – OH groups.
20. (a)
2
3
Hg
3 2 H O
CH CH C CH
tautomerization
3 2 2
OH
|
CH CH C CH
3 2 3CH CH COCH
21. (b) In general,
Acyloins
|||CN
aldehyde Aromatic
Ar
O
CH
OH
CArArCHO
22. (c) Pd
2 5 2 2 6C H Cl H C H
Reduction of alkyl halide by 2H in the presence of Pd
which serves as catalyst. The accumulation of
hydrogen gas occurrs on the surface of active
palladium so that fast reduction may be achieved.
23. (c) Greater the difference of electronegativities, more polar
is the bond.
24. (c) O.N. of Cl in ClO– changes from +1 to +5 and –1.
25. (d) For a first order reaction, we have
0N2.303k log
t N
2.303 0.1
k log
40 min 0.005
 
2.303 2.303
log 20 1.3010
40 min 40
Now rate = k × [reactant]
When [x] = 0.01 M
 rate 
2.303
40
 × 1.3010 × 0.01 M min–1
 = 7.5 × 10–4 M min–1.
26. (c) Correct Nernst equation is
n
o
M
2.303RTE E log a
nF
.
27. (b)
28. (a) This is isocyanide test
NH2
Aniline
 + 3
Chloroform
CHCl 3KOH
 
(Offensive smell)
Phenyl isocyanide
N C
 + 3KCl + 3H2O
29. (a)
30. (b) )g(22ClSO )g(2)g(2 ClSO
% 100
( 1)
D d
d y
 (y = 2)