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NSEC-2011-12 EXAMINATION
CHEMISTRY
CAREER POINT
INDIAN ASSOCIATION OF CHEMISTRY TEACHERS
NATIONAL STANDARD EXAMINATION IN CHEMISTRY 2011-2012
1.
Ans.
Sol.
The number of water molecules present in 0.20 g sample of CuSO4.5H2O (molar mass = 249.7) is (B) 2.14 × 1021
(A) 1.2 × 1021
22
(C) 2.14 × 10
(D) 1.2 × 1023
[B]
249.7 gm of CuSO4 .5H2O contain 5 × 6.023 × 1023 molecule
∴ 0.2g of CuSO4 .5H2O contain
5 × 6.023 × 10 23
× 0.2
249.7
= 2.14 × 1021
2.
Ans.
Sol.
The group that has the species correctly listed in the order of decreasing radius is (B) V, V2+, V3+
(A) Cu2+, Cu+, Cu
(C) F–, Br–, I–
(D) B, Be, Li
[B]
1
Cationic radii ∝
effective nuclear charge
size =
3.
V > V +2 > V +3
z eff ↑
size ↓
Ans.
The number of isomers of dibromobiphenyl (Biphenyl – C6H5-C6H5) is (A) 8
(B) 10
(C) 12
(D) 14
[C]
4.
The enthalpies of decomposition of methane (CH4)(g) and ethane (C2H6)(g) are 400 and 670 kJ mol–1,
Ans.
Sol.
respectively. The ∆HC-C in kJmol–1 is (A) 270
(B) 70
(C) 200
(D) 240
[B]
CH4 → C + 4H
∆H = 400 KJ/mol
∴ ∆H = (Σ B. E)R – (Σ B. E)P
400 = 4 C – H
∴ C–H = 100 kJ/mol
In C2H6 → 2C + 6H
∆H = 670 kJ/mol
∴ ∆H = [C–C + 6 C–H]R – [0]
670 = C–C + 6 × 100
5.
∴ C–C = 70 kJ/mol
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NSEC-2011-12 EXAMINATION
CHEMISTRY
(A) [Co3(NH3)] (NO3)3
(B) [Co3(NH3)6](NO3)3
(C) [Co(NO3)3].6NH3
(D) [Co(NH3)6](NO3)3
Ans.
[D]
Sol.
Formula of hexaammine cobalt (III) nitrate
CAREER POINT
[Co(NH3)6]+3 + NO3– ⇒ [Co(NH3)6](NO3)3
6.
For the reaction PCl3(g) + Cl2(g)
–1
PCl5(g), Kc is 26 at 250ºC Kp at the same temperature is -
–1
(R = 8.314 J K mol )
(A) 4.6 × 10–3
(B) 5.7 × 10–3
(C) 6.0 × 10–3
(D) 8.3 × 10–3
Ans.
[C]
Sol.
kP = kC (RT)∆ng
kP = 26 (8.314 × 523)–1
kP =
7.
26
= 5.9 × 10–3
4348.2
Curved arrows are used in Organic chemistry to show the movements of electrons in the mechanism of a
reaction. The correct product of the following reaction is -
O
O
(B)
(A)
(C)
O
(D)
O
O
Ans.
[C]
8.
Denaturation of protein due to change in pH could be due to (A) loss of van der Waal's interaction
(B) hydrophobic interaction
(C) change in ionic interaction
(D) breaking of covalent bonds
Ans.
[C]
9.
The initial activity of a radionuclide is 9750 counts per min and 975 counts after 5 min. The decay constant
of the radionuclide in min–1 is about (A) 0.23
(B) 0.46
(C) 0.69
(D) 0.99
[B]
Ans.
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2 / 18
NSEC-2011-12 EXAMINATION
CHEMISTRY
Sol.
K=
=
CAREER POINT
2.303
a
log
t
(a − x )
2.303
9750
log
5
975
= 0.46
10.
Ans.
Sol.
According to VSEPR theory the shape of IF5 molecule will be (A) tetrahedral
(B) trigonal bipyramid
(C) square pyramid
(D) trigonal planar
[C]
:IF5
∴ sp3d2
F
F
F
I
F
F
••
Square pyramid
11.
The formal charges on the atoms underlined are -
C6H 5 − C ≡ N − O
(A) C = 0, N = – 1, O = + 1
(C) C = 0, N = +1, O = – 1
Ans.
[C]
Sol.
&& :
C6H5–C≡N– O
..
FC = valance e– –
For C : FC = 4 –
For N : = 5 –
Ans.
Sol.
1
(bonded e–) – (non-bonded)
2
1
×8=0
2
1
(8) = 0 + 1
2
For O : FC = 6 –
12.
(B) C = – 1, N = + 1, O = –1
(D) C = +1, N = 0, O = – 1
1
(2) + 6 = – 1
2
The number of α-particles emitted per second by a radioactive element reduces to 6.25% if the original value
is 48 days. The half-life period of the element in days is (A) 3
(B) 8
(C) 12
(D) 16
[C]
2.303
100
K=
log
48
6.25
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NSEC-2011-12 EXAMINATION
CHEMISTRY
=
2.303
log 24
48
=
2.303 × 4 × 0.3
= 0.0575
48
t1/2 =
13.
Ans.
Sol.
0.693
0.693
=
= 12
k
0.0575
The compound that does not have a π-bond is (A) SO2
(C) O2
[B]
F
F
F
SF6
S
F
F
F
S
SO2
O
O
O
SO3
O
S
(B) SF6
(D) SO3
O
O=O
O2
14.
CAREER POINT
Is mass spectrometry a compound is bombarded with high energy electrons to break it into smaller
fragments, which are recorded in the form of their masses (m/Z). For example butane gives fragments like
m/z 58, 43, 29, 15, etc. The mass spectrum of an unknown compound is shown below -
43
intensity
15
29
57
72
m/z
Ans.
15.
Ans.
Sol.
The likely compound among the following is (A) CH3COCl
(C) CH3CH2COOH
[D]
(B) CH2=CH–CH2CH2OH
(D) CH3COCH2CH3
The solubility of calcium phosphate is S mol dm–3. Hence, the solubility product is (A) S5
(B) 27S3
(C) 54S4
(D) 108S5
[D]
Ca3 (PO4)2
∴ ksp = 108 55
16.
The number of valence electrons in an atom with the configuration 1s2 2s2 2p6 3s2 3p2 is -
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NSEC-2011-12 EXAMINATION
CHEMISTRY
(A) 6
(B) 5
(C) 4
(D) 2
CAREER POINT
Ans.
[C]
17.
The elevation in boiling point of a solution containing 13.44g of CuCl2 in 1 kg of water is (Kb = 0.52 K/kg mol–1)
(A) 0.05
(B) 0.10
(C) 0.16
(D) 0.21
Ans.
[C]
Sol.
∆ Tb = Kb ×
= 0.52 ×
w A × 1000
i
MA × w B
13.44 × 1000
× 0.3
134.5 × 1000
= 0.16
18.
The configuration of the carbon atoms C2 and C3 in the following compound are respectively -
HOOC
OH
H
HO
H
CHO
Ans.
(A) R, R
(B) S, S
(C) R, S
(D) S, R
[A]
COOH
2
1
H
Sol.
3
3
HO1
OH
R
H
R
CHO
2
19.
0.1 dm3 of 0.1 M acetic acid is titrated against 0.1 M NaOH. When 50 cm3 of 0.1 M NaOH are added, the pH
of the solution will be (pKa = 4.74)
Ans.
Sol.
(A) 2.37
(B) 4.74
(C) 1.34
(D) 5.74
[B]
CH3COOH + NaOH → CH3COONa + H2O
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5 / 18
NSEC-2011-12 EXAMINATION
CHEMISTRY
Initial Meq
10
Final
5
Resulting solution is an acidic buffer
∴ pH = pKa + log
5
0
CAREER POINT
0
5
0
–
[Salt]
5
= 4.74 + log
[Acid]
5
pH = 4.74
20.
Ans.
Sol.
21.
Ans.
22.
The IUPAC name of complex [Cu(en)2(H2O)2]+ is (A) ethylene diamine Cu(II) dihydrate
(B) diaquobis(ethylenediamine) Cu(II) ion
(C) diaqobisdiethylamine Cu(II) ion
(D) diaquobis(ethylenediamine) cuprate (II)
[B]
[Co(en)2(H2O)2]+
Diaqua bis(ethylene diamine)Cu(II) ion
Two protein molecules with the same average molecular mass (molecular weight) can absorb different
amount of ultraviolet radiation due to difference in the content of (A) tyrosine
(B) glutamic acid
(C) lysine
(D) methionine
[A]
Each of the following options contains a structure and a description indicating the existence of given
structure. The correct option for methyl 3-hydroxypent-2-enoate is (A)
(C)
COOCH3 YES
OH
OH
NO
(B)
(D)
COOCH3 NO
OH
OH
COOCH3
COOCH3
Ans.
[B]
23.
Major product of mononitration of the following compound is
O
O
(A)
O
YES
O
O
O
(B)
NO2
NO2
O
(C)
O
NO2
O
(D)
O
O 2N
Ans.
[D]
24.
If a ≠ b ≠ c and α = β = γ = 90º, the crystal system is CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-3040000
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NSEC-2011-12 EXAMINATION
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(A) monoclinic
(B) triclinic
(C) hexagonal
CAREER POINT
(D) orthorhombic
Ans.
[D]
25.
The electronic spectrum of [Ni(H2O)6]++ shows a band at 8500 cm–1 due to d–d transition. [Ph4As]2[NiCl4]
will have such a transition in cm–1 at (A) 3778
(B) 8500
(C) 4250
26.
In the conductometric titration of silver nitrate against KCl, the graph obtained is -
(B)
(C)
volume of KCl
Conductance
volume of KCl
(D)
volume of KCl
Conductance
(A)
Conductance
[A]
Conductance
Ans.
(D) 850
volume of KCl
Ans.
[B]
Sol.
AgNO3 + KCl → AgCl(↓) + KNO3
27.
The product obtained from the following sequence of reactions is NaBH
HgSO
H3C–C≡CH  4 → A  
4 → B
H 2SO 4
(A) propanal
(B) 2-propanol
(C) 1-propanol
Ans.
[B]
28.
The compound in which Mn has oxidation state of +3 is
(A) KMnO4
(B) K2[Mn(CN)6]
(C) MnSO4
(D) CsMn(SO4)2.12H2O
Ans.
[D]
29.
The SI units of viscocity is
(A) Nsm2
(B) Ns2m
(C) Nsm–2
(D) propane
(D) Ns–2m
Ans.
[C]
30.
If titration of an amino acid present in the solution yielded pI (isoelectric point) value of 10.80, the amino
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NSEC-2011-12 EXAMINATION
CHEMISTRY
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acid present in the solution may be
(A) glycine
(B) arginine
(C) histidine
Ans.
[B]
31.
In the coordination compound, Na2[Pt(CN)4], the Lewis acid is (A) [Pt(CN)4]2–
(B) Na+
(D) proline
(C) Pt2+
Ans.
[C]
32.
The product (P) of the following reaction is
CH3
(D) CN–
Lewis acid
+ ICl  → P
Cl
I
(A)
(B)
H 3C
CH3
Cl
(C)
(D)
H 3C
I
Ans.
[B]
33.
The correct order of dipole moment for the following molecules is -
OH
(I)
Cl
NO2
(A) I = II = III
(II)
CH3
Cl
(B) I < II < III
Ans.
[C]
34.
Lead dissolves most readily in dilute -
(III)
CH3
(C) I > II > III
(A) acetic acid
(B) sulphuric acid
(C) phosphoric acid
(D) sodium hydroxide
Ans.
[A]
35.
The degrees of freedom for the system CaCO3(s)
(A) 1
(B) 2
(D) II < III < I
CaO(s) + CO2(g) are (C) 3
Ans.
[A]
36.
Semipermeable nature of the cell membrane can be attributed to the presence of CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-3040000
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(D) 4
8 / 18
NSEC-2011-12 EXAMINATION
CHEMISTRY
Ans.
37.
Ans.
Sol.
(A) protein and DNA
(C) polysaccharide and lipid
[B]
(B) lipid and protein
(D) DNA and lipid
The emf of the cell (Zn | ZnSO4 (0.1M) || CdSO4 (0.0.1M) | Cd) is
(E0 Zn2+| Zn = – 0.76 V, E0 Cd2+| Cd = 0.40 V at 298 K)
(A) + 0.33 V
(B) + 0.36 V
(C) + 1.13 V
[C]
E°cell = 0.76 + 0.4 = 4.16
Emf = E°cell –
= 1.16 –
CAREER POINT
(D) – 0.36 V
0.059
[ Zn +2 ]
log
n
[Cd + 2 ]
0.059
0.1
log
2
0.01
= 1.16 – 0.03 log 10 = 1.13 V
38.
Ans.
39.
The nitrogen compound formed when Ca(CN)2 reacts with steam or hot water is
(A) N2O
(B) NO
(C) NO2
(D) NH3
[D]
The order of acidities of the H-atoms underlined in the following compounds is in the order
(I) Ph-CH2–CH3 (II) Ph-C≡CH
Ans.
(III) Ph-CH=CH2 (IV)
(A) IV > II > I > III
(C) III > IV > I > II
[A]
CH2
(B) II > IV > III > I
(D) I > III > II > IV
Ans.
The half time for a second order reaction with equal concentrations of the reactants is 35 seconds. 99%
reaction will be completed in
(A) 69s
(B) 138s
(C) 1733s
(D) 3465s
[D]
Sol.
For second order t1/2 =
40.
Let initial is 100 ∴
∴t=
1
= 35
Ka
1
= 35 × 100
K
1  x 


K  a (a – x ) 
 99 
= 3500 

100 ×1 
= 3465 sec
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NSEC-2011-12 EXAMINATION
CHEMISTRY
41.
CAREER POINT
The 'd' orbitals will be split under square planar geometry into (A) two levels
(B) three levels
Ans.
[C]
42.
Rotational spectra of molecules are observed in -
(C) four levels
(A) UV region
(B) Visible region
(C) Near infrared region
(D) Far infrared region
Ans.
[D]
43.
The pair of cations which cannot be separated by H2S in a 0.3N acid solution is (A) Al+++, Hg++
(B) Bi+++, Pb++
(C) Zn++, Cu++
(D) Ni++, Cd++
(D) five levels
Ans.
[B]
44.
Structural features of proteins secreted outside the cells may be stabilised by presence of (A) hydrogen bond
(B) disulfide bond
(C) hydrophobic force
(D) phospho-diester bond
Ans.
[B]
Sol.
Most secreted protein and extra cellular domains of membrane protein contain disulphide bond. These
covalent bonds are important for the proper folding and stability of secretary protein.
45.
The C–O–C bond angle in dimethyl ether is (A) 109º28'
Ans.
(B) 110º
(C) 120º
(D) 180º
[B]
:O:
CH3 110º CH3
Large size of methyl group. So bond angle greater than 109º 28'
46.
Dimethyl glyoxime forms a square planar complex with Ni2+. This complex should be
(A) diamagnetic
(B) paramagnetic having 1 unpaired electron
(C) paramagnetic having 2 unpaired electrons
(D) ferromagnetic
Ans.
[A]
Sol.
–
Ni+2 + 2DMgH OH

→ [Ni (DMg)2]
Square planar dsp2
Ni+2 = [Ar] 3d8
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NSEC-2011-12 EXAMINATION
CHEMISTRY
3d
4s
CAREER POINT
4p
Ni+2 =
3d
4s
4p
2
dsp
n = 0 ∴ diamagnetic
47.
A 0.056 M solution of benzoic acid, C6H5COOH, is titrated with a strong base. [H+] of the solution when half
of the solution is titrated before the equivalence point is (Ka of benzoic acid = 6.3 × 10–5)
(A) 6.3 × 10–5 M
Ans.
[A]
Sol.
[H+] = Ka ×
(B) 1.8 × 10–3 M
(C) 7.9 × 10–3 M
(D) 2.6 × 10–2 M
(C) ONC–
(D) CN–
[Acid]
[Salt ]
0.056
0.056
= 6.3 × 10–5 ×
= 6.3 × 10–5 M
48.
The formula of the isothiocyanate ion is (A) OCN–
(B) SCN–
Ans.
[B]
49.
The compound that is chiral is (A) 3-methyl-3-hexene
(B) 4-chloro-1-methylcyclohexane
(C) 2-phenylpentane
(D) 1, 3-diisopropylbenzene
Ans.
[C]
50.
The monomer/s of the following polymer is/are
(–CH2–CH(CH3)–CH2–CH(CH3)–CH2–CH(CH3)–)n
(A) ethylene
(B) propylene
(C) 2-butene
(D) ethylene + propylene
Ans.
[B]
51.
Of the interhalogen compounds, ClF3 is more reactive than BrF3, but BrF3 has higher conductance in the
liquid state. The reason is that
(A) BrF3 has higher molecular weight
2–
4–
(C) BrF3 dissociates into BrF and BrF more easily
(B) ClF3 is volatile
(D) ClF3 is most reactive
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NSEC-2011-12 EXAMINATION
CHEMISTRY
Ans.
[C]
Sol.
High E.N difference in Br & F
CAREER POINT
∴ high polarity
∴ Ions form easily
52.
An element X is found to combine with oxygen to form X4O6. If 8.40 g of this element combine with 6.50 g
of oxygen, the atomic weight of the element in grams is (A) 24.0
(B) 31.0
Ans.
[B]
Sol.
6.5 g oxygen combine with 8.4 g
∴ 96 g oxygen combine with
∴ At. wt of one X =
53.
(C) 50.4
(D) 118.7
8.4 × 96
= 124.06
6.5
124.06
= 31.0
4
Synthesis of RNA in a cell would be affected adversely due to shortage of
(A) sulfate
(B) acetate
Ans.
[D]
54.
The most abundant element in the earth's crust is (A) aluminium
(B) oxygen
(C) oxalate
(D) phosphate
(C) silicon
(D) iron
Ans.
[B]
55.
A beaker is heated from 27ºC to 127ºC, the percentage of air originally present in beaker that is expelled is
(A) 50%
(B) 25%
(C) 33%
Ans.
[B]
Sol.
n1T1 = n2T2
56.
The product (C) of the following sequence of reactions is
(D) 40%
i ) AlCl3

→ A Cl
2 (lim)

→ B aq
NaOH
→ C
+ CH2=CH2 (
( ii ) H 2O
hν
OH
(A)
(B)
HO
OH
(C)
Ans.
Cl
HO
(D)
[D]
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NSEC-2011-12 EXAMINATION
CHEMISTRY
CAREER POINT
IAPT has given (c) as the correct answer to this question.
But the most appropriate answer to this question should be (d). For more details refer the solution given below.
OH
Cl
+ CH2 = CH2
57.
(i) AlCl3
(ii) H 2 O
CH—CH3
CH–CH3
CH2–CH3
Cl /
aq
.NaOH
→
2


→
The strongest, but the most reactive bond among the following is (A) C=N
(B) C=C
(C) C≡C
(D) C=O
Ans.
[C]
58.
Ans.
Radioactive inert gas is (A) technetium
(B) radon
[B]
59.
The IUPAC name of the following compound is -
(C) xenon
(D) curium
OC2H5
O
Ans.
60.
Ans.
Sol.
(A) 3-methoxy ethylpropanoate
(C) 1, 4-diethoxybutane
[B]
O
(B) ethyl 4-methoxybutanoate
(D) ethoxy 3-methoxybutyrate
Excess of silver nitrate is added to a water sample to determine the amount of chloride ion present in the
sample. 1.4 g of silver chloride is precipitated. The mass of chloride ion present in the sample is
Molar masses (g. mol–1); AgNO3 169.91, AgCl
143.25
(A) 0.25 g
(B) 0.35 g
(C) 0.50 g
(D) 0.75 g
[B]
143.25 AgCl obtained from 169.91 g AgNO3
then 1.4 g AgCl obtained from 169.91 g Ag NO3
=
169.91
× 1.4 = 1.66 g
143.25
In 169.91 g of sample chlorine is 35.5 g
∴ 1.66 g of sample chlorine
=
61.
The best nucleophile among the following is (A) H2O
Ans.
35.5
× 1.66 = 0.35 g
169.91
(B) CH3SH
(C) Cl¯
(D) NH3
[B]
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NSEC-2011-12 EXAMINATION
CHEMISTRY
62.
CAREER POINT
The wavelength of a moving body mass 0.1 mg is 3.31 × 10–29 m. The kinetic energy of the body in J would
be (A) 2.0 × 10–6
Ans.
[D]
Sol.
λ=
h
2mKE
(B) 1.0 × 10–3
=
6.62 × 10 −34
2 × 0.1 × 10
−6
× KE
(C) 4.0 × 10–3
(D) 2.0 × 10–3
= 3.31 × 10 − 29
2 × 10 −7 × KE = 2 × 10 −5
∴ 2 × 10–7 × KE = 4 × 10–10
KE = 2 × 10–3
Ans.
Secondary structures could be formed in nucleic acid similar to protein due to formation of (A) covalent bond
(B) ionic bond
(C) co-ordinate bond
(D) hydrogen bond
[D]
64.
The following titration curve represents the titration of a _____ acid with a ____ base -
63.
pH 7
mL added
Ans.
65.
Ans.
66.
Ans.
Sol.
67.
Ans.
(A) strong, strong
[A]
(B) weak, strong
The element with the lowest electronegativity is (A) S
(B) I
[C]
(C) strong, weak
(D) weak, weak
(C) Ba
(D) Al
Oxalic acid, H2C2O4, reacts with permanganate ion according to the balanced equation
5H2C2O4(aq) + 2MnO4¯(aq) + 6H+(aq)
2Mn2+(aq) + 10CO2(g) + 8H2O(l).
The volume in mL of 0.0162 M KMnO4 solution required to react with 25.0 mL of 0.022 M H2C2O4 solution
is(A) 13.6
(B) 18.5
(C) 33.8
(D) 84.4
[A]
Equivalent of KMnO4 = Eq. of H2C2O4
N1V1 = N2V2
0.0162 × 5 × V1 = 0.022 × 2 × 25
V1 = 13.6 ml
The element that has the highest tendency to catenate is (A) silicon
(B) germanium
(C) sulphur
[C]
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(D) boron
14 / 18
NSEC-2011-12 EXAMINATION
CHEMISTRY
68.
Ans.
69.
Ans.
70.
Ans.
71.
Ans.
Sol.
CAREER POINT
The isotope of carbon which is used in carbon dating (a method to estimate the age of an ancient sample
containing carbon) is (A) carbon-12
(B) carbon-13
(C) carbon-14
(D) carbon-15
[C]
Electronic configurations for the atoms of four elements are given below. The configuration that indicates
colourless aqueous solution is (A) 2,8,14,2
(B) 2,8,16,2
(C) 2,8,18,2
(D) 2,8,13,1
[C]
The number of stereoisomers of compound CH3–CH=CH–CH(Br)CH3 is (A) 2
(B) 3
(C) 4
(D) 6
[C]
At 445ºC, Kc for the following reaction is 0.020.
H2(g) + I2(g)
2HI(g)
A mixture of H2, I2 and HI in a vessel at 445ºC has the following concentrations :
[HI] = 2.0 M, [H2] = 0.50 M [I2] = 0.10 M. The statement that is true concerning the reaction quotient, Qc is (A) Qc = Kc; the system is at equilibrium
(B) Qc is less than Kc; more H2 and I2 will be produced
(C) Qc is less than Kc; more HI will be produced
(D) Qc is greater than Kc; more H2 and I2 will be produced
[B]
[H 2 ] [I 2 ]
Q=
[HI]2
0.5 × 0.1
= 0.0125
2× 2
∴ Q < KC and forward reaction.
=
72.
Ans.
The order of decreasing stability is -
(I)
(II)
(III)
(IV)
(A) IV > I > II > III
[A]
(B) I > IV > III > II
(C) I > II > IV > III
(D) IV > II > I > III
Ans.
The number of amino acid residues found in a protein that is synthesized from a RNA molecule with 120
nucleotides is (A) 120
(B) 80
(C) 40
(D) 60
[C]
74.
Hypochlorous acid ionizes as -
73.
HOCl (aq)
H+(aq) + OCl¯ (aq).
OCl¯(aq) + H2O(l)
HOCl(aq) + OH¯(aq)
Ka for this reaction at 25ºC is 3.0 × 10–8 (Kw = 1.0 × 10–14 at 25ºC)
Hence, Kh for HOCl is CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-3040000
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15 / 18
NSEC-2011-12 EXAMINATION
CHEMISTRY
(A) 3.3 × 10–7
Ans.
[A]
Sol.
Kh =
(B) 3.0 × 10–8
CAREER POINT
(C) 3.0 × 106
(D) 3.3 × 107
Kw
10
10 −14
1
× 10–7
=
= × 10–6 =
−8
3
ka
3
3 × 10
Kh = 3.3 × 10–7
75.
Einsteinium has 11 electrons in the 4f subshell. The number of unpaired electrons in the subshell is (A) 3
(B) 4
(C) 7
Ans.
[A]
76.
The order of reactivity of ammonia with the following compounds is -
(D) 11
(I) CH2=CHBr
(II) CH3–CH2–COCl
(III) CH3–CH2–CH2–Cl
(IV) (CH3)3C–Br
(A) IV > II > I > III
(B) II > IV > III > I
(C) III > IV > II > I
(D) I > IV > II > III
Ans.
[B]
77.
The freezing point of a solution containing 8.1g of HBr in 100 g of water, assuming the acid to be 90%
ionized is [H=1, Br=80, Kf for water = 1.86 K kg mol–1]
(A) 0.85ºC
Ans.
[B]
Sol.
α=
(B) –3.53ºC
i −1
n −1
0.9 =
(C) 0ºC
(D) –0.35ºC
(∆Tf)obs = i × (∆Tf)theo.
i −1
2 −1
= 1.9 × kf ×
i = 1.9
WA ×1000
M A × WB
= 1.9 × 1.86 ×
8.1 × 1000
81 × 100
(∆Tf)obs = 3.53
∴ T' = –3.53°C
78.
The reaction of 50% aq KOH on an equimolar mixture of 4-methylbenzaldehyde and formaldehyde followed
by acidification gives OH
OH
(A)
OH
(B)
H3C
+HCOOH
H3C
(C
)
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NSEC-2011-12 EXAMINATION
CHEMISTRY
CAREER POINT
COOH
COOH
(D)
+CH3OH
OH
Ans.
H3C
[B]
79.
Iodide ion is oxidized by acidified dichromate ions as shown in this equation.
Cr2O72–(aq) + 9 I¯(aq) + 14 H+(aq) —→ 2Cr3+(aq) + 3I3¯ (aq) + 7H2O (l). These data were obtained when
Ans.
the reaction was studied at a constant pH. The order of the reaction with respect to Cr2O72– (aq) and I¯(aq)
are [I¯] M
Rate, M.s–I
Experiment
[Cr2O72–], M
1
0.0050
0.0125
0.00050
2
0.010
0.0125
0.0010
3
0.0150
0.0250
0.0060
2–
(A) first order with respect to both Cr2O7 and I¯
(B) second order with respect to both Cr2O72– and I¯
(C) second order with respect to Cr2O72– and first order with respect to I¯
(D) first order with respect to Cr2O72– and second order with respect to I¯
[D]
Sol.
r = k[Cr2O7−2 ]m [I − ]n
0.0005 = k[0.005]m [0.0125]n
0.0010 = k [0.01]m [0.0125]n
0.006 = k[0.015]m [0.025]n
equation (1)/(2)
1 1
= 
2 2
….(1)
….(2)
….(3)
m
m=1
equation (2)/(3)
m
1 2 1
=   
6 3 2
1 1
= 
4 2
80.
n
;
1 2 1
= × 
6 3 2
n
n
∴n=2
The number of atoms per unit cell and the number of the nearest neighbours in a body centred cubic structure
are (A) 4, 12
(B) 2, 6
(C) 9, 6
Ans.
[D]
Sol.
1
No. of per unites cell of BCC = 8 × + 1 × 1 = 2 ∴ and C.No. = 8
8
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(D) 2, 8
17 / 18
CHEMISTRY
NSEC-2011-12 EXAMINATION
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