Description
Medical College Admission Test (MCAT) which is accepted by medical schools in US, Australia, Canada and Carribean Islands for making admission decision.The test assesses candidates critical analysis and reasoning skills, biological and biochemical foundations of living systems, psychological, social and biological foundations of behaviour and chemical and physical foundations of biological systems.
In this collection there will be many samples to practice this test
It will be so important to Medical students who wants to prepare for this test
Study Set Content:
Question 11
Which of the following are nucleophiles?
a)
H
+
b)
H
3
O
+
c)
CO
2
d)
AlCl
3
e)
NH
3
Correct Answer:
e) NH
3
Explanation:
Generally, electron pair donors are called as nucleophiles. In the case of ammonia, it has a lone of
pair of electrons. So, it can donate the electron to the other atoms. But, the rest of the atoms can't
donate a pair of electrons instead they can accept a pair of electrons. So, except ammonia all the
other molecules above are known as electrophiles.
Question 12
The ratio of relative abundance of two molecular ion peaks of a chlorine atom in mass
spectroscopy is
a)
M
+
: (M+2) = 1:3
b)
M
+
: (M+2) = 3:1
c)
M
+
: (M+2) = 1:2
d)
M
+
: (M+2) = 2:1
e)
M
+
: (M+2) = 1:1
Correct Answer:
b) M
+
: (M+2) = 3:1
Explanation:
The ratio of relative abundance of two molecular ion peaks of a chlorine atom in mass spectroscopy
is M
+
: (M+2) = 1:3, because the natural abundance of chlorine is higher than other elements. The
natural abundance of chlorine is Cl
35
- 75.77% and Cl
37
t- 24.23%.
Question 13
Compound A reacts by first order kinetics. The rate constant of the reaction is 0.45 sec
-1
. Calculate
the half-life of the compound A in the reaction.
a)
4.62 seconds
b)
3.08 seconds
c)
1.54 seconds
d)
2.25 seconds
e)
0.9 seconds
Correct Answer:
c) 1.54 seconds
Explanation:
Half-life period of the first order reaction can be calculated using the following formula, which is
derived from the first order rate law.
t
1/2
= 0.693/k
1
seconds.
Therefore, t
1/2
= 0.693/ 0.45 sec
-1
= 1.54 seconds.
Question 14
The standard EMF value of a reaction I is E
o
cell
= -1.563 and for a reaction II it is E
o
cell
= +0.86. Predict
the feasibility of these reactions.
a)
Reaction I is not feasible, reaction II is feasible
b)
Reaction I is feasible, reaction II is not feasible
c)
Both the reactions are feasible
d)
Both the reactions are not feasible
e)
Can't be predicted using these values.
Correct Answer:
a) Reaction I is not feasible, reaction II is feasible
Explanation:
In general, if E
o
cell
is positive, the reaction is feasible. If E
o
cell
is negative, the reaction is not feasible.
Here, the E
o
cell
of the reaction I is negative, therefore the reaction is not feasible and E
o
cell
of the
reaction II is positive, therefore the reaction is feasible.
Question 15
“No pairing occurs until all orbitals of a given sub level are half filled”. The statement is known as
a)
Exclusion principle
b)
Uncertainty principle
c)
Hund's rule
d)
Aufbau principle
e)
Bohr's theorem
Correct Answer:
c) Hund's rule
Explanation:
Exclusion principle states that it is impossible for any two electrons in a given atom to have all the
four quantum numbers identical. Hund's rule states, “No pairing occurs until all orbitals of a given
sub level are half filled”. Aufbau principle defines that in the ground state of the atoms the orbitals
are filled in order of their increasing energies. Uncertainty principle states that the position and
momentum of a particle can't be simultaneously measured with high precision.
Sampe paper 3
Question 1
What is the percentage composition of sodium and sulphur and oxygen in sodium sulphate?
a) 30.4, 22, 45
b) 31, 24, 46
c) 31.4, 22.8, 45.7
d) 30.1, 22.8, 46.7
e) 30, 21.8, 46.7
Correct Answer:
c) 31.4, 22.8 & 45.7
Explanation:
Method is as follows:
Molecular formula for sodium sulphate is Na
2
SO
4
Molar mass of sodium sulphate is 44+32+64 = 140
Mass percent of sodium is (2*22) / (140) * 100 = 31.4
Mass percent of sulphur is (1*32) / (140) * 100 = 22.8
Mass percent of oxygen is (4*16) / (140) * 100 = 45.7
Question 2
Identify the anode, cathode, oxidation half-cell and reduction half-cell in the following cell
diagram.
a) A-anode, reduction half-cell; B- cathode, oxidation half-cell
b) A-cathode, reduction half-cell; B- anode, oxidation half-cell
c) A-anode, oxidation half-cell; B- cathode, reduction half-cell
d) A-cathode, oxidation half-cell; B- anode, reduction half-cell
e) A- anode, reduction half-cell; B- cathode, reduction half-cell
Correct Answer:
c) A-anode, oxidation half-cell; B- cathode, reduction half-cell
Explanation:
Explanation: The anodic reaction is always to be written on the left hand of the cell diagram and the
cathodic reaction is to be written on the right hand side. An oxidation reaction always takes place at
the anode and the reduction reaction always takes place at the cathode.
Question 3
Which of the following is an example for an isolated system?
a) A pot of boiling water
b) Boiling a soup in an open sauce pan in a stove
c) Cooking rice in a pressure cooker
d) An open tank of water
e) Hot water in a thermos flask
Correct Answer:
e) Hot water in a thermos flask
Explanation:
Hot water in a thermos flask is an example for an isolated system where neither energy nor matter
can enter or exit. Boiling a soup in an open sauce pan in a stove is an example for open system in
which, it can freely exchange its energy and matter with its surroundings. Cooking rice in a pressure
cooker is an example for a closed system where it can exchange only energy with its surroundings. A
pot of boiling water and an open tank of water is an example for an open system as the matter gets
exchanged with its surroundings.
Question 4
The bond order for helium molecule is
a) 1
b) 2
c) 2.5
d) 3
e) 0
Correct Answer:
e) 0
Explanation:
The
electronic configuration of helium in the ground state is represented as (σ
1s
)
2
and in the excited
state, it is represented as (σ
*
1s
)
2
.
So, number of electrons in bonding molecular orbital (N
b
) is 2 and number of electrons in antibonding
molecular orbital (N
a
) is 2.
The bond order for He
2
is 0, so the molecule does not exist.
Question 5
The order of ionisation energy is
a) s< p < d < f
b) s> p > d > f
c) s> p > d < f
d) s< d < p < f
e) s< p < f < d
Correct Answer:
b) s > p > d > f
Explanation:
The ionisation energy depends upon the atomic radius. As the s-orbital electrons remain closer to the
nucleus, the ionisation energy will be greater for s- orbital than for p-, d- and f- orbitals.
Question 6
Which of the following alkyl halides will undergo faster S
N
2
reactions?
a) CH
3
– X
b) 1
0
alkyl halide
c) 2
0
alkyl halide
d) 3
0
alkyl halide
e) All of these
Correct Answer:
a) CH
3
– X
Explanation:
When the number of R groups in the carbon atoms gets increased, the reactivity of S
N
2
reactions will
be decreased. So, methyl halide will undergo faster S
N
2
reaction than primary, secondary and tertiary
alkyl halides. Tertiary alkyl halide does not undergo S
N
2
reaction because of steric hindrance.
Question 7
Which apparatus is used for mixing of organic chemicals?
a) Erlenmeyer flask
b) Florence flask
c) Beaker
d) Measuring jar
e) Buckner funnel
Correct Answer:
b) Florence flask
Explanation:
Florence flask is used for mixing of organic chemicals. It is commonly called as round-bottom flask or
boiling flask. Its narrow neck prevents the splash exposure.
Erlenmeyer flask is used for volumetric titration. It is also called as conical flask.
Buckner funnel is used for vacuum filtration.
Question 8
The relative intensity of signals in proton NMR is related to
a) Chemical shift and magnetic environment of proton
b) Different number of protons
c) Number of adjacent atoms containing number of protons
d) Total number of protons present in the molecule
e) Coupling constant
Correct Answer:
d) Total number of protons present in the molecule
Explanation:
The relative intensity of signals in proton NMR is proportional to total number of protons present in the
molecule. Number of signals indicates how many different kinds of protons are present in the
molecule. Position of signals indicates the chemical shift and magnetic environment of proton.
Splitting of signals indicates the number of adjacent atoms containing different number of protons.
Question 9
For acid-base equilibrium, the reaction always favours theformation of the
a) Strongest acid and the Strongest base
b) Weakest acid and the strongest base
c) Weakest acid and the weakest base
d) Strongest acid and the weakest base
e) Either a or c
Correct Answer:
c) Weakest acid and weakest base
Explanation:
In an acid-base equilibrium, the reaction always favours the formation of the weakest acid and the
weakest base due to their stability. The weakest acid and the weakest base must always be on the
same side of the reaction.
Question 10
The oxidation state of oxygen in OF
2
is
a) -2
b) +1
c) -1
d) +2
e) 0
Correct Answer:
d) +2
Explanation:
It's an unusual compound, in which oxygen takes the positive oxidation number +2. Because F is
more electronegative than O, therefore F gets the -1 oxidation state and O gets the positive oxidation
state.
Question 11
Calculate the angle at which second order reflection will occur in an X-ray spectrometer when
X-
rays of wavelength 1.54λ are diffracted by atoms of a crystal, with interplanar distance of
4.04 A
0
.
a) 10
0
59'
b) 22
0
24'
c) 24
0
22'
d) 59
0
10'
e) 12
0
50'
Correct Answer:
b) 22
0
24'
Explanation:
Given data: λ = 1.55, d = 4.04 A
0
For second order reflection n=2,
Bragg equation is 2d sin Θ = 2λ
Θ = sin
-1
(λ/d) = sin
-1
(0.381) = 22
0
24'
Question 12
The unit of surface tension is
a) erg cm
-2
b) N
c) dyne cm
-1
d) N m
-1
e) Both c and d
Correct Answer:
e) Both c and d
Explanation:
Surface tension is generally expressed in terms of dyne cm
-1
. In SI unit, surface tension is expressed
in terms of N m
-1
.
Question 13
Pick out the correct statement related to Boyle's law
a) When pressure increases, volume also gets increased
b) Graph of pressure Vs. volume gives straight line
c) P
1
* P
2
= V
1
* V
2
d) P a 1/V
e) None of these
Correct Answer:
d) P a 1/V
Explanation:
When pressure increases, volume gets decreased. Graph of pressure Vs volume gives curve. P
1
*
V
1
= P
2
* V
2
.
Question 14
The defect which is generally found in compounds of transition metals having variable valency
is
a) Schottky defect
b) Frenkel defect
c) Metal excess defect
d) Metal deficiency defect
e) Line defect
Correct Answer:
d) Metal deficiency defect
Explanation:
Metal deficiency defect is due to cation vacancy. Schottky defect is commonly found in ionic crystals
in which cations and anions are of similar size. Frenkel defect is shown by ionic crystals of different
size. Metal excess defect is due to anion vacancy and interstitial cation. In line defect, groups of
atoms are arranged in an irregular position.
Question 15
When a decomposition reaction involves redox reactions, it is called
a) Single displacement reactions
b) Internal redox reaction
c) Simple redox reaction
d) Disproportionation reaction
e) None of these
Correct Answer:
b) Internal redox reaction.
Explanation:
When a decomposition reaction involves redox reactions; it is called asinternal redox reaction,
because the oxidized and reduced elements originate in the same compound. N in NH
4
+
is oxidized
from -3 to 0 and N in NO
2
-
is reduced from +3 to 0. Both redox reactions occur in the same
NH
4
NO
2
molecule.
Ex: NH
4
NO
2(s)
→N
2(g)
+ 2 H
2
O
(g)
In single displacement reactions, atom of one reactant replaces the atom of the other reactant.
Ex: a A + b BC→c AC + d B
In simple redox reaction,oxidation numbers of ionic reactants are changed by the direct transfer of
electrons from one ion to the other.
Ex: 2 Fe
3+
(aq)
+ Sn
2+
(aq)
→ 2 Fe
2+
(aq)
+ Sn
4+
(aq)
In a disproportionation reaction, the same species will simultaneously get oxidised and reduced to
form two different products.
Sample paper 4
Question 1
How many significant figures are there in the following figures?
i. 6*10
4
ii. 0.008320
iii. 4.05*10
-2
iv. 100.0
a) 1, 7, 5, 4
b) 5, 6, 5, 3
c) 1, 4, 3, 4
d) 1, 7, 5, 3
e) 1, 4, 3, 1
Correct Answer:
c) 1, 4, 3, 4.
Explanation:
6*10
4
= 6000 has only one significant figure. Leading zeros are not significant, for 0.008320 it is 4.
Zeros appearing anywhere between two non-zero digits are significant figures, for 4.05*10
-2
=
0.00405, it is 3. Trailing zeros in a number containing a decimal point are significant, for 100.0 it is 4.
Question 2
One mole of any gas at STP occupies
a) 0.224 L
b) 0.022L
c) 2.24 L
d) 22.4 L
e) 23 L
Correct Answer:
d) 22.4 L
Explanation:
By applying Ideal gas equation, V = nRT/P
At STP, P=1 atm, n=1mol, R=0.082 L atm K
-1
mol
-1
, T=273K
V = (1*0.082*273)/273 = 22.38 L = 22.4 L
Question 3
The conversion of liquid to solid is known as
a) Melting
b) Freezing
c) Sublimation
d) Condensation
e) Deposition
Correct Answer :
b) Freezing
Explanation:
Freezing
– liquid to solid, melting – solid to liquid, sublimation – solid to vapour, condensation – gas to
liquid, deposition
– gas to solid.
Question 4
Identify the unit of concentration of the solution (N
A
)/(Kg of solvent).
a) Molarity
b) Molality
c) Normality
d) Mole fraction
e) ppm
Correct Answer:
b) Molality
Explanation:
Molarity (M
A
) = n
A
/ volume in litres.
Normality = Gram equivalent of A/Volume in litres of solution.
Mole fraction (χ
i
) = n
i
/ (n
1
+n
2
+n
3
....).
Parts per million (ppm) = (Mass of A/Total mass) x 10
6
Question 5
In a polyatomic species, the sum of oxidation numbers of the element in the ion _________ the
charge on that species.
a) Is greater than
b) Is lesser than
c) Equals
d) Is either greater or lesser than
e) Is zero to
Correct Answer:
c) Equals
Explanation:
The sum of oxidation numbers in polyatomic ion or species is equal to the charge of the ion. For
example, the sum of the oxidation number for SO
4
2-
is -2.
Question 6
In which of the following processes, is the process always non-feasible?
a)
ΔH>0, ΔS>0
b)
ΔH<0, ΔS>0
c)
ΔH>0, ΔS<0
d)
ΔH<0, ΔS<0
e)
ΔH=0, ΔS=0
Correct Answer:
c) ΔH>0, ΔS<0
Explanation:
For a non-spontaneous or non-
feasible process, ΔH>0 and ΔS<0. For a spontaneous or irreversible
reaction, ΔH<0 and ΔS>0. For an equilibrium or reversible process, ΔH=0 and ΔS=0.
Question 7
The hybridisation in NH
4
+
is
a) sp
b) sp
2
c) sp
3
d) sp
3
d
e) sp
3
d
2
Correct Answer:
c) sp
3
Explanation:
Number of valence electrons in N is 5 and in H it is 4.
So total number of valence electrons = 5 + 4 = 9; Charge = +1.
Therefore, total electrons in NH
4
+
= 9 - 1 = 8
When the total number of electrons is less than 8, divide by 2. If it lies between 9 and 56, divide it by
8.
8/2 = 4; X=4
Therefore, hybridisation in NH
4
+
is sp
3
.
Question 8
Slater's rule is used to calculate the value of
a) Screening constant
b) Electron affinity
c) Ionisation energy
d) Effective nuclear charge
e) Both a and d
Correct Answer:
e) Both a and d
Explanation:
The value of screening constant (S) and effective nuclear charge (Z
*
) can be calculated using Slater's
rule. Effective charge (Z
*
) = Z
– S (where Z- atomic number and S-screening constant).
Question 9
Which of the following solvents is suitable for S
N2
reactions?
a) Ethanol
b) Water
c) Acetonitrile
d) Acetic acid
e) t-butanol
Correct Answer:
c) Acetone
Explanation:
Aprotic solvents do not solvate the anions effectively and it is used for S
N
2
reactions. Acetonitrile is the
only aprotic solvent whereas others are polar protic solvents.
Question 10
Identify the glass equipment with ground-glass joints
a) Graduated pipette
b) Erlenmeyer flask
c) Buckner funnel
d) Separating funnel
e) Funnel
Correct Answer:
d) Separating funnel
Explanation:
Glass equipments are divided into two; with ground-glass joints and without ground-glass joints.
Separating funnel is the only glass equipment with ground-glass joints.
Question 11
The base peak in a mass spectrum is
a) The peak set to 100 % relative intensity
b) The peak set to 0 % relative intensity
c) The peak corresponding to the parent ion
d) The highest mass peak
e) The lowest mass peak
Correct Answer:
a) The peak set to 100 % relative intensity
Explanation:
The most intense peak is called as base peak. It usually corresponds to the molecular ion only, if the
spectra are recorded at low ionization energy.
Question 12
Which of the following is the weakest base?
a) CH
3
b) H-F
c) H-Cl
d) H-Br
e) H-I
Correct Answer:
e) H-I
Explanation:
The electronegativity and atomic size of iodine is larger so there is a weaker bond between hydrogen
and iodine that makes the electron cloud much lesser than H-F bond. So, H-I is the weakest base; in
other words it is the strongest acid.
Question 13
Which of the following shows the increasing order of solubility?
a) KCl<pbs<agcl< li="" style="margin: 0px; padding: 0px;"></pbs<agcl<>
b) KCl<agcl<pbs< li="" style="margin: 0px; padding: 0px;"></agcl<pbs<>
c) PbS<agcl<kcl< li="" style="margin: 0px; padding: 0px;"></agcl<kcl<>
d) AgCl<pbs<kcl< li="" style="margin: 0px; padding: 0px;"></pbs<kcl<>
e) AgCl<kcl<pbs< li="" style="margin: 0px; padding: 0px;"></kcl<pbs<>
Correct Answer:
b) KCl<agcl<pbs< p="" style="margin: 0px; padding: 0px;"></agcl<pbs<>
Explanation:
KCl is highly soluble because its solubility is greater than 0.1M. AgCl is sparingly soluble because its
solubility is less than 0.01 M. PbS is least sparingly soluble becauseits solubility is very much less
than 0.01 M.
Question 14
Calculate the cell potential at 25
0
C for the following cell reaction using Nernst equation.E
o
ox
= -
3.402 V, E
0
red
=0.7996 V
Cu|Cu
2+
(0.024 M)||Ag
+
(0.0048 M)|Ag
1. 0.25 V
2. 0.30 V
3. 0.370 V
4. 0.5 V
5. 0.1V
Correct Answer:
c) 0.370 V
Explanation:
Oxidation: Cu→Cu
2+
+ 2 e
-
E
o
ox
= -(0.340 V) Reduction: Ag
+
+ e
-
→Ag E
o
red
= 0.799 V Overall cell
reaction is Cu(s) + 2 Ag
+
(aq)→ Cu
2+
(aq) + 2 Ag(s) E
0
cell
= E
0
red
+ E
0
ox
= 0.799 V + (-0.340 V) = 0.459
V Nernst equation, E
cell
= E
0
cell
– (0.0256/n) (In
ox
/In
red
) = 0.459
–(0.0256/2) * In [0.024 / (0.0048)
2
] =
0.459
– 0.0128 * In (1043) = 0.459 – 0.0128 * 6.95 E
cell
= 0.370 V
Question 15
An ideal gas can be defined thermodynamically, when,
I. PV = constant
II. (∂U/∂V)
p
= 0
III. (∂U/∂V)
T
= 0
0
I only
1
I & II
2
I & III
3
II & III
4
II
Correct Answer:
c) I & III
Explanation:
For an ideal gas, PV = constant, at constant temperature. The internal energy of a given quantity of
an ideal gas at a constant temperature is independent of its volume, thus (∂U/∂V)T = 0.
Sample paper 5
Question 1
The boiling point of benzene is 80OC. Estimate its molar heat of vaporization. Assume that it
obeys Trouton's rule.
a) 25.64 kJ mol
-1
b) 31.064 kJ mol
-1
c) 29.96 kJ mol
-1
d) 39.54 kJ mol
-1
e) 40.67 kJ mol
-1
Correct Answer:
b) 31.064 kJ mol
-1
Explanation:
From the Trouton's law ΔH/T
b
= 88 J mol
-1
K
-1
The given data is T
b
= 80
O
C = 80+273 = 353 K
Therefore, ΔH = (88J mol
-1
K
-1
) (353 K) = 31064 J mol
-1
= 31.064 kJ mol
-1
Question 2
The force of attraction between gaseous particles is
a) Strong
b) Weak
c) Very strong
d) Very weak
e) Negligible
Correct Answer:
e) Negligible
Explanation:
Due to very high kinetic energy, gaseous particles easily escape from the attractions of other particles
and tend to maintain very long distances from each other. So, the attraction between gaseous
particles is negligible.
Question 3
The coordination number of hcp structure is
a) 6
b) 4
c) 8
d) 12
e) 2
Correct Answer:
d) 12
Explanation:
Each sphere in hcp structure is in contact with 12 neighbouring spheres. Six spheres are on its own
layer, three on the above layer and remaining three on the below layer. So, the coordination number
of hcp structure is 12. For BCC structure it is 8. For ZnO it is 4.
Question 4
Diamond is an example for
a) Ionic crystals
b) Metallic crystals
c) Covalent crystals
d) Molecular crystals
e) None of these
Correct Answer:
c) Covalent crystals
Explanation:
In diamond, atoms are linked together by continuous system of covalent bonds. In ionic crystals,
atoms are held together by ionic bonds. In metallic crystals, atoms are held together by metallic
bonds. In molecular crystals, molecules are held together by weak Van der Waals forces.
Question 5
Select the incorrect phrase(s) about the significance of salt bridge
I. Connects the solutions of two half-cell reactions.
II. It keeps the solution of two half-cells electrically charged.
III. Prevents liquid-liquid junction potential.
a) I & II
b) I & III
c) II & III
d) II only
e) III only
Correct Answer:
d) II only.
Explanation:
It keeps the solution of two half-cells electrically neutral. In anodic half-cell, positive ions will
accumulate around the anode due to deposition of negative ions by oxidation. To neutralise these
positive ions, sufficient number of negative ions are provided by salt bridge. Similarly for cathodic half-
cell, it is vice-versa.
Question 6
Hess's law is related to ________ of the system.
a) Free energy change
b) Entropy change
c) Enthalpy change
d) Internal energy
e) None of these
Correct Answer:
c) Enthalpy change
Explanation:
Hess's law states that the enthalpy or heat energy change accompanying a chemical reaction is
independent of the pathway between the initial and final states. ΔH for a single reaction can be
calculated from the difference between the enthalpy (heat) of formation of the product and reactant.
ΔH
0
reaction = Σ ΔH
f
0
(products)
– Σ ΔH
f
0
(reactant)
Question 7
7) Which one of the following molecules is held together by dative bond?
a) AlBr
3
b) NaCl
c) Al
2
Cl
6
d) C
2
H
6
e) H
2
O2
Correct Answer:
c)Al
2
Cl
6
Explanation:
In Al
2
Cl
6
lone pairs of electron from chlorine are donated to electron deficient aluminium in such a
way that it is held together by dative bond, where aluminium acts as an electron acceptor and chlorine
acts as an electron donor.
AlBr
3
is a neutral ionic molecule. NaCl- ionic bond, C
2
H
6
- covalent bond,H
2
O
2
– Hydrogen bond.
Question 8
Arrange the following atoms and ions in the increasing order of atomic size
Mg, Mg
2+
, Al, Al
3+
a) Al
3+
> Al > Mg
2+
> Mg
b) Mg
2+
Mg > Al
3+
> Al
c) Mg > Mg
2+
> Al > Al
3+
d) Mg > Al > Mg
2+
> Al
3+
e) Al
3+
> Mg
2+
> Al > Mg
Correct Answer:
d) Mg > Al > Mg
2+
> Al
3+
Explanation:
Atomic radii decrease across the period. Cations are smaller than their parent atoms. Among these
isoelectronic ions, the one with larger positive nuclear charge will have the smaller radius. So, here
the largest atom is Mg and the smallest one is Al
3+
.
Question 9
Which of the following groups has
–I effect?
a) -CH
3
b) -C
2
H
5
c) -C(CH
3
)
3
d)
–C
6
H
5
e) Both a and c
Correct Answer:
d)
–C
6
H
5
Explanation:
The polarisation of the bond is due to electron withdrawing or electron donating effect of adjacent
atoms or groups. Such a type of electron displacement along a carbon chain is called Inductive effect.
Eg: C
–>--- C –>--- C –>--- C
6
H
5
The electron withdrawing nature of groups or atoms is called
negative inductive effect. C
6
H
5
is the only group, which has
–I effect. Since --C
6
H
5
is electron-
withdrawing group, it pulls up the electrons towards itself. Thereby it creates a partial positive charge
in adjacent carbon atoms and partial negative charge in phenyl group (C
6
H
5
).
Question 10
An example for interpolation error is
a) Incorrect identification of indicator's color change in titration
b) Guessing the correct value between two calibrated marks on the metre scale
c) Zero setting of the needle in analog display
d) Calibration of measured instrument
e) None of these
Correct Answer:
b) Guessing the correct value between two calibrated marks on the metre scale
Explanation:
Guessing the correct value between two calibrated marks on the metre scale is an example for
interpolation error. It is one of the two types of human or personal errors.
Incorrect identification of indicator's color change in titration
– Operative error. Zero setting of the
needle in analog display
– static error. Calibration of measured instrument - Instrument error.
Question 11
The one which is most commonly used as a detection of developed colorless chromatogram
spots in T.L.C plate is
a) Iodine
b) Phosphorus
c) Water
d) Copper salts
e) Ammonia
Correct Answer:
a) Iodine
Explanation:
The spots of colorless compounds can be detected by placing the T.L.C plate in a closed jar
containing few crystals of iodine. Spots of compounds, which absorb iodine, will indicate as a brown
color.
Question 12
The Henderson-Hasselbalch equation for acid is
a) pH = pK
a
- log ([A
-
] / [HA])
b) pH = pKa + log ([HA] / [A
-
])
c) pH = pKb + log ([B] / [HB
+
])
d) pH = pK
a
+ log ([A
-
] / [HA])
e) pH = pK
a
+ 2log ([A
-
] / [HA])
Correct Answer:
d) pH = pK
a
+ log ([A
-
] / [HA])
Explanation:
The Henderson-Hasselbalch equation explains whether a compound will exist in its acidic form or in
its basic form at a particular pH. [A
-
] = Molar concentration of a conjugate base. [HA] = Molar
concentration of an undissociated weak acid (M).
Question 13
An azeotropic mixture is also called
a) Increased boiling mixture
b) Decreased boiling mixture
c) Constant boiling mixture
d) Either a or b
e) None of these
Correct Answer:
c) Constant boiling mixture
Explanation:
A pure chemical compound boils at a constant temperature and distils over completely at the same
temperature without any change in composition. It is also known as constant boiling mixture.
Question 14
A system is said to be in equilibrium at all the times. Such a process is called
a) Irreversible process
b) Equilibrium process
c) Static process
d) Quasi-static process
e) Reversible process
Correct Answer:
d) Quasi-static process
Explanation:
A quasi-static process is the one in which the system may be considered to be in equilibrium at all
times.
Question 15
Which one of the following is an intensive property?
a) Volume
b) Density
c) Mass
d) Energy
e) None of these.
Correct Answer:
b) Density
Explanation:
Density is the only intensive property of the above-mentioned properties. All other properties are
extensive properties. Intensive property of a system is that which is independent of the amount of the
substance present in the system. Extensive property of a system depends upon the amount of
substance or substances present in the system.
