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28801.

What is the function of a salt bridge?

Answer»

To allow IONS to go from one cell to another
To PROVIDE link between two HALF cells
To KEEP the emf of the cell positive
To MAINTAIN electrical neutrality of the solution in two half cells

Answer :D
28802.

The main function of roasting is :

Answer»

To REMOVE the VOLATILE matter
Oxidation
Reduction
To MAKE slag

Answer :B
28803.

The main force(s) which stabilise the 2^@ and 3^@ structures of proteins is/are

Answer»

hydrogen BONDS
disulphide linkages
van der WAALS' and electrostatic forces of attraction
All of the above

Solution :The MAIN forces which stabilise the `2^@` and `3^@` . structures of proteins are hydrogen bonds, d1sulph1de linkages, van der Waals' and electrostatic forces of attraction.
28804.

The main factors which affect corrosion are:

Answer»

position of METAL in electrochemical series
presence of `CO_(2)` in water
presence of IMPURITIES in metal
preence of PROTECTIVE coating

Answer :A::B::C::D
28805.

The main factor (s) which affect corrosion is/are

Answer»

POSITION of metl in electrochemical series
presence of `CO_(2)` in water
presence of IMPURITIES in metal
presence of protective coating

Solution :a,C,d
28806.

The main function of haemoglobin is to :

Answer»

help in muscular MOVEMENT.
STORE OXYGEN until it is needed for reproduction.
catalyse biochemical processes.
transport oxygen from lungs to various TISSUES through BLOOD stream.

Answer :D
28807.

The main factor responsible for weak acidic nature of B-F bonds in BF_(3) is :-

Answer»

LARGE electronegativity of F
Three CENTRED two electron BONDS in `BF_(3)`
`p pi-d pi` BACK BONDING
`p pi-p pi` back bonding

Answer :D
28808.

The main factor responsible for weak acidic nature of B-F bonds in BF_3 is :

Answer»

Large electronegativity of F
Three CENTRED two electron bonds in `BF_3`
`ppi-dpi` back BONDING
`ppi-ppi` back bonding

Answer :D
28809.

The main factor of shorter B-F bonds in BF_(3) is :

Answer»

LARGE electronegativity of fluorine
three-centred TWO electron bonds in `BF_3`
`p pi = p pi` back bonding
`p pi - d pi` back bonding

ANSWER :C
28810.

The main difference between conventional electro-chemical cell and fuel cell is the materials undergoing oxidation at the anode or reduction at the cathode in fuel cell

Answer»

are not STORED
are stored INSIDE the cell
are stored OUTSIDE the cell
reactions at ANODE and cathode

Solution :Storage of `H_(2) "and" O_(2)` is outside.
28811.

The main difference between bathing and washing soap is

Answer»

bathing SOAP are potassium salts of FATTY acids washing soaps are SODIUM salt of fatty acids .
bathing soaps are sodium salt of fatty acids while washing soapsare potassium salt of fatty acids
bathing soaps are cationic in nature while washing soaps are anionic
bathing soaps are calciumsalts of fatty acids while washing soapsare anionic

ANSWER :A
28812.

Themain cuplrit for the depletion of protective ozonelayer is

Answer»

OXYGEN
CFCS
OXIDES of S
Particulatematter.

ANSWER :B
28813.

The main couse of acidity in the stomach is

Answer»

release of extra gastric acids which decrease the pH LEVEL
indigestion and PAIN in LARGE intestine
increase the pH level in the stomach
release of extra bile JUICE which increases ALKALINE medium in stomach.

Answer :A
28814.

How is acid rain caused ?

Answer»

`H_(2) SO_(4)`
`CaCO_(4)`
`CH_(3)COOH`
`PAN`

Answer :A
28815.

The main constituents of dettols are

Answer»

CHLORAMPHENICOL + glycerol
2-3% solution of IODINE in alcohol
0.2% solution of PHENOL.
CHLOROXYLENOL and terpineol.

Answer :D
28816.

The main constituents of coal gas are :

Answer»

`CH_4+CO+H_2`
`CO_2+CO+H_2`
`CO+CO_2`
`CO+N_2`

ANSWER :A
28817.

The main constituent of steel in india are:

Answer»

NI and Mg
V andCo
Al and Zn
Mn and Cr

Answer :D
28818.

The main constituent of most natural fibres is

Answer»

GLYCOGEN
STARCH
CELLULOSE
all of these

ANSWER :C
28819.

The main constituent of egg-shells is:

Answer»

`CaCO_3`
`CaSiO_3`
`CaSO_41/2H_2O`
`CaSO_42H_2O`

ANSWER :A
28820.

The main constituent of bones is:

Answer»

`CaCO_3`
`CaF_2`
`CaSO_4`
`Ca_3(PO_4)_2`

ANSWER :D
28821.

The main component/compounds provided by crushed grapes for wine making is/are

Answer»

sugars
sugars and yeast
sugars and tartaric acid
sugars, yeast and tartaric acid.

Solution :The two most IMPORTANT components NEEDED for fermentation, sugars (in the fruit juice) and yeast (present on the SKIN of grapes). Both of these are provided by crushed groups.
28822.

The main cause of acidity in the stomach is

Answer»

releaseof EXTRA gastric acidswhich decrease the pH level
indigestionand pain in larger intestine
increase the pH level in the STOMACH
releaseof extra bile juice which INCREASES alkaline MEDIUM in stomach .

ANSWER :A
28823.

The main cause of depression is

Answer»

LOW LEVEL of noratronaline
high livel of noradronaline
low livel of HCI in stomach
high livel of HCI in stomach

ANSWER :A
28824.

The main application of osmotic pressure measurement is in the determination of the molar mass of a substance which is either slightly soluble or has a very high molar mass such as proteins, polymers of various types and colloids.This is due to the fact that even a very small concentraion of the solution produces fairly large magnitude of osomotic pressure.In the laboratory the concentrations usually employed are of the order of 10^(-3) to 10^(-4) M.At concentration of 10^(-3) mol dm^(-3) , the magnitude of osmotic pressure of 300 K is : P=10^(-3)xx0.082xx300=0.0246 atm or 0.0246xx1.01325xx10^5=2492.595 Pa At this concentration, the values of other colligative properties such as boiling point elevation and depression in freezing point are too small to be determined experimentally. Further polymers have following two types of molar masses : (A) Number average molar mass (barM_n), which is given by (undersetisumN_iM_i)/(undersetisumN_i) where N_i is the number of molecules having molar mass M_i. (B) Molar average molar mass (barM_m), which is given by (undersetisumN_iM_i^2)/(undersetisumN_iM_i) Obviously the former is independent of the individual characteristics of the molecules and gives equal weightage to large and small molecules in the polymer sample.On the other hand later gives more weightage to the heavier molecules.Infact with the help of a colligative property only one type of molar mass of the polymer can be determined. What will be the observed molecular weight of the above polymer sample generating osomotic pressure equal to the answer in the previous questions at 27^@C?

Answer»

4000
3000
2500
1200

Solution :`0.0164xx1=2/mxx0.082xx300`
`:. M=3000`
28825.

The main application of osmotic pressure measurement is in the determination of the molar mass of a substance which is either slightly soluble or has a very high molar mass such as proteins, polymers of various types and colloids.This is due to the fact that even a very small concentraion of the solution produces fairly large magnitude of osomotic pressure.In the laboratory the concentrations usually employed are of the order of 10^(-3) to 10^(-4) M.At concentration of 10^(-3) mol dm^(-3) , the magnitude of osmotic pressure of 300 K is : P=10^(-3)xx0.082xx300=0.0246 atm or 0.0246xx1.01325xx10^5=2492.595 Pa At this concentration, the values of other colligative properties such as boiling point elevation and depression in freezing point are too small to be determined experimentally. Further polymers have following two types of molar masses : (A) Number average molar mass (barM_n), which is given by (undersetisumN_iM_i)/(undersetisumN_i) where N_i is the number of molecules having molar mass M_i. (B) Molar average molar mass (barM_m), which is given by (undersetisumN_iM_i^2)/(undersetisumN_iM_i) Obviously the former is independent of the individual characteristics of the molecules and gives equal weightage to large and small molecules in the polymer sample.On the other hand later gives more weightage to the heavier molecules.Infact with the help of a colligative property only one type of molar mass of the polymer can be determined. One gram each of polymer A (molar mass=2000) and B(molar mass=6000) is dissolved in water to form one litre solution at 27^@C.The osmotic pressure of this solution will be :

Answer»

0.0164 ATM
1862 PA
Both are CORRECT
NONE of these

Solution :`Pxx1=(1/2000+1/6000)xx0.082xx300`
`:.` P=0.0164 atm
28826.

The main bonding responsible for blockgae of enzyme is

Answer»

HYDROGEN BOND
ionic bond
VAN der Waals attractions
strong COVALENT bond

Answer :D
28827.

The main agent for polluting the environment is

Answer»

gas
plant
ANIMAL
man.

Solution :animal
28828.

The magnitude of work done (in kcal) when 120 gm of Mg (s) is reacted with excess of hydrochloric acid in an open vessel at 27^@C is

Answer»


Solution :MAGNITUDE of W=3 kcal
`Mg(s)+2HCL(AQ)toMgCl_2(s)+H_2(G)uarr`
`W= -P_(ed)(V_2-V_1)= -Deltan_(g)RT= - 5XX(2xx300)/1000=-3` kcal
28829.

The magnitude of work done by a gas that undergoes a reversible expansion alog the path ABC shown in the figure is __________.

Answer»


ANSWER :62
28830.

The magnitude of work done by a gas is a reversible expansion is greater than that in an irreversible expansion provided the initial and final state are identical in both processes-explain.

Answer»

Solution :Suppose, an ideal gas expands isothermally from its initial state `P_(1) V_(1)` to final state `P_(2)V_(2)`.

A reversible process consists of an infinite number of small steps, and at each STEP, the driving from is infinitesimally greater than the OPPOSING force. suppose, in a reversible expansion of an ideal gas, the external pressure of the gas at a step of the process is decreased by a very small amount dP, CAUSING the increase in VOLUME of the gas by dV. the work done by the gas will be, dw=`-(P_(ex)-dP)dV=-PdV`[`becauseP_(ex)-dP=P` (Pressure of the gas) and `dPxxdV` is very small as in the P-V diagram the point, m and n are very close and pressure in these two points are not much different]. in the above diagram, PdV is givenn by the area of the strip, mnqp. total work done in reversible process will be equal to the sum of the areas of all such strips. thus, in such a process, work done,
`-w=underset(V_(1))overset(V_(2))intPdV=`area ABNM
Absolute value of work done, `|w|=` area ABNM
Now, in isothermal irreversible expansion, suppose, the gas is expanded from volume`V_(1)` to `V_(2)` against an external pressure which is equal to the final pressure of the gas. in this process work done,
`w=underset(V_(1))overset(V_(2))intP_(2)dV=-P_(2)underset(V_(1))overset(V_(2))intdV` [`P_(2)=`constant during expansion]
Absolute value of work done, `|w|=(V_(2)-V_(1))`=area. ABNE. From the graphs, it is clear than work done is reversible isothermal expansion is greater than that in irreversibleisothermal expansion, since the arae of ABNM is greater than that of ABNE. this also indicates that work involved in a process depends UPON the path (or nature) o the process.
28831.

The magnitude of thelattice enthalpy of a solidincreases if-

Answer»

the IONS are large
the ions are small
the ions are of EQUAL SIZE
CHARGES on the ions are small

Answer :B
28832.

The magnitude of spin angular momentum of an electron is given by

Answer»

`s=sqrt(s(s+1))`
`s=sxxh/(2pi)`
`s=(sqrt(3))/2xxh/(2pi)`
`s=+-1/2xxh/(2pi)`

SOLUTION :SPIN angularmomentum `=sqrt(s(s+1))H/(2pi)`
ALSO `s=sqrt(1/2(1/2+1))xxh/(2pi)=(sqrt(3))/2xxh/(2pi)`
28833.

The magnitude of crystal stabilization energy (CFSE or Delta_(t)) in tetrahedral complexes is considerably less than in the octahedral field. Because

Answer»

There are only four ligands instead of six so the ligand field is only 2/3 the size hence the `Delta_(t)` is only 2/3 the size
The direction of the ORBITALS does not coincide with the direction of the ligands. This reduces the CRYSTAL field stabilization energy `(Delta_(t))` by further 2/3
Both points (a) and (b) are correct
Both points (a) and (b) are wrong

Solution :CFSE in case of TETRAHEDRAL complexes is less than the OCTAHEDRAL field due to the points (a) and (b).
28834.

The magnitude of crystal field stabilisation energy in octaheral field depends on: (I) the nature of the ligand (II) the charge on the metal ion. (III) whether the metal is in the first, second or third row of the transition elements

Answer»

I, II, III only correct
I, II only correct
II, III only correct
III only correct

Answer :A
28835.

The magnitude of crystal field stabilization energy (CFSE or Deltat) in tetrahedral complexes is considerably less than in octahedral field. This is because

Answer»

There are FOUR ligands instead of six so the ligand field is only 2/3 the size hence , the `Delta`t is only 2//3 the size
The direction of the orbitals does not coincide with the direction fo the ligands . This reduces the CRYSTAL field stabilisation ENERGY `(Deltat)` by further 2/3
Both points (A) and (B) correct
Both points (A) and (B) are wrong .

Solution :CFSE in case of tetrahedral complexes is less than the octahedral field due to both the points A and B.
28836.

The magnitude of colligative properties in all colloidal dispersions is …..than solution

Answer»

LOWER
HIGHER
Both
NONE

ANSWER :A
28837.

The magnetic property, dipole moment, plane of symmetry, colour and absorption band cann be helpful in structure elucidation of complex compounds: Q. Complex compound() having even number of space (stereo) isomers is/are: (Where AA-symmetrical bidentate ligand ad a,b,c,d,e-monodentate ligands)

Answer»

`[M(A A)_(2)b_(2)]^(n+-)`
`[Ma_(3)b_(3)]n^(+-)`
`[Ma_(3)BCD]^(n+-)`
`[Ma_(2)bcde]^(n+-)`

ANSWER :B
28838.

The magnetic quantum number for valency electron of sodium is:

Answer»

3
2
1
Zero

Answer :D
28839.

The magnetic property, dipole moment, plane of symmetry, colour and absorption band cann be helpful in structure elucidation of complex compounds: Q. Compound 'x' has molecular formula CrCl_(2)Br*6H_(2)O can show type of isomerism. (i) hydrate isomerism (ii) ionization isomerism (iii) geometrical isomerism (iv) optical isomerism

Answer»

(i), (II) and (III) only
(i) and (ii) only
(i), (ii), (iii) and (IV)
(i) and (iii) only

Answer :A
28840.

The magnetic property, dipole moment, plane of symmetry, colour and absorption band cann be helpful in structure elucidation of complex compounds: Q. Which of the following complex ion is expected to absorb light in 4000 Å to 7800 Å region?

Answer»

`[Ti(en)_(3)]^(4+)`
`[CR(H_(2)O)_(6)]^(3+)`
`[Sc(NH_(3))_(4)(H_(2)O)_(2)]^(3+)`
`Zn(en)_(2)(NH_(3))_(2)]^(2+)`

Answer :B
28841.

The magnetic properties of oxygen and ozone are different. Explain.

Answer»

SOLUTION :Oxygen is a MOLECULE with even number of ELECTRONS. But it is paramagnetic . Its magnetic moment is 2.83BM corresponding to two unpaired electrons. Molecular orbital THEORY explains the presence of unpaired electrons.
Ozone is diamagnetic. Its magnetic moment is ZERO. Ozone molecule has no unpaired electrons.
28842.

The magnetic properties of NO_(2) and N_(2)O_(4) are different. Why?

Answer»

SOLUTION :`NO_(2)` MOLECULE has unpaired ELECTRON. Hence it is paramagnetic and reactive. `NO_(2)` dimerises to GIVE `N_(2)O_(4).N_(2)O_(4)` molecule has no unpaired electrons. Hence it is diamagnetic.
28843.

The magnetic nature of elements drpends on the presence of unpaired electrons. Identify the configuration of transition element, which shows highest magnetic moment.

Answer»

`3d^(7)`
`3d^(5)`
`3d^(8)`
`3d^(2)`

Solution :`d^(5)` has HIGHEST number of unapired electron. HENCE, it has highest MAGNETIC moment.
28844.

The magnetic nature of elements depends on the presence of unpaired electrons. Identify the configuration of transition elements, which shows highest magnetic moment.

Answer»

`3D^(7)`
`3d^(5)`
`3d^(8)`
`3d^(2)`

SOLUTION :`3d^(5)` has MAXIMUM unpaired electrons `( = 5)`.
28845.

The magnetic ore is :

Answer»

Fe3O4
Fe2O3
Fe(OH)3
FeCO3

Answer :D
28846.

The magnetic nature of elements depends on the presence of unpaired electrons. Identify the configuration of transition elements which shows highest magnetic moment ?

Answer»

`3d^7`
`3d^8`
`3d^5`
`3d^2`

ANSWER :C
28847.

The magnetic nature of elements depends on the presence of unpaired electrons. Identify the configuration of transition element which shows highest magnetic moment.

Answer»

`3d^7`
`3d^5`
`3d^8`
`3d^2`

Solution :CONFIGURATION `3d^5` is associated with maximum number of unpaired electrons (5) and thus has maximum MAGNETIC MOMENT.
28848.

The magnetic nature of elements depends on the presence of unpaired electrons. Identify the configuration of transition elements which shows highest magnetic moment?

Answer»

`3d^7`
`3d^5`
`3d^8`
`3d^2`

Solution :The LARGER the number of unpaired electrons in a element, the greater is the paramagnetic character and larger is the magnetic MOVEMENT, this can be calculated from the relation.
`MU = SQRT(N(n+2)) B.M`
Where, n is the number of unpaired electrons and u is magnetic moment. `3d^(5)` contain highest unpaired electron so have highest magnetic moment.
28849.

The magnetic moments of few transition metal ions are given below : {:("Metal ion",,,"Magnetic moment (BM)"),(Sc^(3+),,,0.00),(Cr^(2+),,,4.90),(Ni^(2+),,,2.84),(Ti^(3+),,,1.73):} (At. no. Sc = 21, Ti = 22, Cr = 24, Ni = 28) Which of the given metal ions : (i) has the maximum number of unpaired electrons ? (ii) forms colourless aqueous solution ? (iii) exhibits the most stable +3 oxidation state ?

Answer»

Solution :(i) As `Cr^(2+)` SHOWS the maximum magnetic moment, it has the maximum number of unpaired electrons.
(II) The configuration of `Sc^(3+)` ion is `1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6)`. As all the ORBITALS are completely filled, there is no possibility of excitation of electrons. It forms a colourless AQUEOUS solution.
(iii) Again because of completely filled orbitals, `Sc^(3+)` shows the most STABLE +3 oxidation state.
28850.

The magnetic nature of elements depend on the presence of unpaired electrons. Identify the configuration of transition element,which shows highest magnetic moment.

Answer»

`3d^7`
`3d^5`
`3d^8`
`3d^2`

ANSWER :B