Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

If soap has high alkali content it irritates skin. How can the amount of excess alkali be determined ? What can be the source of excess alkali ?

Answer»

Solution :Acid-base titration tie, titrating a solution of soap with STANDARD hydrochloric acid using phenolphthalein as indicator) can be used to DETERMINE the excess amount of alkali present in soap. The source of this excess alkali is the alkali left unused when the soap is PREPARED by hydrolysis of OILS.
2.

If{:(Sn^(2+) + 2e^(-) rarr Sn(s), E^(o) = - 0.14 V),(Sn^(4+)+2e^(-) rarr Sn^(2+),E^(o)= + 0.13 V):} then which of these is true?

Answer»

`Sn^(2+)` is unstable and disproportionates to form `Sn^(4+)` and Sn
`Sn^(2+)` is stable and DISPROPORTIONATION reaction is not SPONTANEOUS
`Sn^(4+)` is easily REDUCED to Sn
None of these is correct

Answer :B
3.

Name the product discharged at the anode during the electrolysis of an aqueous solution of sodium chloride.

Answer»

Solution :II) Chlorine or `Cl_(2)` is the PRODUCT DISCHARGED.
4.

If screening does not split the subshell(l) of a given shell(n), how many elements would occur in 5th period.

Answer»

50
32
18
64

Solution :then no of ELECTRONS in 5TH shell=no. of ELECTRON in 5th pd=2xxn^(2)=2xx5^(2)=50`
5.

Mention one advantage of H_(2) - O_(2) fuel cell.

Answer»

Solution :II) The advantage of `H_(2)-O_(2)` fuel CELL is efficiency is more than thermal power PLANTS.
6.

If same type of atoms are packet in hexagonal close packing and cubic close packing separately, then

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Density of hcp will be greater than CCP
Density of hcp will be smaller than ccp
Density of hcp will be equal to ccp
Density of hcp and ccp will depend UPON the temperature of the system.

Solution :The percentage of TOTAL space OCCUPIED by the particles in ccp and hcp is same ( 74% )
7.

ii) Name the element of lanthanide with maximum paramagnetic property.

Answer»

SOLUTION :II) Neodynium is the element of LANTHANIDE with MAXIMUM paramagnetic property.
8.

If S+O_(2)rarrSO_(2),(DeltaH=-298.2)SO_(2)+(1)/(2)O_(2)rarrSO_(3),(DeltaH=-98.2)H_(2)+(1)/(2)O_(2)rarrH_(2)O, (DeltaH=-287.3)then the enthalpy of formation of H_(2)SO_(4) at 298 K will be

Answer»

`-433.7 KJ`
`-650.3 kJ`
`+320.5 kJ`
`-813.9 kJ`

ANSWER :D
9.

Mention any one application of Kohlrausch law.

Answer»

SOLUTION :ii) It is USED to DETERMINE the molar conductivity of a weak electrolyte which cannot be determined experimentally.
It is used to determine the degree of dissociation a and the dissociation constant Ka for a weak electrolyte ( mention any 1 )
10.

If same amount of electricity is passed through aqueous solutions of AgNO_(3) and CuSO_(4) and the number of Ag and Cu atoms deposited are'x' and 'y' respectively . Then

Answer»

`x = y`
`x LT y`
`y = 2x`
`x = 2y`

Solution :`AG^(+) + e^(-) to Ag , Cu^(2+) + 2 e^(-) to Cu` .
THUS , same quantity of ELECTRICITY will deposit double the no. of Ag ATOMS than Cu, i.e., x = 2y .
11.

Based on theraupatic action, to which class of drug does ranitidine belong to?

Answer»

SOLUTION :II) Based on theraupatic action, RANITIDINE belong to class of antacid.
12.

If, S+O_2rarrSO_2, triangleH=-298.2 kJ (i)SO_2+1/2O_2rarrSO_3, triangleH =-98.7 kJ (ii)SO_3+H_2OrarrH_2SO_4, triangleH=-130.2 kJ (iii)H_2+1/2O_2rarrH_2O, triangle=-287.3 kJ The enthalpy of formation of H_2SO_4 at 298 K will be:

Answer»

`-754.4 KJ`
`-814.4 kJ`
`-650.3 kJ`
`-433.7 kJ`

ANSWER :A
13.

Ignoring the water lost by evaporation, some water has still to be added periodically into the battery used in an invertor or car. Why? Why this is not requried in the maintenance free batteries?

Answer»

Solution :During discharge, electrolysis of water also takes place. HENCE, some water is lost due to electrolysis of water into `H_(2) and O_(2)`. Hence, water has to be added periodically into the ACID of the battery to have desired concentration. Now a days, int he maintenance free batteries, no water NEEDS to be added as they USE electrodes of Ca/Pb ALLOY which prevents electrolysis of water.
14.

(i)How do antiseptics differ from disinfectants ? Give one example of each . (ii)Name a substance which can be used as an antiseptic as well as disinfectant ?

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SOLUTION :Antiseptics are chemical substances which prevent the growth of microorganisms and may even kill them but are not harmful to human or animal tissues. For example, dettol and savlon. They are generally APPLIED on wounds, cuts, ulcers and DISEASED skin surfaces. Furacin and soframycin are well known antiseptic creams
Disinfectants are chemical substances which kill microorganisms but are not safe to be applied to the living tissues. These are generally used to kill microorganisms present in the drains, toilets. Doors, etc Some common examples of disinfectants are PHENOL (`GE`1% solution) and chlorine (0.2 to 0.4 ppm).
15.

(i)"Graphene"""(a)"Honeycomb crystal" (ii)"Diamond"""(a)"Aromatic character"(iii)"Fullerene"""(c)"Lubricant" (iv)"Graphite"""(d)"Very hard"

Answer»


ANSWER :A::B::C::D
16.

If S + O_(2) rarr SO_(2), Delta H = -298.2 kJ SO_(2) + (1)/(2) O_(2) rarr SO_(3), Delta H = -98.2 kJ SO_(3) + H_(2)O rarr H_(2)SO_(4), Delta H = -130.2 kJ H_(2) + (1)/(2) O_(2) rarr H_(2)O, Delta H = -287.3 kJ the enthalpy of formation of H_(2)SO_(4) at 298 K will be

Answer»

`- 433.7 kJ`
`- 650.3 kJ`
`+320.5 kJ`
`- 813.9 kJ`

Solution :The required EQUATION is
`H_(2) (G) + S + 2O_(2) rarr H_(2)SO_(4) DELTA H = ?`
(i) `(S + O_(2) rarr SO_(2) Delta H = - 298.2 kJ mol^(-1)`
(ii) `SO_(2) + (1)/(2) O_(2) rarr SO_(3) Delta H = - 98.2 kJ mol^(-1)`
(iii) `SO_(2) + (1)/(2) O_(2) rarr SO_(3) Delta H = - 130.2 kJ mol^(-1)`
(iv) `H_(2) + (1)/(2) O_(2) rarr H_(2)O Delta H = - 287.3 kJ mol^(-1)`
ADD eqns. (i), (ii), (iii) and (iv),
`H_(2) + S + 2O_(2) rarr H_(2)SO_(4) Delta H = -813.9 kJ`
17.

If S^@ for H_2, Cl_2 and HCl are 0.13, 0.22 and 0.19 kJ K^-1 mol^-1 respectively. The total change in standard entropy for the reaction, H_2+Cl_2rarr2HCl is:

Answer»

`30 JK^-1 mol^-1`
`40 JK^-1 mol^-1`
`60 JK^-1 mol^-1`
`20 JK^-1 mol^-1`

ANSWER :A
18.

If S^(@) for H_(2), CI_(2) and HCI are 0.13 , 0.22 and 0.19 kJ K^(-1) mol^(-1) respectively. The total change in standard entropy for the reaction H_(2) + CI_(2) to 2 HCI is

Answer»

`20 JK^(-1) MOL^(-1)`
`40 JK^(-1) mol^(-1)`
`60 JK^(-1) mol^(-1)`
`30 JK^(-1) mol^(-1)`

ANSWER :D
19.

Igniting MnO_(2) coverts it quantitatively to Mn_(3)O_(4). A sample of pyrolusite is of the following composition : MnO_(2)=80%,SiO_(2) and other inert contents =15%, rest being water. The sample is ignited in air to constant weight. What is the percentage of Mn in the ignited sample? (Atomic weight of Mn = 55)

Answer»

Solution :`undersetunderset("=261 g")(3(55+2xx16))(3MnO_(2))rarrundersetunderset("= 229 g")(3xx55+4xx16)(Mn_(3)O_(4))+O_(2)`
Suppose the amount of pyrolusite ignited = 100 g
`therefore""MnO_(2)" present in it = 80 g"`
`SiO_(2)" and other inert contents = 15 g"`
`"Water "=100-(80+15)=5g`
`"Now,261 g of "MnO_(2)" give "Mn_(3)O_(4)=229g`
`therefore"80 g of "MnO_(2)" will give "Mn_(3)O_(4)=(229)/(261)xx80g=70.19g`
An on ignition, water is REMOVED but `SiO_(2)` and other inert contents remain, therefore TOTAL mass of the residue after ignition `=70.19+15g=85.19g`
`underset(=229g)underset(3xx55+4xx16)(Mn_(3)O_(4))-=underset(=165g)underset(3xx55)(3Mn)`
`"229 g "Mn_(3)O_(4)" CONTAIN Mn"=165 g`
`therefore""70.19 Mn_(3)O_(4)" contain Mn"=(165)/(229)xx70.19g=50.57g`
This is present in 85.19 g of the total residue
`therefore""%" of Mn in the residue "=(50.57)/(85.19)xx100=59.36`
20.

If s and S are respectively solubility and solubility product of a sparingaly soluble binary electrolyte then:

Answer»

s=S
s=`S^12`
s=`s^(1/2)`
s=1/2S

Answer :C
21.

If S and K_(sp) are respectively solubility and solubility product of a sparingly soluble binary electrolyte, then

Answer»

`S = K_(SP)`
`S = K_(sp)^(2)`
`S = SQRT(K_(sp))`
`S = (1)/(2) K_(sp)`

Solution :For BINARY electrolyte, so that `K_(sp) = S XX S = S^(2)`
`S = sqrt(K_(sp))`.
22.

Igniting MnO_(2) inair converts it quantitativelyto Mn_(3) O_(4) . A sample of pyrolusite isof the followingcomposition : MnO_(2) = 80%, SiO_(2) and otherconstituents - 15% restbeingwater. Thesample is ignitedin air to constantweight,what is thepercentageof Mnin theignited smaple ?

Answer»

Solution :Suppose the weight of pyrolusite is x g
wt. Of `MnO_(2)""( 80)/( 100) XX x = 0.8 x `
Wt. of `SiO_(2) ` etc. = `(15)/(100) xx x = 0.15 x`
Wt. of WATER ` = (5)/( 100) xx x = 0.05 x`
Whenpyrolusite isignited, `MnO_(2)` changes to `Mn_(3)O_(4)and H_(2)O` evaporates.
Theresidue CONTAINS, therefore , `SiO_(2), ` etc., and `Mn_(2) O_(4)`
Now , we know
`MnO_(2) to Mn_(2) O_(4)`
0.8x g
ApplyingPOAC for Mn atoms,
moles of Mn is `MnO_(2)` = molesof Mnin `Mn_(3) O_(4)`
`1 xx` moles of `MnO_(2) = 3 xx ` moles of `Mn_(3) O_(4)"" . . . (i)`
`(0.8)/( 87) = 3 xx ("wt.of "Mn_(3) O_(4))/( 229) [{:(MnO_(2)=87),(Mn_(3)O_(4)=229):}] `
Wt. of `Mn_(3)O_(4) = 0.702` x g
`:.` wt. of the residue= wt. of `Mn_(3)O_(4)` + wt. of `SiO_(2)`, etc.
`= 0.702 x + 0.15 x = 0.852 x g `
Now, since Mn atoms areconserved,
moles of Mn= moles of Mn = `Mn_(3) O_(4)`
= moles of Mn in `MnO_(2)`
`= 1 xx ` moles of `MnO_(2)`
`= (0.8x)/( 87) (MnO_(2) = 87)`
`:. ` wt. of Mn = molesof Mn ` xx ` at . wt. of Mn
`= (0.8 x)/( 87) xx 55 g`
`%`of Mn in residue = `("wt. of Mn")/( "wt. of residue ") xx 100`
`= (0.8 x 55)/(87) xx (100)/( 0.852 x) = 59.37% `
23.

If r.m.s. speed of gaseous molecules is x cm s^(-1) at a pressur eof p atm, their r.m.s. at a pressure of2 atmand constant temperature will be

Answer»

x
2x
4x
`x//4`

Solution :`u = sqrt((3RT)/( M)) = sqrt((pV)/( M))`
As temperature is CONSTANT, pV is constant. THEREFORE, r.m.s. speed is constant EVEN if PRESSURE is doubled.
24.

Ignition mixture used in aluminothermic process is ………….

Answer»

`MG + BaO_2`
`MGO + BAO`
`Al_2O_3 + Mg`
`Al_2O_3 + BaO_2`

ANSWER :A
25.

If rho pis the resistance in ohm of a centimeter cube, generally called the specific resistance of the substance constituting the conductor, the resistance r of the layer containing "a" cubes is given by

Answer»

`1/R = (1)/(RHO) + (1)/(rho)+….`
`1/r =(1)/(rhoa)+ (1)/(rhoa)+…`
`r=a//rho`
`r= rho + rho +….`

SOLUTION :Reciprocal of resistance is conductance. We can add the conductance and not resistance.
Hence `1/tau = 1/rho + 1/rho +……..`
26.

If(X) turns acidified K_(2)Cr_(2)O_(7) solution green , then then X may be

Answer»

`SO_(2)`
`CO_(2)`
`NO_(2)^(Θ)`
`FE^(2+)`

ANSWER :a,C,d
27.

If relative decrease in vapour pressure is 0.4 for a solution containing 1 mole of NaCl in 3 moles H_(2)O, NaCl is. . . Ionised.

Answer»

0.6
0.5
1
0.4

Answer :C
28.

Iftheta = 45^@ in the figure given below. Calculate the value of 'n' in Freundlich adsorption isotherm.

Answer»


SOLUTION :TAN `45^@`
29.

If reduction of silver ion on copper electrode is carried out at 298K temperature and having standard potential is +0.46 V, then what is the value of standard free energy (DeltaG^(@)) ?

Answer»

`-44.5kJ`
`-98.0kJ`
`-89.0kJ`
`-89.0J`

SOLUTION :`CU+2Ag^(+)to Cu^(2+)+2Ag`
`DELTAG^(@)=-nFE^(@)`
`=-2xx96500xx0.46`
`=-88780J`
`=-89kJ`
30.

IfN_(1),N_(2),N_(3),……are the number of molcules with molecular masses M_(1),M_(2),(M_(3),…respectively ,then mass average molar mass is expressed as:

Answer»

`(SigmaNiMi^(2))/(SigmaNiMi)`
`(SigmaNiMi)/(SigmaNi)`
`(SigmaMi^(2))/(SigmaNi)`
`(SigmaNiMi)/(SigmaMi)`

ANSWER :A
31.

If red hot steel is suddenly immersed in water, the steel becomes:

Answer»

SOFT and malleable
hard and brittle
tough and DUCTILE
fibrous

Answer :2
32.

If rate of reaction becomes double when temperature is increases from 27^(@)C"to"37^(@)C. Then find activation energy of reaction in calorie. (R = 2 "cal"//"mol"-K) I"in" 2=0.7) [Fill your answer after dividing by 10.]

Answer»


ANSWER :1302
33.

IfK_e of the reaction,2HI rarr H_2+ I_2is 0.25, the equilibrium constant of the reaction H_2+ I_2 rarr 2HI would be :

Answer»

1
2
3
4

Answer :D
34.

IfGennanium crystallises in the same way as diamond, then which of the following statement is not correct?

Answer»

Every atom in the structure is tetrahedrally BONDED to 4 ATOMS.
Unit cell consists of 8 Ge atoms and co-ordination numbcr is4
All the octahedral voids are occupied
All the octahedral voids and 50% tetrahedral voids remain unoccupie

Answer :C
35.

If Raoult's law is obeyed, the vapour pressure of the solvent in a solution is directly proportional to

Answer»

The VOLUME of the solution
Mole FRACTION of the solute
Mole fraction of the SOLVENT and solute
Mole fraction of the solvent

Answer :D
36.

IF_(7) exists but BrF_(7) does not, why ?

Answer»

Solution :The ATOMIC size of I (133 pm) is much bigger than that of Br (114 pm). Therefore, 7 SMALL sized F atoms can be arranged around big sized I ATOM but not around small sized Br atom.
37.

If rate of reaction between a and B is expressed as k[A][B]^2 , the reaction is

Answer»

FIRST ORDER in A
second order in B
overall having THIRD order
All are correct

Answer :D
38.

If Z_("eff") of sodium is .x. then Z_("eff") of calcium is :-

Answer»

`x+0.35`
`x+0.65`
`x-0.35`
`x-0.65`

SOLUTION :
39.

If ratio of the 1st three Bohr orbits in H is 1:4:9, then the ratio of 1st three orbits in He^(+) will be

Answer»

1 `0.5:2:4.5`
2. `2:4:8`
3. `2:3:4`
4. `1:4:9`

ANSWER :D
40.

If Zn^(2+)//Zn electrode is diluted 100 times, then the change in emf is

Answer»

increase of 59 mV
decrease of 59 mV
increase of 29.5 mV
decrease of 29.5 mV

Solution :`E_(cell)^(@)=E_(cell)^(@)-(0.0591)/(2)"LOG"(1)/([ZN^(2+)])=E_(cell)^(@)+0.02955logC`
`E_(cell)'=E_(cell)^(@)+0.02955"log"(C)/(100)=E_(cell)^(@)+0.02955(logC-2)`
`thereforeE_(cell)` will decrease by 0.02955`xx2V=0.059V=59mV`.
41.

If Z is the compressibility factor, van der Waals equation at low pressure can be written as

Answer»

`Z = 1+ ( RT)/( Pb)`
`Z = 1- ( a) /( VRT)`
`Z = 1- ( Pb)/( RT)`
`Z = 1+ ( Pb)/( RT)`

Solution :Compressibility FACTOR ( Z ) = `( PV)/( RT)`
For one mole of a REAL gas van DER Waals EQUATION
`( P + ( a)/( V^(2))) ( V - b) = RT`
At LOW pressure
`V -b = V ` ( b can be neglected )
`( P + ( a)/( V^(2))) V = RT `
`pV + ( a)/( V ) = RT `
`pV = RT - ( a)/( V )`
`( pV)/( RT) = 1- ( a)/( VRT)`
`:. Z = 1- ( a)/( VRT)`
42.

If Raoult's law is obeyed, the vapour pressure of the solution is directly proportional to:

Answer»

MOLE FRACTION of SOLVENT
Mole fraction of the solute
Mole fraction of the solvent and solute
The VOLUME of the solution

Answer :A
43.

If z-axis is the molecular axis, then the p-mole-cular orbitals are formed by the overlap of :

Answer»

1. 2S and `2p_(x)`
2. `2p_(x)` and `2p_(z)`
3. `2p_(x)` and `2p_(x)`
4. `2p_(x)`and`2p_(z)`

Answer :C
44.

If you have a mixture of CO and CO_(2), how would you know about the relative proportions of the two gases in the given mixture ?

Answer»

Solution :Pass mixture through the `CA(OH)_(2)` solution, `CO_(2)` is absorbed by `Ca(OH)_(2)`.The residual volume will be that of `CO`
`Ca(OH)_(2)+CO_(2)toCaCO_(3)+H_(2)O`
(ii)Pass mixture through `I_(2)O_(5),CO` reduces `I_(2)O_(5)` to `I_(2)`.
`5CO + l_(2)O_(5) to I_(2)+5CO_(2)`
`l_(2)` thus liberated is determined is determined by titration with `Na_(2)S_(2)O_(3)`.
`2Na_(2)S_(2)O_(3)+l_(2)to2Nal+Na_(2)S_(4)O_(6)`
This is the quantitative METHOD of estimation of `CO`.
45.

If radius ratio (r^+//r^-) is in the range 0.414 to 0.732, the possible co-ordination number is ______and structural arrangement is ______

Answer»

SOLUTION :6, OCTAHEDRAL.
46.

If X^(a) species emit firstly a positron then two alpha andbeta last one alpha is also emitted and finally convert in Y^(c ) species so correct the relation is

Answer»

a=c+12,d=b-5
a=c-8,d=b-1
a=c-6,d=b-0
a=c-4,d=b-2

Solution :`_(92)U^(235)` NUCLEUS ABSORBS a neutron and then DISINTEGRATE in `_(54)XE^(139),38Sr^(94) and X.`
Thus
`_(92)U^(235)+_(0)n^(1)rarr_(54)Xe^(139)+_(38)Sr^(94)+3_(0)n^(1)`
47.

If radius of Bohr first orbit is x, then de Broglie wavelength of electron in 3rd orbit is nearly

Answer»

`2pi x`
`9pix`
`9x`
`6pix`

Solution :`mvr_(N)=(ny)/(2pi)` or `MV((n^(2)a_(0))/Z)=(nh)/(2pi)`
Bohr.s RADIUS ofatom `a_(0)=x,=3`
`Z=1`
`:.mv(9x)/1=(2h)/(2pi)` or `mv=h/(6pix)`
`:. lamda=h/(mv)=6pix`
48.

If x_(1) and x_(2) represent the mole fraction of a component A in the vapour phase and liquid mixture respectively and p_(A)^(@) and p_(B)^(@) represent vapours pressures of pure A and pure B. then total vapour pressure of the liquid mixture is

Answer»

`(p_(A)^(@)-x_(1))/(x_(2))`
`(p_(A)^(@)-x_(2))/(x_(1))`
`(p_(B)^(@)x_(1))/(x_(2))`
`(p_(B)^(@)x_(2))/(x_(1))`

SOLUTION :Mole fraction in the VAPOUR phase `(x_(1))=(p_(A))/(P_("total"))`
But `p_(A)=x_(A)xxp_(A)^(@)=x_(2)xxp_(A)^(@)`
Hence, `x_(1)=(x_(2)p_(A)^(@))/(P_("total"))or P_("total")=(p_(A)^(@)x_(2))/(x_(1))`
49.

If radium and chlorine combine to form radium chloride the compound is

Answer»

No LONGER RADIOACTIVE
Twice as radioactive as radium
Half as radioactive as radium
Thrice as radioactive as radium

Answer :D
50.

Mole fraction of component A in vapour phase is chi_(1) and that of component A in liquid mixture is chi_2, then (p_(A)^@)= vapour pressure of pure A, p_(B)^@ = vapour pressure of pure B), the total vapour pressure of liquid mixture is

Answer»

<P>`(P_(A)^(@)x_(1))/(x_(2))`
`(P_(A)^(@)x_(2))/(x_(1))`
`(P_(B)^(@)x_(1))/(x_(2))`
`(P_(B)^(@)x_(2))/(x_(1))`

ANSWER :B