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 the correct matches are A-p,A-s,B-q,B-r,C-p,C-q, and D-p then correctly labelled 4 x 4 matrix should book like the following

Answer»


ANSWER :A::B::C::D
2.

If the correct matches are A-p,A-s,B-q,B-r,C-p,C-q, and D-p then correctly labelled 4 x 4 matrix should book like the following

Answer»


ANSWER :A::B::C::D
3.

If the critical frequency (v_(0)) for emission of photoelectrons from a metal is 9.62xx10^(14) s^(-1), then light that can emit photoelectons should have a wavelength equal to

Answer»

6000Å
5000Å
4500Å
3000Å

Solution :The light with lowest `lambda` shall have LARGEST ENERGY.
`therefore` it can CAUSE the emission . (In this EQUATION calculation is not required).
4.

If the concentration of OH^- ions in the reaction Fe(OH)_(3(s)) hArr Fe_((aq))^(3+) + 3OH_((aq))^(-)is decreased by 1/4 times,then equilibrium concentration of Fe^(3+) will increase by:

Answer»

8 times
16 times
64 times
4 times

Solution :For this reaction `K_(eq)` is GIVEN by
`K=([Fe^(3+)][OH^-]^3)/([Fe(OH)_3])`
`=[Fe^(3+)][OH^-]^3` [ `because` [solid]=1]
If `[OH^+]` is decreased by `1/4` times then for reaction equilibrium constant to remain constant, we have to increase the CONCENTRATION of `[Fe^(3+)]` by a factor of `4^3` i.e. 4 x 4 x 4 =64 .
5.

If the concentration of glucose (C_(6)H_(12)O_(6)) in blood is0.9 g L^(-1) what will be the molarity of glucose in blood ?

Answer»

5 M
50 M
0.005 M
0.5 M

Solution :`0.9 G L^(-1)` means that 1000 mL (or 1 L) solutioncontains 0.9 g of glucose.
`:.` Number of moles `= 0.9g` glucose
`= (0.9)/(180)` mol glucose `= 5 xx 10^(-3)` mol glucose (molecular mass of glucose `(C_(6)H_(12)O_(6)) = 12 xx 6 + 12 xx 1 + 6 xx 16=180 U`)
i.e., 1 L solution contains 0.05 mole glucose or the molarity of glucose is 0.005 M.
6.

If the concentration of dissolved oxygen of water is below 6 ppm then the growth of fish gets inhibited.

Answer»

SOLUTION :TRUE STATEMENT
7.

If the components of air areN_(2), 78%, O_(2), 21%, Ar, 0-9% and CO_(2), 0-1%by volume, what would be the molecular weight of air?

Answer»

Solution :The volume RATIO of the gases will be the same as their mole ratio (Avogadro.s principle)
`THEREFORE` mol. Wt of air `=(78 xx 28 + 21 xx 32 + 0.9 xx 40 + 0.1 xx 44)/(78 + 21 + 0.9 + 0.1)`
(`N_(2) = 28, O_(2) = 32, Ar = 40` and `CO_(2) = 44`)
8.

If the combustion of 1 g of graphite produces 20. 7 kJ of heat, what will be molar enthalpy change ? Give the significance of sign also.

Answer»

SOLUTION :Enthalpy of combustion of 1 G graphite= 20.7 kJ
Molar enthalpy change for the combustion of graphite `DELTA H = ` enthalpy of combustion of 1g graphite `xx` molar mass
`Delta H = - 20.7 "kJ g"^(-1) xx 12 "g mol"^(-1)`
`Delta H = - 2.48 xx 10^(2) "kJ mol"^(-1)`
Negative sign in the value of `Delta H` indicates that the reaction is exothermic.
9.

If the combustion of1g of graphite produces 20.7 kJ of heat, whatwill be molar enthalpy change ? Give the significance of sign also.

Answer»

Solution :Molar enthalpy CHANGE of combustion of graphite `=` Enthalpy of combustion of 1 g `xx` Molar mass
`= 20.7 kJ MOL^(-1) xx 12 g mol^(-1)`
i.e., `DeltaH = - 2.48 xx 10^(2)kJ mol^(-1)`
Negative value of `DeltaH` shows that the reaction is exothermic .
10.

If the combustion of 1 g of graph produces 20.7 kJ of heat, what will be molar enthalpy change ? Give the significance of sign also.

Answer»

Solution :GIVEN that, ENTHALPY of combustion of 1 g graphite = 20.7 kJ
Molar enthalpy CHANGE for the combustion of graphite, `Delta H =` enthalpy of combustion of 1 g graphite `XX` molar mass
`Delta H = - 207 kJg^(-1)xx 12 g MOL^(-1)`
`Delta H = - 2.48 xx 10^(2) kJ mol^(-1)`
Negativve sign in the value of `Delta H` indicates that the reaction is exothermic.
11.

If the collision frequency of a gas at 1 atm pressure is Z, then its collision frequency at 0.5 atm is

Answer»

0.25 Z
0.50 Z
Z
2Z

Solution :At CONSTANT `T,Z PROP P^(2)`
`:. (Z_(1))/(Z_(2))=(P_(1)^(2))/(P_(2)^(2)), i.e., (Z)/(Z_(2))=((1)^(2))/((0.5)^(2))" or " Z_(2)=0.25" Z"`
12.

If the close packed cations in an AB type solid with NaCl structure have a radius of 75 pm, what would be the maximum and minimum sizes of the anions filling the voids ?

Answer»


SOLUTION :For close packed AB type solid with NaCl structure , `r_+/r_(-)`=0.414-0.732
`THEREFORE` MINIMUM value of `r_(-)=r_(+)/0.732=75/0.732` pm =102.5 pm , MAXIMUM value of `r_(-)=r_(+)/0.414=75/0.414` =181.2 pm .
13.

If the CO bond length is 121 pm, what is the distance between the nuclei of oxygen atoms in carbondioxide molecule?

Answer»

SOLUTION :Carbon dioxide is a LINEAR molecule with two C=O bonds opposite to each other.
`O=C=O`
121 PM`""` 121 pm
The DISTANCE between the nuclei of OXYGEN atoms in `CO_(2)` molecule is `121+121=242` pm.
14.

If the close packed cations in an AB type solid with NaCl structure have a radius of 75 pm,what would be the maximum and mimimumsizes of the anions filling the voids ?

Answer»


Solution :For close PACKED AB type solid with NaCl structure , ` ( r_(+))/(r_(-)) = 0.414 - 0.732`
MINIMUM value of ` r_(-)= ( r_(+))/ 0.732 = 75/0.732 "pm" ` = 102 . 5 pm , MAXIMUM value of ` r_(-)= ( r_(+))/ 0.414 = 75/0.414 = 181 .2`pm .
15.

If the electronegativity of two atoms is low, then expected bond between the elements is

Answer»

IONIC Bond
Covalent Bond
DATIVE bond
METALLIC Bond

ANSWER :D
16.

If the change in pH value is 2 unit then what is the change in concentration of H^+ ?

Answer»

SOLUTION :`10^(-2)` M MEANS 0.01 M CHANGE.
17.

If the change in entropy at 353 K is 0.087 kJ mol""^(-1) for benzene. Then calculate the heat of vapourisation.

Answer»

`-6.96 "KJ MOL"^(-1)`
`6.96 kJ mol"^(-1)`
`-30.711 "kJ mol"^(-1)`
`30.711 "kJ mol"^(-1)`

ANSWER :D
18.

If the buffer capacityof a buffer solutions is x, the volume of 1M NaoH added to 100mL of this solutionsto change the pH by 1 is

Answer»

`0.1 xmL `
` 10 xmL `
` 100x ML `
` x mL`

Solution :Buffer capacity`= ( N )/(Delta PH) `
` x = ( V xx 1)/(0.1 xx 1 ), V =0.1 x ` litres = 100 x ml
19.

If the boundary of system moves by an infinitesimal amount, the work involved is given by dw= -P_("ext") dV, for irreversible process W= -P_("ext") Delta V (where Delta V= V_(f) - V_(i)). For reversible process. P_("ext") = P_("int") +- dP ~= P_("int"), so for reversible isothermal process W = -nRT "In" (V_(f))/(V_(i)) 2 mole of an ideal gas undergoes isothermal compression along three different paths: (i) reversible compression from P_(i) = 2 bar and V_(i) = 8L " to" P_(f) = 20 bar (ii) a single stage compression against a constant external pressure of 20 bar (iii) a two stage compression consisting initially of compression against a constant external pressure of 10 bar until P_("gas") = P_("ext"), followed by compression aganist a constant pressure of 20 bar until P_("gas") = P_("ext") Order of magnitude work is

Answer»

`w_(1) gt w_(2) gt w_(3)`
`w_(3) gt w_(2) gt w_(1)`
`w_(2) gt w_(3) gt w_(1)`
`w_(1) = w_(2) = w_(3)`

SOLUTION :`W_("REV")` compression `lt W_("irr")`
Less the STEPS, HIGHER the W
20.

If the boundary of system moves by an infinitesimal amount, the work involved is given by dw= -P_("ext") dV, for irreversible process W= -P_("ext") Delta V (where Delta V= V_(f) - V_(i)). For reversible process. P_("ext") = P_("int") +- dP ~= P_("int"), so for reversible isothermal process W = -nRT "In" (V_(f))/(V_(i)) 2 mole of an ideal gas undergoes isothermal compression along three different paths: (i) reversible compression from P_(i) = 2 bar and V_(i) = 8L " to" P_(f) = 20 bar (ii) a single stage compression against a constant external pressure of 20 bar (iii) a two stage compression consisting initially of compression against a constant external pressure of 10 bar until P_("gas") = P_("ext"), followed by compression aganist a constant pressure of 20 bar until P_("gas") = P_("ext") Work done (in bar -L) on the gas in reversible isothermal compression is:

Answer»

<P>9.212
36.848
18.424
none of these

Solution :Work done `= -2.303 n TR "long" (V_(2))/(V_(1))`
`= -2.303 xx 0.083 xx 2 xx ((P_(i)V_(i))/(NR)) ."LOG" (V_(F))/(V_(i))`
`= -2.303 xx 0.083 xx 2 ((2 xx 8)/(2 xx 0.083)) "log" (2)/(20)`
= 36.848 bar Lt
21.

If the both compound (A) and (B) are optically active, the possible structrue of (A) is:

Answer»

`cis-I`
`trans-I`
`cis-II`
`trans-II`

ANSWER :C::D
22.

If the bond length and dipolemoment of a diatomic molecule are 1.25A^@and 1.0D respectively, what is the percent ionic charcter of the bond ?

Answer»

`10.66`
`12.33`
`16.66`
`19.33`

ANSWER :C
23.

The dipole moment of HX molecule is 1.92 D and bond distance is 1.2 A'. What is the percentage ionic character of HX ?

Answer»

`10.66`
`12.33`
`16.66`
`19.33`

ANSWER :C
24.

If the bond energies of H-H, Br- Br and H -Br are 433, 192 and 364kJ mol^(-1) respectively, DeltaH^(@) for the reaction H_(2)(g) + Br_(2)(g) rarr 2HBr(g) is

Answer»

`- 261kJ`
`+103kJ`
` + 261kJ`
`-103kJ`

SOLUTION :`Delta_(r)H= SigmaBE` ( REACTANTS)`- SIGMA` B.E.( PRODUCTS)
25.

If the bond energies of H-H, Br-Br and HBr are 433, 192 and 364 kJ mol^(-1) respectively, then DeltaH^@ for the reaction : H_(2(g)) + Br_(2(g)) rarr 2HBr_((g)) is

Answer»

`-261kJ`
`+103kJ`
`+261kJ`
`-103kJ`

ANSWER :D
26.

If the bond dissociation energies of XY , X_(2) and Y_(2) ( alldiatomic molecules )are in the ratio1:1 : 0.5 and DeltaH _(f)for the formation of XY of - 200 kJ mol^(-1) , the bond dissociation energy of X_(2) will be

Answer»

`100 kJ mol^(-1)`
`200 kJ mol^(-1)`
` 400 kJ mol^(-1)`
` 800 kJmol^(-1)`

Solution :Suppose BOND dissociation energy of XY `=akJ mol^(-1) , i.e., `BE (XY )= a . Then BE `9X_(2)0= a, BE(Y_(2))= 0.5 a`
Aim `:(1)/(2) X_(2)+ (1)/(2) Y_(2) rarr XY`
`Delta_(r)H = BE` ( Reactants ) - BE ( Products )
`= [ (1)/(2) BE (X_(2)) + (1)/(2) BE(Y_(2)) ] = BE(XY)`
`:.- 200 =((a)/(2) + (0.5a)/( 2)) -a`
or` - 200 = - 0.25 a` or `a= 800 kJ mol^(-1)`
27.

If the binding energy of electrons in a metal is 250 kJ mol^(-1), what should be threshold frequency of the striking photons ?

Answer»

Solution :BINDING energy of 1 mole of ELECTRONS = 250 kJ
`:.` Binding energ of 1 electron `= (250)/(6.022 XX 10^(23)) kJ = 4.15 xx 10^(-22) kJ = 4.15 xx 10^(-19) J`
Threshold energy `(hv_(0))` = Binding energy
`:. hv_(0) = 4.15 xx 10^(-19) J`
or `v_(0) = (4.15 xx 10^(-19)J)/(h) = (4.15 xx 10^(-19) J)/(6.626 xx 10^(-34) Js) = 6.26 xx 10^(14) s^(-1)`
28.

If the atomic weight of carbon were set at 24 amu, the value of the Avogadro constant would be

Answer»

`6.022 XX 10^(23)`
`12.044 xx 10^(23)`
`3.011 xx 10^(23)`
none of these

ANSWER :B
29.

If the atomic radius of non-metal bromine is 1.14 Å, its covalent radius is

Answer»

1.14 Å
`1.12 Å`
`1.16 Å`
`0.57 Å`

ANSWER :A
30.

If the atomic number of an element is 33, it will be placed in the periodic table in the

Answer»

FIRST GROUP 
THIRD group
Fifth group 
Seventh group 

ANSWER :C
31.

If the absolute temperature of a sample of gas is increased by a factor of 1.5, by what ratio does the average molecular speed of the molecules increases?

Answer»

1.2
1.5
2.2
`3.0`

ANSWER :a
32.

If tempetature changes form 27^(@)C to 127^(@)C, the relative percentage change in rms velocity is

Answer»

1.56
2.56
15.6
82.4

Solution :`c=sqrt((3RT)/(M))`
At `27^(@)C, ""c=sqrt((3Rxx300)/(M))=sqrt(300)X`
`=17.3x"" (x=sqrt((3R)/(M)))`
At `127^(@)C`
`c=sqrt((3Rxx400)/(M))=sqrt(400)x=20x`
`:.` Increase`=20x-17.3x=2.7x`
% increase`=(2.7x)/(17.3x)xx100=15.6`
33.

If ten volumes of dihydrogen gas react with five volumes of dioxygen gas, how many volumes of water vapour would be produced?

Answer»

Solution :`underset("2 volumes")(2H_(2)(g)) + underset("1 VOLUME")(O_(2)) to underset("2 volumes")(2H_(2)O) (g)`
Two volumes of `H_2` react with one volume of `O_(2)` to produce two volumes of WATER vapour. Hence, 10 volumes of `H_2` will react completely with 5 volumes of `O_(2)` to produce 10 volumes of water vapour.
34.

If temperature of CO_(2) will be increased or decreased to 30.98^(@)C then what changes can be observed ?

Answer»

Solution :If TEMPERATURE is increased by `30.98^(@)C, CO_(2)` VAPOUR is obtained. If temperature is decreased by `30.98^(@)C`. LIQUID `CO_(2)` is obtained.
35.

If temperature and volume of an ideal gas is increased to twice its values, the initial pressure P becomes

Answer»

<P>4P
2P
P
3P

Solution :PV is equal to nRT and Boyle's law P is indirectly proportional to VALUME at CONSTANT temperature
36.

If T is the time required by electron in taking one round in an orbit, n represents the number of waves in an orbit, r represents the radius of orbit, then which have the value of 1/2 for 2^(nd) orbit of H and 4^(th) orbit of He^(+) ?

Answer»

`(r_(2(H)))/(r_(4(He^(+))))`
`(T_(2(H)))/(r_(4(He^(+))))`
`(n_(2(H)))/(n_(4(He^(+))))`
`(E_(2(H)))/(E_(4(He^(+))))`

Answer :A::B::C
37.

If system moves from ordered state to disordered state, its entropy ____

Answer»

SOLUTION :INCREASES
38.

If substance is completely pure crystalline at 273 K, which of the following state function will be zero at 273 K ?

Answer»

FREE energy
Entropy
Enthalpy
All the given

Answer :B
39.

If standard enthalpy change Delta_(r )H^(Θ) + -2.05 xx 10^(3) kJ mol^(-1) calculate the energy of oxygen-oxygen bond in O_(2) molecules and compare the calculate value with the value given in the table.

Answer»

Solution :`-205xx10^(3)=(694+3312+5x)=(4446+3712)`
`4006+5x-8158=-4152+5x`
`:. 5x=-2050+4152`
`:. x=420.4`
`DeltaH_(O=O)RARR 420.4 kJ mol^(-1)`
40.

If spheres of radius 'r' are arranged in ccp fashion (ABC ABC…)the vertical distance between any two consecutive A layers is

Answer»

`4rsqrt(2/3)`
`4r sqrt(3/2)`
6 r
`rsqrt6`

SOLUTION :Distance between two consecutive A layers (d)
` 2 sqrt(2/3)a `
Buta = 2 r. ` d = 2sqrt(2/3) XX 2r = 4r sqrt(2/3)`
41.

If spheres of radius 'r' are arranged in ccp fashion (ABC ABC…) , the vertical distance between any two consecutive A layers is

Answer»

`4rsqrt(2/3)`
`4rsqrt(3/2)`
6 r
`r sqrt6`

Solution :Distance between two consecutive A LAYERS (d)=`2sqrt(2/3)a`
But a=2r `therefore d=2sqrt(2/3) xx2r =4rsqrt(2/3)`
42.

If solubility product for CaF_(2) is 1.7xx10^(-10) at 298 K, calculate the solubility in mol L^(-1).

Answer»


ANSWER :`3.5xx10^(-4)`
43.

If silver iodide crystallizes in a zinc blende structure with I^- ions forming the lattice, then calculate fraction of the tetrahedral voids occupied by Ag^+ ions.

Answer»

Solution :In AgI, if there are N `I^-` IONS, there will be n `Ag^+` ions. As `I^-` ions from the lattice, number of tetrahedral voids=2n. As there are n `Ag^+` ions to occupy these voids, therefore, fraction of tetrahedral voids OCCUPIED by `Ag^+` ions =n/2n=1/2=50%
44.

If silver iodide crystallizes in a zinc blende structure withI^(-)ions forming the lattice, then calculate fraction of the tetrahedral voids occupied byAg^(+) ions.

Answer»

SOLUTION : In AGL, If there are n`I^(-)`ions, there will ben ` Ag^(+)`ions. As ` I^(-)`ions form the lattice , number of tetrahedral voids = 2N, As there are n ` Ag^(+)`ions to occupy these voids, therefore, fraction of tetrahedralvoids occupied by `Ag^(+) " ions"= n //2n `= 1/2 = 50 % .
45.

If same mass of liquid water and a piece of ice is taken, then why is the density of ice less than that of liquid water ?

Answer»

Solution :The mass per unit volume (i.e. mass volume) is called density since water EXPANDS on freezing, therefore, volume of ice for the same mass of water is more than liquid water. In other words, the density of ice is LOWER than liquid water and HENCE ice floats on water.

HEXAGONAL honey comb STRUCTURE of ice.
46.

If same mass of liquid water and a piece of ice is taken, then why is the denisty of ice less than that of liquid water ?

Answer»

Solution :The mass per unit volume (i.e., mass/volume) is called density. Since WATER on FREEZING , THEREFORE , volume of ICE for the same mass of water is more than LIQUID water. In other words, density of ice is lower than liquid water and hence ice floats on water.
47.

If salt bridge is removed from two half cells, the voltage

Answer»

DROPS to zero
does not changes
increase gradually
increases rapidly

Solution :Salt bridge permits the FLOW of current by COMPLETING the circuit. No current will flow and the voltage will DROP to zero if the salt bridge is REMOVED.
48.

Mg^(2+) and Al^(3+) have same

Answer»

`+505.6`
`-505.6`
`-1011.2`
`+1011.2`

ANSWER :B
49.

If S + O_(2) rarr SO_(2), DeltaH= -398.2kJ SO_(2) + (1)/(2) O_(2) rarr SO_(3), DeltaH= -98.7kJ , SO_(3) + H_(2)O rarr H_(2)SO_(4), DeltaH= -130.2 kJ H_(2) + (1)/(2) O_(2) rarr H_(2)O, DeltaH= -227.3kJ The enthalpy of formation of sulphuric acid at 298K will be

Answer»

`-854.4 K.J `
`-754.4 KJ `
`-650.3 KJ `
`433.7 KJ`

ANSWER :A
50.

If S + O_(2) rarr SO_(2), Delta H = -398.5 kJ SO_(2) + (1)/(2)O_(2) rarr SO_(3), Delta H= -98.7kJ , SO_(3) + H_(2)O rarr H_(2)SO_(4), Delta H= - 130.5kJ H_(2) + (1)/(2) O_(2) rarr H_(2)O, Delta H= -227.3kJ If magnitude of enthalpy of formation of sulphuric acid at 298K is 95x, x= ?

Answer»

`-854.4 K.J`
`-754.4K.J`
`-650.3 K.J`
`-433.7 K.J`

ANSWER :A