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.

For which of the following gaseous mixtures, Dalton's law of partial pressure is not applicable-

Answer»

`SO_(2),He,NE`
`NH_(3),HBr,HCl`
`O_(2),N_(2),CO_(2)`
`N_(2),H_(2),O_2`

Answer :B
Discussion
2.

For which of the following elements, the standard enthalpy is not zero?

Answer»

C (DIAMOND)
C (GRAPHITE)
Liquid mercury
Rhombic SULPHUR

ANSWER :A
Discussion
3.

For which of the following compounds will Lassaigne's test for nitrogen fail ?

Answer»

`NH_(2)CONH_(2)`
`CH_(3)CONH_(2)`
`NH_(2)NH_(2)`
`C_(6)H_(5)NH_(2)`

SOLUTION :`NH_(2)NH_(2)` (hydrazine) does not RESPOND to LASSAIGNE's test.
Discussion
4.

For which of the following compounds theLassaigne's test of nitrogen will fail?

Answer»

` H_(2)NCONH_(2)`
` H_(2)"NNH"_(2) 2HCI`
`C_6H_(5)NH_2`
`CH_3CONH_2`

ANSWER :B
Discussion
5.

For which of the following compound,the Lassaigne's test for N will fail ?

Answer»

`NH_2 CONH NH_2 HCI`
`NH_2 NH_2. 2HCI`
`NH_2 CONH_2`
`C_6 H_5 N = N- C_6 H_5`

ANSWER :B
Discussion
6.

For which of the following cases DeltaS = (DeltaH)/T

Answer»

A PROCESS for which `DeltaC_p =0` but `DeltaC_v = 0` 
An adiabatic process 
An isobaric or ISOTHERMAL process 
An isothermal REVERSIBLE TRANSITION process

Answer :D
Discussion
7.

For which of the element, law of octave is not applicable?

Answer»

SOLUTION :HEAVY METAL.
Discussion
8.

For which hybridization, there are two unequal bond angles

Answer»

`sp^3`
`sp^2`
sp
`sp^3d`

Solution :`sp^3d` has BOND ANGLES `90^@ & 120^@`
Discussion
9.

For which compound are the empirical and molecualr formular the same?

Answer»

`C_(6)H_(5)COOH`
`C_(6)H_(4)(COOH)_(2)`
HOOCCOOH
`CH_(3)COOH`

Answer :A
Discussion
10.

For which crystal anion-anion contact is valid ?

Answer»

NaF
NaI
CsBr
KCl

Solution :Iodide ION is LARGEST in size and `Na^+` ion is small in size. Hence, `I^(-) - I^-` CONTACT is possible in `Na^+ I^-` (`Na^+` ion PRESENT in the VOID)
Discussion
11.

For what purpose are the black diamonds used ?

Answer»

SOLUTION :`For CUTTING glass and for making ROCK DRILLS
Discussion
12.

For weak acid (alphais very small)

Answer»

`K_a = C. alpha ^(@) `
` alpha = SQRT((K_a)/( c) `
` pH^(+) ] =C. alpha `
All the above

SOLUTION :`K_a =(CALPHA ^(2))/( 1- alpha )~~ Calpha ^(2) `
Discussion
13.

For vaporization of water at 1 atmospheric pressure, the values of Delta H and Delta S are 40.63 KJ/mol""^(1) and 108.8 JK^(-1)/mol^(-1), respectively. The temperature when Gibbs energy change(Delta G) for this transformation will be zero, is :

Answer»

293.4 K
273.4 K
393.4 K
373.4 K

Solution :`H_(2) O_((L)) overset(1 "atm")(hArrr) H_(2) O_((g))`
`Delta H = 40630 "J mol"^(-1)`
`Delta S = 108.8 "JK mol"^(-1)`
`Delta G = Delta H - T Delta S`.
(When `Delta G = 0, Delta H - T Delta S= 0`)
`T= (Delta H)/( Delta S)= (40630 "J mol"^(-1) )/( 108.8 "J mol"^(-1) )`
`= 373.4` K
`THEREFORE` Correct answer is (D)
Discussion
14.

For types of reaction mechanism…. (i) reaction of propene with HBr in presence of peroxide is not a type of free radical addition reaction. (ii) chlorination of methane in presence of sunlight is a kind of free radical substitution reaction. (iv) dehydration of propanol is a kinf of beta-elemination reaction.

Answer»


ANSWER :(i-F), (ii-T), (iii-T), (iv-T)
Discussion
15.

For two ionic solids CaOand KI , identify the wrong statement among the following ?

Answer»

LATTICE energy of CaO is much HIGHER than that of KI
KI is SOLUBLE in benzene
CaO has high m.p.
KI has high m.p.

Solution :Due to lower lattice energy of KI as compared to that of CaO , the m.p. of KI is much lower than that of CaO .
Discussion
16.

For two gases A and B with molecular masses M_(A) and M_(B), it is observed that at a certain temperature T, the mean velocity of A is equal to the root mean square velocity of B. What should be done to the temperature of A so that its mean velocity becomes equal to the mean velocity of B at temperature T ?

Answer»

Solution :`BAR(c_(A))=sqrt((8RT)/(piM_(A))),"" c_(B)=sqrt((3RT)/(M_(B)))`
GIVEN`bar(c_(A))=c_(B)`.HENCE,`(8RT)/(piM_(A))=(3RT)/(M_(B))"or"(M_(A))/(M_(B))=(8)/(3pi)`
Now, if B at TEMPERATURE T and A is at T', then
`bar(c_(A))=bar(c_(B))""` (Required CONDITION)
`:. "" sqrt((8RT')/(piM_(A)))=sqrt((8RT)/(piM_(B)))"or"(T')/(M_(A))=(T)/(M_(B))"or"(T')/(T)=(M_(A))/(M_(B))=(8)/(3pi)`
or`T'=(8)/(3pi)xxT`. As`(8)/(3pi)lt1,T'ltT`. Hence, temperature of B should be lowered to `(8)/(3pi)` or 0.85 of temperature T.
Discussion
17.

For two gases A and B with molecular weights M_A and M_B, it is observed that at a certain temperature T, the mean velocity of A is equal to the root mean square velocity of B. Thus the mean velocity of A can be made equal to the mean velocity of B if

Answer»

A is increased to a temperature `T_2 = (3 PI)/8 T`
A is lowered to a temperature `T_2 = (8T)/(3pi)`
B is increased to a temperature `T_2 = (3pi T)/8`
B is lowered to a temperature `T_2 = (8T)/(3pi)`

Solution :`(barC)_A = ( C)_B`
`implies sqrt((8RT)/(pi M_A)) = sqrt((3RT)/(M_B)) implies 8/(pi M_A) = 3/(M_B) `
Case (i) changing T of A
`(barC_A) = (barC)_B implies sqrt((8RT)/(pi M_A)) = sqrt((8RT_2)/(pi M_B))`
`implies (T)/(M_A) = (T_2)/(M_B) implies T_2 = (M_B)/(M_A) T implies T_2 = (3pi)/(8) T > T.`
Discussion
18.

For two-dimensional hexagonal lattice, thee unit cell is _______

Answer»


ANSWER :RHOMBUS with an ANGLE of `60^@`
Discussion
19.

For two- dimensional hexagonal lattice, the unit cell is ______

Answer»


ANSWER :TRICLINIC
Discussion
20.

For tow ionic solids CaO and KI, identify the wrong statement among the following ?

Answer»

Lattice energy of CaO is much higher than that of KI
KI is soluble in BENZENE
CaO has HIGH m.p.
KI has high m.p.

Solution :DUE to lower lattice energy of KI as COMPARED to CaO, the m.p. of KI is much lower than that of CaO.
Discussion
21.

For three different metals A,B,C photoemission is observed one by one. The graph of maximum kinetic energy versus frequency f incident radiation are sketched as:

Answer»




ANSWER :D
Discussion
22.

For this process (overall change) which is correct

Answer»

q= 0, W= 0
`q = +ve, W = -ve`
`q = -ve, W_("on the SYSTEM") = +ve`
`DELTA S_("sys") = 0, Delta U_("sys") = 0, Delta H_("sys") = 0`

SOLUTION :Cyclic `RARR Delta u = 0, Delta S = 0, Delta H = 0`
Clockwise `rArr W_("on") = -Ve rArr q = +ve`
Discussion
23.

For the transition C_(("diamonid")) rarr C_(("graphite")) , Delta H = -1.5KJ. It follows that

Answer»

GRAPHITE is stabler than DIAMOND 
Diamond is stabler than graphite 
Graphite is ENDOTHERMIC SUBSTANCE
Diamond is EXOTHERMIC substance 

Answer :A
Discussion
24.

For the third electron of Lithium atom moving in its premissible orbit, the values of angular momentum and energy are...... Respectively

Answer»

Angular momentum `(h)/(2PI)` energy `=(-9pi^(2)E^(4)m)/(2h^(2)`
Angular momentum `(3h)/(2pi)` energy `=(+3pi^(2)em^(4))/(4h^(2)`
Angular momentum `(h)/(2pi)` energy `=(3pi^(2)me^(4))/(2h^(2)`
Angular momentum `(h)/(2pi)` energy `=(-6pi^(2)e^(4)m)/(3h^(2)`

Solution :`Li 1S^(2)2S^(1)`
third ELECTRON of Li means `2s^(1)`
`therefore` angular momentum quantum no = l =0 angularmomentum of electron in any orbital
accordingto bohrequation`E=(2e^(4)pi^(2)z^(2)m)/(n^(2)h^(2))`
for third electronof li `1s^(2) 2s^(1)`
`=-9/2 (e^(4)pi^(2)m)/(h^(2))`
Discussion
25.

For the test of halides, the soda extract is acidified with :

Answer»

dil. `H_(2)SO_(4)`
dil. `HNO_(3)`
dil. `HCl`
any of the three

Solution :SODA extract is ACIDIFIED with dilute `HNO_(3)`
Discussion
26.

For the system 3A+2BhArrC, the expression for equilibrium constant K is

Answer»

`([3A]XX[2B])/([C])`
`([A]^(3)xx[B])/([C])`
`([C])/([A]^(3)xx[B]^(2))`
`([C])/([3A]xx[2B])`

Answer :C
Discussion
27.

For the synthesis of ammonia by the reaction N_(2) + 3 H_(2) hArr 2 NH_(3)in the Haber's process, the attainment of equilibrium is correctly predicted by the curve

Answer»




Solution :(a) is correct because CONCENTRATIONS of `H_(2) and N_(2)` decrease with time while that of `NH_(3)` INCREASES with time and after equilibrium , all of them reamain constant.
Discussion
28.

For the study of catalystic reaction who was awarded noble prize

Answer»

OSTWALD
Berzilius
Vant Hoff
Werner

Solution :ostwald was AWARDED 1909 Chemistry Noble prize in the FIELD of catalysis.
Discussion
29.

For the spontaneous reaction at each temperature .......... .

Answer»

`DELTA G -ve, Delta H +ve and Delta S +ve`
`Delta G +ve, Delta H -ve and Delta S +ve`
`Delta G -ve ,Delta H -ve and Delta S -ve`
`Delta G - ve, Delta H -ve and Delta S +ve`

ANSWER :D
Discussion
30.

For the spontaneous reactions ........... .

Answer»

`DELTAH = -ve , DELTA S = +ve`
`Delta H = + ve , Delta S = -ve`
`Delta H = +ve, Delta S = -ve`
`Delta H = -ve , Delta S = +ve`

ANSWER :A
Discussion
31.

For the sparingly soluble salts, an equilibrium is established between the undissolved solid unit and ions of the dissolved salt. For a solution of the salt like A_x B_y. A_x B_y =xA^(y+)+yB^(x-) K_(sp)=[A^(y+)]^x [B^(x-)]^y For precipitationto occur , the ionic product must be greater than K_(sp). What is the minimum concentration of OH^- ions required to precipitate Fe(OH)_3 from a 0.001 M solution of FeCl_3 (K_(sp) of Fe(OH)_3 =1xx10^(-36) )

Answer»

`1xx10^(-12)`
`1xx10^(-11)`
`1xx10^(-18)`
`1xx10^(-16)`

Solution :`K_(sp)=[Fe^(3+)][OH^-]^3`
`[Fe^(3+)]=[FeCl_3]`=0.001 M
`[OH^-]^3 =(1xx10^(-36))/0.001 =1xx10^(-33)`
`THEREFORE [OH^-]=1xx10^(-11)`
Discussion
32.

For the same increase in volume,why work done is more if the gas is allowed to expand reversibly at higher temperature?

Answer»

SOLUTION :For isothermal reversible expansion, `w=-P_("INT") xxDeltaV`. At HIGHER temperature, INTERNAL pressure of the gas is more.
Discussion
33.

For the reversible reaction N_(2(g))+3H_(2(g))hArr2NH_(3(g))+"Heat". The equilibrium shifts in forward direction.

Answer»

by INCREASING the concentration of `NH_(3(g))`
by increasing the pressure and DECREASING the temperature.
by decreasing the pressure and decreasing the temperature
by decreasing the concentration of `N_(2(g))` and `H_(2(g))`.

Answer :B
Discussion
34.

For the reversible reaction N_(2)(g) + 3H_(2)(g) hArr2NH_(3)(g) " at " 500^(@) C, the value of Kp is 1.44 xx 10^(-5)" atm"^(-2) . Find the K_(c) value.

Answer»

SOLUTION :`1.44 XX 10^(-5)//(0.082 xx 773)^(-2) `
Discussion
35.

For the reversible reaction ,N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) (g) " at " 500^(@)C , " the value of " K_(p) " is "1*44 xx 10^(-5)when partial pressure is measured in atmospheres. The corresponding value of K_(c) , with concentration in mole litre^(-1), is

Answer»

` 1*44 xx 10^(-5) //(0* 082 xx 500) ^(-2)`
` 1*44 xx 10^(-5) //(8*314 xx 773 )^(-2)`
` 1*44 xx 10^(-5) //(0*082 xx 773)^(2)`
` 1*44 xx 10^(-5) // (0*082 xx 7773 )^(-2)`

Solution :` K_(C) = K_(p) //(RT)^(Delta N) = 1*44 xx 10^(-5) //(0*082 xx 773 )^(-2)`
Discussion
36.

For the reversible reaction, N_(2) (g) + 3 H_(2) (g) hArr 2 SO_(3) (g) The equilibrium shifts in the forward direction

Answer»

by increasing the CONCENTRATION of `NH_(3)`
by decreasing the pressure
by decreasing the pressure
by decreasing the concentration of `N_(2) (g) and H_(2)` (g)

Solution :As forward reaction is accompanied by decrease in the number of moles, applying Le Chatelie'rs principle , increses of pressure will shift the equilibrium in the forward DIRECTION, Also , as forward reaction is exothermic , increases of TEMPERATURE will shift the equilibrium in the forward direction .
Discussion
37.

For the reversible reaction N_(2 (g)) + 3 H_(2 (g)) hArr 2 NH_(3 (g)) at 500^(@) C . The value of K_(p) is 1.44 xx 10^(-5) , when partial pressure is measured in atmosphers . The corresponding value of K_(c) with concentration in mol L^(-1) is

Answer»

`1.44 XX 10^(-5) l (0.082 xx 500)^(-2)`
`1.44 xx 10^(-5) (8.314 xx 773)^(-2)`
`1.44 xx 10^(-5) l (0.082 xx 500)^(2)`
`1.44 xx 10^(-5) l (0.082 xx 773)^(-2)`

ANSWER :D
Discussion
38.

For the redox recation MnO_(4)^(-)+C_(2)O_(4)^(2-)+H^(+)toMn^(2+)+CO_(2)+H_(2)O the correct coefficients of the reactants for the blalanced equation are

Answer»

`{:(MnO_(4)^(-),C_(2)O_(4)^(2-),H^(+)),(2,16,5):}`
`{:(MnO_(4)^(-),C_(2)O_(4)^(2-),H^(+)),(2,5,16):}`
`{:(MnO_(4)^(-),C_(2)O_(4)^(2-),H^(+)),(16,5,2):}`
`{:(MnO_(4)^(-),C_(2)O_(4)^(2-),H^(+)),(5,16,2):}`

Answer :B
Discussion
39.

For the redox reaction : MnO_4^(-) +Fe^(2+) +H^(+) rarr Mn^(2+)+Fe^(3+)+H_2O The correct coefficients of the reactants in the balanced reaction are :

Answer»

`{:(MnO_4^(-), FE^(2+),H^+),(1,5,8):}`
`{:(MnO_4^(-), Fe^(2+),H^+),(2,5,8):}`
`{:(MnO_4^(-), Fe^(2+),H^+),(4,5,16):}`
`{:(MnO_4^(-), Fe^(2+),H^+),(2,5,16):}`

ANSWER :A
Discussion
40.

For the redox reaction MnO_(4)^(-)+C_(2)O_(4)^(2-)+H^(+)toMn^(2+)+CO_(2)+H_(2)O the correct coefficients of the reactants for the balanced equation are

Answer»

`{:(MnO_(4)^(-),C_(2)O_(4)^(2-),H^(+)),("5","16","2"):}`
`{:(MnO_(4)^(-),C_(2)O_(4)^(2-),H^(+)),("16","5","2"):}`
`{:(MnO_(4)^(-),C_(2)O_(4)^(2-),H^(+)),("2","16","5"):}`
`{:(MnO_(4)^(-),C_(2)O_(4)^(2-),H^(+)),("2","5","16"):}`

Solution :
n = 5 of `MnO_(4)^(-)`, n = 2 of `C_(2)O_(4)^(-2)`, Ration = 5 : 2
So, Balanced equation :
`2MnO_(4)^(-)+5C_(2)O_(4)^(2-)+16H^(+)to2Mn^(2+)+10CO_(2)+8H_(2)O`
Discussion
41.

For the redoxreaction MnO_(4)^(-)+C_(2)O_(4)^(2-)+H^(+)rarrMn^(2+)+CO_(2)+H_(2)O The correct coefficients of the reactants for he balanced equation are MnO_(4)^(-)C_(2)O_(4)^(2-) H^(+)

Answer»

16 5 2
2 5 16
2 16 5
5 16 2

Solution :`2MnO_(4)^(-)+5C_(2)O_(4)^(2-)+16H^(+)rarr2Mn^(2+)+10CO_(2)+8H_(2)O`
Thus option (B) is correct
Discussion
42.

For the redox reaction, MnO_(4)^(-)+C_(2)O_(4)^(2-)+H^(+) to Mn^(2+)+CO_(2)+H_(2)O the correct coefficients of the reactants for the balanced reaction are :

Answer»

`{:(MnO_(4)^(-),C_(2)O_(4)^(2-),H^(+)),(" "2," "5,16):}`
`{:(MnO_(4)^(-),C_(2)O_(4)^(2-),H^(+)),(" "16," "5,2):}`
`{:(MnO_(4)^(-),C_(2)O_(4)^(2-),H^(+)),(" "5," "16,2):}`
`{:(MnO_(4)^(-),C_(2)O_(4)^(2-),H^(+)),(" "2," "16,5):}`

SOLUTION :N//A
Discussion
43.

For the redox reaction, MnO_4^- + C_2 O_4^(2-) + H^+ rarr Mn^(2+) + CO_2 + H_2 O the correct coefficients of the reactants for the balanced reaction are respectively MnO_4^-, C_2 O_4^-, H^+ :

Answer»

`{:(MnO_(4)^(-),C_(2)O_(4)^(2-),H^(+)),(2,5,16):}`
`{:(MnO_(4)^(-),C_(2)O_(4)^(2-),H^(+)),(16,5,2):}`
`{:(MnO_(4)^(-),C_(2)O_(4)^(2-),H^(+)),(5,16,2):}`
`{:(MnO_(4)^(-),C_(2)O_(4)^(2-),H^(+)),(2,16,5):}`

Answer :A
Discussion
44.

For the redox reaction: MnO_(4)^(-)+C_(2)O_(4)^(2-)+H^(+)rarrMn^(2+)+CO_(2)+H_(2)O The correct coefficeents of the reactants for the balanced reaction are: MnO_(4)^(-)C_(2)O_(4)^(2-) H+

Answer»

`2" "5" "16`
`16" "5" "2`
`5" "16" "2`
`2" "16" "5`

SOLUTION :The BALANCED equation is:
`2Mn_(4)^(-)+5C_(2)O_(4)^(2-)+16H^(+)rarr2Mn^(2+)+10CO_(2)+8H_(2)O`
Discussion
45.

For the real gases reaction 2CO_((g)) + O_(2(g)) rarr 2CO_(2(g)), Delta H = - 560kJ. In a 10L rigid vessel at 500K, the initial pressure is 70 bar and after reaction it becomes 40 bar. The change in internal energy is:

Answer»

<P>`-557 kJ`
`-530 kJ`
`-563 kJ`
NONE of these

Solution :`Delta E = Delta H - V (Delta P)` `= -560 + 30 = -530KJ`
Discussion
46.

For the reactions N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) (g), " at " 400 K, K_(p) = 41." Find thevalue of "K_(p) for each of the followingreactions at the same temperature : (i)2 NH_(3) (g) hArr N_(2) (g) + 3 H_(2) (g)(ii) 1/2 N_(2) (g) + 3/2H_(2) (g) hArr NH_(3) (g)(iii) 2 N_(2) (g) + 6 H_(2) (g) hArr 4 NH_(3) (g)

Answer»

Solution :(i) It is the reverse of the GIVEN reaction. Hence , `K_(p) = 1/41`
(ii) It is obtained by DIVIDING the given equation by 2. Hence, `K_(p) = sqrt(41)`
(iii) It is obtained by MULTIPLYING the given equation by 2. Hence, `K_(p) = (41)^(2)`.
Discussion
47.

For the reaction,HC_(2)H_(2)ClO_(2)(aq) Leftrightarrow H^(+)+C_(2)H_(2)ClO_(2)^(-)(aq) HCH,C//O (aq) =H+CHCO_2 (aq), the equillibrium constant at 25^(@)C is 1.35xx10^(-3)M. Calculate DeltaG^(@).

Answer»


ANSWER :16.3 KJ
Discussion
48.

For the reaction, Zn + Cu^(2) rarr Zn^(2+) + Cuwhich of the following is the CORRECT statement ?

Answer»

ZN is REDUCED to `Zn^(2+)`
Zn is oxidized to `Zn^(2+)`
`Zn^(2+)` is oxidized to Zn.
`CU^(2+)` is oxidized to Cu.

Answer :B
Discussion
49.

For the reaction, X_(2)O_(4)(l) to 2XO_(2)(g) DeltaU = 2.1 kcal, DeltaS=20 cal K^(-1) at 300 K hence, DeltaG is

Answer»

`2.7` kcal
`-2.7` kcal
`9.3`kcal
`-9.3` kcal

Solution :`X_(2)O_(4)(l) to 2XO_(2)(g)`
`DeltaH = 2.1 kcal = 2100 cal, DeltaS = 20 cla K^(-1)`
`DeltaH = DELTAU + Deltan_(g) RT`
`Deltan_(g) = 2-0 = 2 ,R = 2 cal K^(-1) mol^(-1), T = 300 K`
`DeltaH = 2100 + (2 xx 2 xx 300) = 3300 cal`
`DeltaG = DeltaH-TDeltaS` = `3300 - 300(20) = -2700 cal = -2.7 kcal`
Discussion
50.

For the reaction X_(2)O_(4)(l) rarr 2XO_(2)(g), DeltaU = 2.1 kcal , DeltaS =20 calK^(-1) at300K.Hence,DeltaGis

Answer»

2.7 kcal
`-2.7 kcal`
9.3 kcal
`-9.3kcal`

Solution :`DeltaH=DeltaU+Deltan_(G) RT`
Given`:DeltaU =2.1 kcal ,Deltan_(f) = 2-0=2`,
`R = 2 xx 10^(-3) kcal, T= 300K`
`:. DeltaH =2.1+ 2 xx 2XX 10^(-3) xx 300 = 3.3 kcal`
Again, `DeltaG= DeltaH - T DELTAS`
Given `: DeltaS =20 xx 10^(-3) kcal K^(-1)`
`DeltaH= 3.3 kcal` ( calculateabove)
`:. DeltaG = 3.3 - 300 xx (20 xx 10^(-3))`
`=3.3-6 = - 2.7 kcal`
Discussion