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 an ideal gas ____

Answer»

`C_P-C_V=0`
`C_P - C_V =R`
`C_V -C_P=R`
`C_R-C_P GT R`

ANSWER :B
2.

For an exothermic reaction, what happens to the equilibrium constant if temperature is increased?

Answer»

Solution :`K=K_(f)//K_(b)`. In exothernic REACTION, with increases of TEMPERATURE, `K_(b)` increases much more than `K_(f)`. Hence `K` DECREASES.
3.

For an exothermic reaction, what happens to the equilibrium constant if temperature is incresed ?

Answer»

Solution :`K= k_(f)//k_(B)." In EXOTHERMIC reaction, with increase of TEMPERATURE , " k_(b)" increases much more than " k_(f) " Hence".`
4.

For an equilibrium reaction, the rate constants for the forward and the backward reaction are 2.38 xx 10^(-4) and 8.15 xx 10^(-5)respectively . Calculate the equilibrium constant for the reaction.

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Solution :Equilibrium CONSTANT `K= (k_(f))/(k_(b)) = (2.38 xx10^(-4))/(8.15 XX 10^(-5))=2.92`
5.

For an equilibrium reaction K_(p) = 0.0260 at 25^(@)C and DeltaH = 32.4 kJ mol^(-1) . Calculate K_(p) at 37^(@)C.

Answer»

<P>

Solution :`T_(1) = 25 + 273 = 298 K , T_(2) = 37 + 273 = 310 K` ,
`Delta H = 32.4 KJ mol^(-1) = 32400 J mol^(-1) , R = 8.314 JK^(-1) mol^(-1) , K_(P_(1)) = 0.0260 , K_(P_(2)) = ?`
`log (K_(2))/(0.0260)= (Delta H^(@))/(2.303 xx 8.314) [(T_(2) - T_(1))/(T_(2) T_(1))]`
`log (K_(2))/(0.0260) = (32400)/(2.303 xx 8.314) ((310 - 298)/(310 xx 298))`
`= (32400 xx 12)/(2.303 xx 8.314 xx 310 xx 298) = 0.2198`
log `(K_(2))/(0.0260)` = anti log 0.2198 = 1.6588
`K_(2) = 1.6588 xx 0.026 = 0.0431`
6.

For an equilibrium H_(2)O (s) hArrH_(2)O (l) which of the following statement is true:

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the pressure changes do not affect the equilibrium
more of ICE metls if pressure on the system is increase
more of liquid freezes if pressure on the system is increase
the pressure changes may increase or decrease the DEGREE of advancement of the REACTION depending upon the TEMPERATURE of the system

Solution :NA
7.

For an endothermic reaction energy of activation is E_(a) and enthalpy of reaction is DeltaH (both in kJ "mol"^(-1) ) Minimum value of E_(a) will be

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LESS than `DELTA H `
EQUAL to `Delta H `
more than `Delta H`
equal to ZERO

Answer :C
8.

For an electron v= 300 ms^(-1) andcertaintyin velocityis 0.001 % whatis theuncertainty in position

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`5.76xx 10^(2) m`
`1.92 xx 10^(2) m`
`3.84 xx 1^(2) m`
`19.2 xx 10^(-2) m`

Solution :`Delta Xxxm Delta v = (H)/(4PI)`
`Delta x= (h)/(4pim Delta v )`
`Delta v = 0.001 %of300 `
`=(300 xx0.001)/(100)`
`m=9.1 xx 10^(31) kg`
`=(6.626 xx 10^(34))/(4xx 3.14 xx 9.1xx 10^(31)xx 300 xx 0.001 xx 10^(2))`
`=1.932 xx 10^(2) m`
9.

For an electrophilic substitution reaction, the presence of a halogens atom in the benzene ring.

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Deactivates the ring by inductive EFFECT
Deactivates the ring by resonance
Increases the charge density at ortho and PARA position relative to meta position by resonance
Directs the incoming electrophilic to meta position by increasing the charge density relative to ortho and para position.

Solution :Halogen atom present on benzene ring shows +R effect and increases electron density on ortho and para position and direct electrophilic on ortho and para POSITIONS. Halogen being more electronegative pulls the electron from benzene DUE to -I effect and DECREASES electron density but due to resonance the electron density on ortho and para positions is greater than meta position.
10.

For an electron to have the same de Broglie wave length as that of a Deutron, its velocity should be times that Deutron

Answer»

1836
`1//1836`
3672
`1//3672`

ANSWER :C
11.

For an electron to have the same de Broglie wave length as that of a Deuteron, its velocity should be --- times that of Deuteron

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ANSWER :`V_(E) = 3672 V_(D)`
12.

For an e in a hydrogen atom, the wave function Psi is proportional toe^(-r//a_0) where a_0 as Bohr.s radius, what is the ratio of probability of finding the e^(-) at the nucleus to the probability of finding it at a_0, the wave function is Psi = 1/(sqrt(pi)) ((1)/(a_0))^(3/2) e^(-r//a_0)

Answer»

e
`e^2`
`1//e^2`
Zero

Solution :HINTS: `Psi^2 = 1/(pi) ((1)/(a_0))^3 e^((-2r)/(a_0)) `
At nucleus `R=0` and in `1^(st)` orbit `r= a_0`
`Psi^2 = 1/(pi) (1/(a_0))^3 e^(-3) , (Psi_n^2)/(Psi_0^2) = e^2`
13.

For an aqueous solution of NH_(4)Cl, provethat [H_(3)O^(+)]=sqrt(K_(h)C).

Answer»

Solution :For salt of STRONG acid and weak BASE, `[H_(3)O^(+)]=sqrt((K_(w)C)/(K_(b)))`
SUBSTITUTING `(K_(w))/(K_(b))=K_(h)`, we get `[H_(3)O^(+)]=sqrt(K_(h)C)`.
14.

For alkali metals, which one of the following trends is incorrect ?

Answer»

Hydration energy : `Li GT Na gt K gt Rb`
IONISATION energy : `Li gt Na gt K gt Rb`
Density : `Li gt Na gt K gt Rb`
Atomic size : `Li gt Na gt K gt Rb`

ANSWER :A::B::C::D
15.

For alkali metals, which one of the following trends is incorrec

Answer»

Hydrogen ENERGY : LI gt NA gt K gt Rb
Ionization energy : Li gt Na gt K gt Rb
Density : Li lt Na lt K lt Rb
Atomic SIZE : Li lt Na lt K lt Rb

Answer :C
16.

For alkali metals, which one of the following trends is incorrect?

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Hydration ENERGY: `LigtNagtKgt Rb`
 lonization energy : `LigtNagtKgt Rb`
DENSITY : `LiltNaltKltRb`
Atomic size : `LI lt NaltKltRb`

Solution :Density : `Li lt Na ltK ltRb`
Potassium is lighter than sodium. The correct order of density is
`Li lt Na ltK ltRb ltCs`
`0.54lt0.86lt.0.97lt1.53lt1.90(in G CM^(-3))`
17.

For alkali metal , which of the following trends is incorrect ?

Answer»

Hydration energy : `LI gt Na gt K gt Rb`
IONIZATION energy : `Li gt Na gt K gt Rb`
DENSITY : `Li lt Na lt K lt Rb`
Atomic size : `Li lt Na lt K lt Rb `

Solution :The density of K is lower than that of Na . THEREFORE , trend (c) is incorrect .
18.

For ag, Cp (JK^(-1mol^(-1) is given be 24+0.006 T//K. Calculate DeltaH (in KJ) if 3 mole of silver are raided from 27^(@)C to its meltiong point 927^(@)C under 1 atm pressure.

Answer»


ANSWER :77
19.

For the reaction at 298K:A_((g))+B_((g)) hArr C_((g)) + D_((g)) Delta H^(@) + 29.8kcal and Delta S^(@) = 100cal K^(-1). Find the value of equilibrium constant.

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SOLUTION :`LOG(K_(2))/(K_(1))=(DELTAH)/(2.303R)[(1)/(T_(1))-(1)/(T_(2))]`
20.

For Ag^(+)(aq) + 2NH_(3(aq)) harr Ag(NH_(3))_(2(aq))^(+) k = 1.7 xx 10^(7) at 25^(@)C, for this equilibrium state, which is correct?

Answer»

`DeltaG^(@) = 4-1.2KJ`
`DeltaG=0`
`DeltaG = -41.2KJ`
`DeltaG^(@) = 0`

Solution :At EQUILIBRIUM `Delta G=0` `DeltaG^(@) = -RT LNK= 41.2kJ`
21.

For adsorption of a gas on a solid, the plot of log x/m vs log P is linear with slope equal to: (n being whole number )

Answer»

k
log k
n
1/n

Answer :D
22.

For adiabatic expansion of an ideal gas

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`PV ^(GAMMA) =`constant
`TV^(gamma -1) = ` constant
`TP ^(1- gamma) = ` const.
`TP^( 1//gamma) = `constant

Solution :(c ) and (d) are wrong. The correct relation is`TP ^((1-gamma)/(gamma))= `constant
23.

For acidic buffer pH and pK_a an same? Why ?

Answer»

SOLUTION :For BUFFER solution PH `APPROX pK_a`
because, `pH=pK_a + "log" "[Salt]"/"[Acid]"`
Here [Salt]=[Acid]
e.q., `[CH_3COONa] approx [CH_3COOH]`
For preparation of buffer solution such acid is used WHOSE `pK_a` and pH are almost same.
24.

For adiabatic change ..........

Answer»

`Delta U NE w_(AD)`
`Delta U gt w_(ad)`
`Delta U = w_(ad)`
`Delta U LT w_(ad)`

Answer :C
25.

For acetic acid and sodium acetate buffer, addition of which of the following increases the P^(H)?

Answer»

`CH_3 COONA`
` H_2O `
` CH_3 COOH`
none of these

SOLUTION :PH increases on adding BASE
26.

For above transitions in hydrogen like atoms, select the incorrect relation(s).

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`v_3 = v_1 + v_2`
`v_3 = (v_1 v_2)/(v_1 + v_2)`
`lambda_3 = lambda_1 + lambda_2`
`lambda_3 = (lambda_1 lambda_2)/(LAMBDA+ lambda_2)`

SOLUTION :`E_3 = E_1 + E_2 implieshv_3 = hv_1 + hv_2 impliesv_3 = v_1 + v_2 `
`C/(lambda_3) =c/(lambda_2) , 1/(lambda_3) , 1/((lambda_1) , 1/((lambda_2) , lambda_3 = (lambda_1 lambda_2)/(lambda_1 + lambda_2)`
27.

For above transition in hydrogen like atoms, select the correct relations(s).

Answer»

`v_(3) = v_(1) + v_(2)`
`v_(3) = (v_(1)v_(2))/(v_(1) + v_(2))`
`lambda_(3) = lambda_(1) + lambda_(2)`
`lambda_(3) = (lambda_(1)lambda_(2))/(lambda_(1) + lambda_(2))`

Answer :A::D
28.

For a van der Waal's gas with a = 0.2463 atm lit^(2) mol^(-1) and b=0.01 lit mol^(-1) subjected to adiabatic free exapansion at an initial temperature of 650 K will show which of the following characterstics?

Answer»

`q=0, w=0, DeltaT=0`
`q=0, w=0, DeltaTgt0`
`q=0, w=0, DeltaTlt0`
`QNE0, w=0, DeltaT=0`

ANSWER :B
29.

For a water gas reaction, C_((s)) + H_(2)O_((g)) hArr CO_((g)) + H_(2(g)) at 1000 K, the standard Gibb's energy change is -8.1 kJ mol^(-1). Calculate the value of equilibrium constant.

Answer»

SOLUTION :`Delta G^(@) = -2.303 RT log K`
`log K = (-Delta G^(@))/(2.303 xx RT)`
`= (-(-8.1 xx 10^(3)J))/2.303 xx 8.314 xx 1273 = 0.3323` , TAKING antilogs of both sides.
`:. K = "ANTILOG" (0.3323) = 2.14 g`
30.

For a van der Waal's gas, determine Boyle Temperature (given a = 4.5 atmL^2 mol^(-2) , b = 0.9 L mol^(-1)and R = 0.082 L atm K^(-1)mol^(-1)]

Answer»

`609.8 K`
`6.09 K`
`273K`
`60.98 K`

ANSWER :D
31.

For a system at equilibrium some changes are made which is reported by a graph (shown below). Changes has been made at constant temperature. Choose the correct options:

Answer»




Solution :`P_("inert")` REMAINS same as number of moles of inert gas & volume remains same after changes been done. `toP_("total")` increased more than INCREASE in `P_("inert")` so clearly there has been reduction in volume, by Le chatelier's PRINCIPLE PRESSURE will decrease but will remain more than initial.
32.

For a substance more internal energy is observed in [ same quantity ]

Answer»

SOLID STATE
Liquid state
Gaseous state
All have same

Answer :C
33.

For a stable molecule, the value of bond order must be ……………….

Answer»

SOLUTION :POSITIVE
34.

For a spontaneous reaction, the DeltaG, equilibrium constant (K) and E_(cell)^(@) will be respectively,

Answer»

`- ve, gt 1, + ve`
`+ve, gt 1, -ve`
`-ve, LT 1, -ve`
`-ve, gt 1, -ve`

Solution :For a spontaneous reaction,
`DELTAH = -ve K gt 1 and E_("cell")^(@)= + ve`
35.

For a solution the plot of osmotic pressure(pi )verses the concentration( c "in" mol L^(-1)) gives a straight line with slope 310 R where 'R' is the gas constant The temperature at which osmotic pressure measured is

Answer»

`310 XX 0.082 K `
`310^(@)C`
`37^(@)C`
`(310)/(0.082)K`

Solution :`PI CRT, y = x ( m), m =RT`
`310 R = RT`
`THEREFORE T = 310 K = 37^(@)C`
36.

For a solution formed by mixing liquids L and M, the vapour pressure of L plotted against the mole fraction of M in solution is shown in the following figure, Here x_(L) and x_(M) represent mole fraction of L and M, respectively in the solution, the correct statement(s) applicable to this system is (are) :

Answer»

ATTRACTIVE intramolecular interactions between L-L in pure liquid L and M-M in pure liquid M are stronger than those between L-M when MIXED in solution
The point Z represents vapour pressure of pure liquid M and Raoult's law is obeyed when `x_(L) rarr 0`
The point Z represents vapour pressure of pure liquid L and Raoult's law is obeyed when `x_(L) rarr1`
The point Z represents vapour pressure of pure liquid M and Raoult's is law is obeyed from `x_(L) = 0` to `x_(L) =1`

Solution :Graph shows positive deviation from Raoult's law hence INTERMOLECULAR force must be less in solution as compared to pure liquids
`F_(L-L),F_(M-M)gt F_(L-M)` (SEQUENCE of intermolecular force)
At point Z
`p_(L) = p_(L)^(0)x_(L)`
When `x_(M)=0`, then `x_(L)=1`
`:. p_(L) = p_(L)^(0)`= pure liquid L]
37.

For a series of indicators the following colours and pH range over which colour change takes place are as follows : Indicator W would be suitable for use in the determination of the concentration of acetic acid in white vinegar by base titration : (a)True (b) False

Answer»


SOLUTION :N//A
38.

For a series of indicators the following colours and pH range over which colour change takes place are as follows : Indicator U could be used to distinguish between 0.1 M and 0.01 M solution of sulphuric acid : (a)True (b) False

Answer»


SOLUTION :N//A
39.

For a series of indicators the following colours and pH range over which colour change takes place are as follows : Indicator X could be used to distinguish between the solution of ammonium chloride and sodium acetate solution : (a)True (b) False

Answer»


SOLUTION :N//A
40.

For a sequential reaction. NH_(3) rightarrow N_(2) + H_(2) H_(2) + O_(2) rightarrow H_(2)O What will be the amount of water which will be obtained if 5 moles of NH_(3) is mixed with 3 moles of O_(2) and % yield of 1^(st) and 2^(nd) reaction is 50% and 80% respectively ?

Answer»

3 moles
2.5 mole
2 mole
2.4 mole

Answer :A
41.

For a series of indicators the following colours and pH range over which colour change takes place are as follows : Indicator V could be used to find the equivalence point for 0.1 M acetic acid and 0.1 M ammonium hydroxide solution : (a) True (b) False

Answer»


SOLUTION :N//A
42.

For a series of indicators the following colours and pH range over which colour change takes place are as follows : Indicator Y could be used to distinguish between the solutions of ammonium chloride and sodium acetate solution : (a)True (b) False

Answer»


SOLUTION :N//A
43.

For a sample of perfect gas when its pressure is changed isothermally from p_(i) to p_(f) , the entropy change is given by

Answer»

`DeltaS= nRln((p_(F))/( p_(i)))`
`DeltaS= nRln((p_(i))/( p_(f)))`
`DeltaS= nRTln((p_(f))/( p_(i)))`
`DeltaS= nRTln((p_(i))/( p_(f)))`

Solution :For n moles of an ideal gas UNDERGOING reversible EXPANSION when temperature changes from`T_(1)` to`T_(2)`and pressure changes from `p_(1)` to `p_(2)`, ENTROPY change is given by
`DeltaS = n C_(p) ln. (T_(2))/(T_(1)) + nRln. (P_(1))/(P_(2)) ` ( Refer to page)or `DeltaS = nC_(p)ln. (T_(f))/(T_(i))+ nRTln. (p_(i))/(p_(f))`
For isothermal expansion,
`T_(i) = T_(f)` so that `ln. (T_(f))/(T_(i)) = ln 1= 0`. Hence, `DeltaS = nR ln. (p_(i))/( p_(f))`
44.

For a sample of perfect gas when its pressure is changed isothermally from P_(i) to P_(f) the entropy change is given by

Answer»

`DELTA S = NRT "In" ((P_f)/( P_i))`
`Delta S = RT "In" ((P_i)/( P_f))`
`Delta S = nR "In" ((P_f)/( P_i))`
`Delta S= nR "In" ((P_i)/( P_f))`

Solution :`Delta S = NC "In" (T_i)/( T_f) + nR "In" (P_i)/( P_f)`
For isothermal `T_i = T_f, "In" 1=0`
`therefore Delta S = nR "In" (P_i)/( P_f)`
45.

For a sample of perfect gas when its pressure is changed isothermally form p_(i) to p_(p) the entropy change is given by

Answer»

`DeltaS = nR in ((p_(f))/(p_(i)))`
`DeltaS = nR in ((p_(i))/(p_(f)))`
`DeltaS = nRT in ((p_(f))/(p_(i)))`
`DeltaS = nRT in ((p_(i))/(p_(f)))`

Solution :For an ideal gas UNDERGOING reversible expansion when temperature changes from `T_(i) `to `T_(f)` and pressure changes from `p_(i)` to `p_(f)`
`DeltaS = nC_(p) In((T_(f))/(T_(i))) + nR"In((p_(i))/(p_(f)))`
SINCE the process is isothermal , `T_(i) = T_(f)` and In 1 = 0
`THEREFORE DeltaS = nR"In ((p_(i))/(p_(f)))`
46.

For a salt AB_(2)(s) solution if Ionic product (I.P) gt K_("sp"),then precipitation will take place.

Answer»


ANSWER :TRUE
47.

For a reversible reaction alphaA+betaBhArr^.cC+^.dD,the variation of K with temperature is given by log(K_(2))/(K_(1))=(-DeltaH^circ)/(2.303R)[(1)/T_(2)-(1)/T_(1)]then,

Answer»

`K_(2) GT K_1if T_2 gt T_1` for an endothermic CHANGE
`K_(2) LT K_1if T_2 gt T_1` for an endothermic change
`K_(2) gtK_1if T_2 gt T_1` for an endothermic change
`K_(2) lt K_1if T_2 gt T_1` for an endothermic change

ANSWER :A::D
48.

For a reversible reaction at 298 K, if the concentration of reactants is doubled, the value of equilibrium constant will be ………

Answer»

DOUBLED
ONE fourth
Same
Halved

Answer :C
49.

For a reversible reaction aA+bBhArrcC+dD, the variation of K with temperature is given by log (K_(2))/(K_(1))=(-DeltaH^(@))/(2.303R) [(1)/(T_(2))-(1)/(T_(1))] then,

Answer»

`K_(2)gtK_(1) "if" T_(2)gtT_(1)` for an endothermic change
`K_(2)ltK_(1) "if" T_(2)gtT_(1)` for an endothermic change
`K_(2)gtK_(1) "if" T_(2)gtT_(1)` for an endothermic change
`K_(2)gtK_(1) "if" T_(2)gtT_(1)` for an endothermic change

Solution :N//A
50.

For a reversible reaction A underset(K_(2))overset(K_(1))(hArr) B rate constant K_(1) (forward) = 10^(15)e^(-(200)/(T)) and K_(2) (backward) = 10^(12)e^(-(200)/(T)). What is the value of (-Delta G^(@))/(2.303 RT) ?

Answer»


SOLUTION :`K=A.e^(-(E_(o))/(RT))`
Equilibrium constant `(K_(P))=(K_(1))/(K_(2))`
`K_(1)=10^(15)e^((-2000)/(T)), K_(2)=10^(12)e^((-2000)/(T))`
Thus `(K_(1))/(K_(2))=10^(3), DELTAG^(@)= -2.303 RT log K_(P)`
`-((DeltaG^(@))/(2.303 RT))=log KP=(K_(1))/(K_(2))=log 10_(10)^(3)=3`