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 reaction : A+2Bunderset("yield")overset(40%)rarr4C If 10 moles of A and B each are taken calculate number of moles of 'C' formed are:

Answer»

20
4
16
8

Solution :`A+2Brarr4C`
10 10
B is LR which is 40% react
`0.4xxn_(B)=(n_(C)"formed")/(2)RARR n_(C)"formed"=0.4xx10xx2=8` Ans
2.

For reaction 2NOCl_((g))hArr2NO_((g))+Xl_(2(g)), K_(c)at 427^(@)Cis 3xx10^(-6)Lmol^(-1). The value of K_(p) is nearly

Answer»

<P>`7.50xx10^(-50)`
`2.50xx10^(-50)`
`2.50xx10^(-4)`
`1.72xx10^(-4)`

Solution :`2NOCl_((G))=2NO_((g))+Cl_(2(g))`
`K_(p)+K_(c)(RT)^(DELTAN)`
`K_(p)=3xx10^(-6)(0.0821xx700)=172.41xx10^(-6)=1.72xx10^(-4)`
3.

For reaction , 3A toproducts, it is found that the rate of reaction increases 4- fold when concentration of A is increased 16 times keeping the temperature constant. The order of reaction is

Answer»

`2`
`1`
`1`
`0.5`

ANSWER :D
4.

For reaction, 2A_((g))hArr3C_((g))+D_((s)), the value of K_(c) will be equal to

Answer»

`K_(p)(RT)`
`K_(p)//RT`
`K_(p)`
None of these

Solution :`2A_((g))hArr3C_((g))+D_((s))`
For this reaction, `Deltan_(g)=3-2=1`
`THEREFORE K_(p)=K_(C)[RT]^(1)or (K_(p))/(K_(c))=RTor K_(c)=(K_(p))/(RT)`
5.

For reaction 2Cl(s)rarrCl_(2)(s), the signs of DeltaH and DeltaSrespectively are

Answer»

`+, -`
`+, +`
`-, -`
`-, +`

SOLUTION :Because SOLID `RARR` solid, `DELTAS` is same and `DELTAH` is -ve.
6.

For radioactive decay:

Answer»

`t_(3//4) = 2 t_(1//2)`
`t_(7//8) = 3 t_(1//2)`
`t_(99%) = 2 t_(90%)`
`t_(90%) = (10)/(3) t_(50%)`

ANSWER :a,B,C,d
7.

For raction : mA+nB rarr pC+ qDthe rate law given by dx/dt = K[A]^x[B]^y The correct statement for this raction if

Answer»

the molecularityh of the raction is m+n
the overall order of the raction is m + n
both are correct
none of corrcet

Solution :For the reaction `m A+ n B to p C + q D ` , the rate law , `(dx)/(DT) = k [A]^(x) xx [B]^(y)`
Molecularity of the reaction = Total NUMBER of reactant molecules in the reactant molecules in the reaction = m+ n
Order of the reaction = SUM of the powers of the conc. of reactants in the rate law = x+ y
8.

For Ra^(266 t_((t)/(2))) yrs is 1620 yrs from 0.001 g of Ra how many alpha particles relaease per min?

Answer»

SOLUTION :`2.17xx10^(9) MIN ^(-1)`
9.

For pure water on small increase in temperature:

Answer»

PH decreases
POH decreases
pH increases
pOH increases

ANSWER :a,B
10.

For production of steel ________ is used .

Answer»

AOP
BOP
COP
DOP

Solution :BOP-basic OXYGEN PROCESS
11.

For producing the effective collisions the colliding molecules must have

Answer»

a certain MINIMUM AMOUNT of energy
energy equal to or greater than THRESHOLD energy
proper orientation
both threshold energy and proper orientation

Answer :D
12.

For producing the effective collisions thecolliding molecules must have :

Answer»

A CERTAIN MINIMUM amount of energy
Energy lesser than threshold energy
Improper ORIENTATION
Proper orientation and energy equal or greater than threshold energy

Answer :D
13.

For prevention of rusting of iron,which is used in paints

Answer»

PbO
`PbO_(2)`
`Pb_(3)O_(4)`
`PbSO_(4)`

ANSWER :C
14.

For preparing one litre N/10 solution of H_(2)SO_(4), we need H_(2)SO_(4)

Answer»

98 gms
10 gms
100 gms
4.9 gms

Solution :`(1)/(10)=(Wxx1000)/(EQ." "wtxxvol.)=(Wxx1000)/(49xx1000)impliesW=4.9gm`
15.

For prevention of rusting of iron, which is used in paints ?

Answer»

PBO
`PbO_2`
`Pb_3O_4`
`PbSO_4`

SOLUTION :`Pb_3O_4`
16.

For preparing monohalogen derivative of phenol, halogenation is carried out

Answer»

at HIGH temperature
at LOW temperature
inpresence of non-polar solvents
both 'B' and 'c

Answer :D
17.

For preparing methyl acetylene, we take -

Answer»

`CH_(3)-UNDERSET(CHCOOK)underset(||)(C)-COOK`
`underset(CHCOOK)OVERSET(CHCOOK)(||)`
`CH_(3)-CH_(2)-underset(CHCOOK)underset(||)(C COOK)`
All of these

Answer :A
18.

For preparing M/10 solution of H_2 SO_4 in one litre we need H_2 SO_4

Answer»

0.009 g
49.0 g
4.8 g
9.8 g

Answer :D
19.

For preparing H_(2)O_(2)in the laboratory

Answer»

`PbO_(2)`is added to ACIDIFIED solution of `KMnO_(4)`
`BaO_(2)` is added to `CO_(2)`bubbling through cold water
`MnO_(2)`is added to dilute cold `H_(2)SO_(4)`
`Na_(2)O_(2)`is added to BOILING water

Solution :`BaO_(2)+H_(2)O+CO_(2)rarr BaCO_(3)+H_(2)O_(2)`
20.

For preparing a salt bridge, gelatin or agar-agar is dissolved in a hot aqueous solution of

Answer»

`K_(2)CO_(3)`
`K_(2)SO_(4)`
`NH_(4)NO_(3)`
`(NH_(4))_(3)PO_(4)`

ANSWER :C
21.

For preparing an alkane, a saturated solution of sodium or potassium salt of a carboxylic acid is subjected to

Answer»

Hydrolysis
Oxidation
Hydrogenation
Electrolysis

Solution :ALKANES are prepared from carboxylic ACIDS byelectrolysis. In electrolysis , and AQUEOUS solution of sodium or potassium salt of carboxylic ACID is electrolysed and alkane is evolved at the anode (Kolbe's electrolysis)
22.

For preparing a buffer solution of pH 6 by mixing sodium acetate and acetic acid, the ratio of the concentration of salt and acid should be (K_(a) = 10^(-5))

Answer»

`1:10`
`10:1`
`100:1`
`1:100`

SOLUTION :`K_(a) = 10^(-5) , PH = 6`
`pH = -LOG K_(a) + log .(["SALT"])/(["Acid"]), 6 = -log 10^(-5) + log .(["Salt"])/(["Acid"])`
`6 = 5 log 10 + log .(["Salt"])/(["Acid"]), 6 = 5 + log .(["Salt"])/(["Acid"])`
`log .(["Salt"])/(["Acid"]) = 6 =5 = 1, (["Salt"])/(["Acid"]) = (10)/(1)`.
23.

For preparing 0.1 N solution of a compound fromits impure sample of which the percentage purity is known, the weight of the substance required will be

Answer»

More than the theroretical WEIGHT
LESS than the THEORETICAL weight
Same as the throretical weight
None of these

Answer :A
24.

For preparing a buffer solution of pH 6 BY mixing sodium acetate and acetic acid the ratio of concentration of salt and acid (K_a=10^(-5))Should be :

Answer»

1:10
10:1
100:1
1:100

Answer :B
25.

For preparing 0.1 N solution of a compound from its impure sample, of which the percentage purity is known, the weight of the substance required will be

Answer»

More than the THEORETICAL weight
Less than the theoretical weight
Same as theoretical weight
None of these

Answer :A
26.

For precipitation reaction of Ag^(+) ions with NaCl, which of the following statements is correct

Answer»

`DELTAH` for the reaction is ZERO
`DeltaG` for the reaction is zero
`DeltaG` for the reaction is NEGATIVE
`[DeltaG]=[DeltaH]`

Solution :Because the reaction is SPONTANEOUS.
27.

For precipitation of silver from the complex [Ag(CN)_(2) ]^(-), zinc is used but not copper. Why?

Answer»

Solution :copper is ALSO more electropositive than SILVER. Thuspractically copper can also precipitatesilver from`[Ag (CN)_(2)]^(-)`.
ButZincis more electropositivewhencompared with copper and ZINC is cheaper than copper .
Thus zincis USED commonly to precipitate silver from `[Ag(CN)_(2)]^(-)`
28.

For positive sol arrange the coagulation power of the following active ions in the decreasing order, Cl-, PO_(4)^(3-), [Fe(CN)_(6)]^(4-), SO_(4)^(2-)

Answer»

SOLUTION :Decreasing ORDER : `[Fe(CN)_(6)]^(4-) gt PO_(4)^(3-) gt SO_(4)^(2-) gt Cl^(-)`
29.

For per gram of reactant, the maximum quantity of N_(2) gas is produced in which of the following thermal decomposition reactions? (Given : Atomic wt. - Cr = 52u, Ba = 137u)

Answer»

`2NH_(4)NO_(3)(s) to 2N_(2)(g) + 4H_(2)O(g)+O_(2)(g)`
`BA (N_(3))_(2)(S) to Ba(s) + 3N_(2)(g)`
`(NH_(4))_(2)Cr_(2)O_(7)(s) to N_(2)(g) + 4H_(2)O(g)`
`2NH_(3)(g) to N_(2)(g) + 3H_(2)(g)`

Answer :D
30.

For per gram of reactant, the maximum quantity of N_(2) gas is produce in which of the following thermal decomposition reactions ? (Given : Atomic wt : Cr = 52 u, Ba = 137 u)

Answer»

`(NH_(4))_(2) Cr_(2)O_(7) (s) rarr N_(2)(g) + 4H_(2) O (g) + Cr_(2)O_(3) (s)`
`2NH_(4)NO_(3) (s) rarr 2N_(2) (g) + 4H_(2) O (g) + O_(2) (g)`
`Ba(N_(3))_(2) (s) rarr Ba (s) + 3N_(2) (g)`
`2NH_(3) (g) rarr N_(2) (g) + 3H_(2) (g)`

ANSWER :D
31.

For PCl_5 hArr PCl_3 + Cl_2, DeltaH = 22 kcal the dissociation of PCl_5 will be more on:

Answer»

INCREASING temperature
Decreasing pressure
Increasing pressure
Increasing the CONCENTRATION of chlorine

Answer :A
32.

for oxyacid HClO_(x), if x=y=z (x,y and are natural numbers), then calculate the value of |x+y+z|. Where x= Number of 'O' atoms y= Total number of Ione pair at central atom z= Total number of pi (pi) electrons in the oxyacid

Answer»


SOLUTION :6
The COMPOUND is `HCIO_(3)`
33.

For oxidation of iron, 4Fe(s) + 3O2(g) to 2Fe2O3(s) entropy change is – 549.4 JK​ ^(–1 ) ​mol^(​–1)at 298K. Deltar H^(@) for this reaction is – 1648 xx 10​^( 3)​ J mol^(​–1)Above reaction is :-

Answer»

Spontaneous
Non-spontaneous
At equilibrium
Cant predict

SOLUTION :`Delta G =Delta H – T.DeltaS `
34.

For osazone formation, the effective structural unit necessary is

Answer»

`UNDERSET(|)underset(CO)underset(|)(""CH_2OCH_3)`
`underset(|)underset(CO)underset(|)(""CH_2OH)`
`underset(|)underset(""CHOCH_3)underset|(""CH_2OH)`
`underset|underset(""CHOCH_3)underset|(""CHO)`

ANSWER :B
35.

For ......... ores froth floatation process is used.

Answer»

BAUXITE 
Cinnabar 
Haematite 
Horn SILVER 

Answer :B
36.

For one of the element various successive ionization enthalpy (in kJ "mol"^(-1) ) are given below: The element is

Answer»

Al
Si
Mg
P

Answer :A
37.

For one of the element varioussuccessiveionizationethalpies ( in kj "mol"^(-1)) are givenbelow : The elementis

Answer»

A) P
B) Mg
C) Si
D) Al

Solution :LARGE jump between `IE_(3)` and `IE_(4)` suggest that the element has three valence electrons.
38.

For one mole of NaCl(s) the lattice enthalpy is : Na(s) + (1)/(2)Cl_(2)(g) overset(+108.4 kJ//mol)rarr Na(g) + (1)/(2)Cl_(2)(g) overset(+495.6 kJ//mol)rarr Na^(+)(g) + (1)/(2)Cl_(2)(g) overset(121 kJ//mol)rarr Na^(+)(g) + Cl(g) overset(-348.6 kJ//mol)rarr Na^(+)(g) + Cl^(-)(g) overset(Delta H^(theta)" lattice")rarr NaCl("solid") underset(-411.2 kJ/mol)larr Na(s) + (1)/(2)Cl_(2)(g)

Answer»

`- 788 KJ//mol`
`+ 878 kJ//mol`
`+ 788 kJ//mol`
`- 878 kJ//mol`

Solution :`Delta_(f) H = - 411.2 kJ mol^(-1)`
`- 411.2 = 108.4 + 495.6 + 121 - 348.2 + Delta H_("LATTICE")^(@)`
:. `Delta H_("lattice")^(@) = - 411.2 - 108.4 - 495.6 - 121 + 348.2`
`= - 788 kJ mol^(-1)`
39.

For one of the element various successive ionization enthalpies (in kJ mol^(-1)) are given below : The element is

Answer»

`Mg`
`AL`
`P`
`Si`

Solution :LARGE jump between `IE_3 and IE_4` suggests that the element has THREE VALENCE electorns.
40.

For one molecule reaction at higher pressure or higher concentration what is order of reactant?

Answer»

0
1
`1(1)/(2)`
2

Solution :The order of ONE molecule reaction is 1.
41.

For one mole of glycerol, how many mole of acetyl chloride are required for complete acetylation:

Answer»

One
Two
Three
Four

Answer :C
42.

For one mole of an ideal gas the slope of V vs. T curve at constant pressure of 2 atm is X lit "mol"^(-1)K^(-1). The value of the ideal universal gas constant R in termsof X is

Answer»

X lit ATM `"mol"^(-1)K^(-)`
X/2 lit atom `"mol"^(-1)K^(-1)`
2X lit atm `"mol"^(-1)K^(-1)`
2X atm `"lit"^(-1)"mol"^(-1)K^(-1)`

Solution :We can TAKE T to be the independent variable and V tobe the dependent variable. Now, for ONE mole of an ideal GAS ,
`PV=RT`
`impliesV/T=R/Pimplies` slope `=R/P=X` lit `"mol"^(-1)K^(-1)`
`impliesR=PxxX` lit `"mol"^(-1)K^(-1)`
`R=23` atm `xx X` lit `"mol"^(-1)K^(-)` ( `:. P=2` atm)
`impliesR=2` X lit atm `"mol"(-1)K^(-1)`
43.

For one mole of electrolyte which of the following increases with dilution?

Answer»

Resistance
Specofic conductance
Molar conductance
None of these

Answer :C
44.

For one mole of an ideal gas, increasing the temperature from 10^(@)C to 20^(@)C

Answer»

A) increases the average kinetic energy by TWO TIMES.
B) increases the rms velocity by `sqrt(2)` times
C) increases the rms velocity by two times
D) increases both the average kinetic energy and rms velocity, but not significantly.

Solution :Initial temperature `=283K`
Final temperature `=293K`
`KE_(1)=(3RT_(1))/(2),KE_(2)=(3RT_(2))/(2)`
`(KE_(1))/(KE_(2))=(T_(1))/(T_(2))=(283)/(293)=0.96`
`KE_(2)=1.04KE_(1)`
`v_(rms)=sqrt((3RT)/(M))`
`(v_(1))/(v_(2))=sqrt((T_(1))/(T_(2)))rArrsqrt((283)/(293))`
`v_(2)=v_(1)sqrt((293)/(283))=1.02v_(1)`
Both average K.E. and rms velocity increase but not significantly.
45.

For one mole of an ideal gas ((delta P)/(delta T))_(V) ((delta V)/(delta T))_(P) ((delta V)/(delta P))_(T)=

Answer»

`-1`
`(R^(2))/(P^(2))`
`(R^(2))/(P^(2))`
`+1`

Answer :A::B::C
46.

For one mole of a van der Waal's gas when b = 0 and T = 300K, the PV us 1//V plot is shown ahead. The value of the van der Waal's constant a(atm. "litre"^(2) mol^(-2)) is:

Answer»

`1.0`
`4.5`
`1.5`
`3.0`

ANSWER :C
47.

For one mole of a gas, the total kinetic energy is equal to :

Answer»

RT
`(3)/(2) RT`
`( 2)/( 3) RT`
`( 3)/( 2) ( RT)/( N_(0))`

Answer :B
48.

For one first order reaction 60 minutes are required to decrease the initial concentration from 0.8 M to 0.1 M,determine the half reaction time (t_((1)/(2)))

Answer»

20 min
30 min
40 min
15 min

Answer :A
49.

For O_(2) gas at T_(1) and T_(2) following Maxwell speed distribution is observed [not to be scale] Ratio A_(1)//A_(2) is equal to

Answer»


ANSWER :1
50.

For O_3(g) + OH(g)iffH(g) + 2O_2(g), K=0.096 at 298 K and K=1.4 at 373 K. Above what temperature will the reaction become thermodynamically spontaneous?

Answer»

SOLUTION : For spontaneous process `K_p GT 1`
`T gt 361 K`