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 a certain oxide of nitrogen weighing 0.11 g gives 56 mL of nitrogen and another oxide of nitrogen weighing 0.15 g gives the same volume of nitrogen (both at sTP), show that these results support the law of multiple proportions.

Answer»

Solution :22400 mL of `N_(2)` at STP weigh = 28 G. CALCULATE the mass of nitrogen. Then SUBSTRACT from the mass of oxide of nitrogen to calculate the mass of oxygen.
2.

If a certain buffer solution contains equal concentration of X^- and HX. Then the pH of buffer will be (K_b for X^- is 10^-10)

Answer»

10
4
5
11

Answer :B
3.

If a centi normal solution of NH_4OH has molar conductivity equal to 9.6 Omega^(-1) cm^(2) mol^(-1). what will be the per cent dissociation of NH_4OH at this dilution.

Answer»


ANSWER :0.04
4.

If a carbanion is bonded to hydrogen or an alkyl group then, the shape of methyl carbainon is likely to be

Answer»

tetrahedral
planar
PYRAMIDAL
linear

Solution :They are formed by heterolytic fission. Due to presence of one LONE pair of electron on carbn atomis one of tetrahedral oriented, four `sp^(3)` HYBRIDE orbitals,the geometry of carbanion BECOME pyramidal.
5.

If a be the edge length of the unit cell and r be the radius of an atom, then for fcc arrangement, the correct relation is

Answer»

`4A = sqrt( 3) R`
`4r = sqrt( 3) . a `
`4r = sqrt( 2) a `
` 4 r = ( a)/( sqrt( 2))`

ANSWER :C
6.

If A, B, C and D are element's of 3^(rd) period of p-block in modern Periodic Table. Among these 'A' is a metal, 'B' is metalloid and 'C', 'D' are non metals and their order of electronegativity is given as rarr A lt B lt C lt D Which element is expected to form Amphoteric Oxide?

Answer»

A
B
C
D

Solution :`._(25)Mn-1s^(2), 2s^(2),2p^(6),3S^(2),3p^(6),3d^(5),4s^(2)`
`l+m=0implies""i.e "s-subshell"`
`l=1, m=-1 ""i.e. "ONE orbital of p."`
`l=2, m=-2 ""i.e. "one of d-orbitals."`
Hence there are total of `13 e^(-)` having `l+m=0`
7.

If a 6.84% (wt/vol.) solution of cane sugar (mol.wt.=342) is isotonic with 1.52%(wt/vol.) solution of thiocarbamide, then the molecular weight of thiocarbamide is:

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152
76
60
180

Answer :B
8.

If a 0.1 M solution of HCN is 0.01 % ionised , the K_(a) for HCN is :

Answer»

`10^(-9)`
`10^(-7)`
`10^(-5)`
`10^(-3)`

Answer :A
9.

If a 0.1 M solution of glucose ( mol.wt. 180 ) and 0.1molarsolution of urea ( mol. wt. 60 ) are placedon the two sides of a semipermeable membrane to equal heights, then itwill be correct to say

Answer»

There will be on net MOVEMENT across the MEMBRANE
Glucose will FLOW across the membrane into urea SOLUTION
Urea will flow across the membrane into glucose solution
Water will flow from urea solution two flucose solution.

Solution :Equimolar solution are isotonic.
10.

If ._(92)U^(235) nucleus absorbs a neutron and disintegrates in ._(54)Xe^(139), ._(38)Sr^(94) and X, then what will be the product X

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`alpha`-particle
`beta-`particle
`2-`neutrons
`3-`neutrons

Solution :`.^(92)U^(235) + ._(0)N^(1) rarr ._(54)Xe^(139) + ._(38)Sr^(94) + 3 ._(0)n^(1)`
11.

If 9650 coulombs of electricity is passed through a copper sulphate solution, the number of moles of copper deposited will be

Answer»

3.15
2
0.05
0.01

Solution :`CU^(2+) + 2e^(-) to Cu THEREFORE 2F =2 XX 96500= 1` mole of Cu
`therefore N= (1xx9650)/(2xx96500) =0.05` mole
12.

If 965 coulombs of electricity is passed through a metal cup dipped in silver(l) salt solution, in order to plate it with silver. Then the amount of silver deposited on its surface is (Given : the molar mass of Ag = 108 g mol^(-1), 1F = 96500 coulombs)

Answer»

`1.08 G`
`1.002 g`
`108 g`
`9.89 g`

ANSWER :A
13.

If 9650 coulombs of electricity is passed through a copper sulphate solution , the number of moles of copper deposited will be _________.

Answer»

0.05
0.01
2
3.15

Answer :A
14.

If ._(92)U^(236) nucleus emits one alpha-particles, the remaining nucleus will have

Answer»

119 neutrons and 119 protons
142 neutrons and 90 protons
144 neutrons and 92 protons
146 neutrons and 90 protons

Solution :`._(92)U^(236) rarr ._(90)X^(232) + ._(2)He^(4)`
`._(90)X^(232)` have 90 protons and 142 neutrons
15.

If 900 J/g of heat is exchanged at boiling point of water, then what is increase in entropy

Answer»

`43.4 J//mol`
`87.2 J//"mole"`
`900 J//"mole"`
Zero

Solution :`DeltaS_("vap")=((900xx18))/(373)=43.4 JK^(-1) mol^(-1)`.
16.

If 90 g of water is electrolysed completely with 50% current efficiency 10 faraday of electricity will be consumed 20 faraday of electricity will be consumed c) 168 L (STP) of gases will be produced d) 84 L (STP) of gases will be produced. Correct statements are

Answer»

B, C
a, b, c
a, c, d
All

ANSWER :A
17.

IF 900 J/g of heat is exchanged at boiling point of water then increase in entropy is

Answer»

43.4 J/MOL K
87.2 J/mol K
900 J/mol K
zero

Solution :`DeltaS_(VAP)=((900xx18))/(373)=43.4JK^(-1)mol^(-1)`
18.

If 8.3mL of a sample of H_(2)SO_(4) (36 N) is diluted by 991.7 mL of water required 19.8 mL of 0.1 N HCl .The value of x is

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`0.4`
`0.2`
`0.1`
`0.3`

ANSWER :d
19.

If 772 mL of SO_(2) gas at STP is adsorbed on 2 g of charcoal at an equlibrium pressure of 16 atmospheres and the value of the constant 'k' in the Freundlich equation is 0.48. the value of the constant 'n' will be

Answer»


Solution :Molar mass of `SO_(2)=64 G mol^(-1)`
Moar volume at `STP=22400mL`
`THEREFORE` ML of `SO_(2)` at STP well have mass
`=(64)/(22400)xx772=1.92g`
`therefore x/m=(1.92)/(2)=0.96`
But `x/m=kP^(1//n)`
`therefore 0.96 =0.48 xx(16)^(1//n) or 2=(16)^(1//n)`
`therefore n=4.`
20.

If 8.2g of anhydrous sodium acetate is added to 11 of 0.l(M) acetic acid solution, then what will be the change in degree of ionisation of the acid? [Given: K_1(CH_3COOH)=1.8xx10^(-5)]

Answer»

SOLUTION :If the degree of dissociation of monobasic acid, HA in its aqueous solution `=prop`, MOLAR concentration=c(M) and dissociation constant of the acid`=K_a`, then `K_a=(prop^2c)/(1-prop)`
Let the solution is x times diluted. Hence, `c=(0.1)/(x)(M)`
As given, degree of ionisation of HA in the dilute solution`=2prop`
`:.2prop=sqrt((K_a)/((0.1//x)))=sqrt((x xxK_a)/(0.1))`
From [1] & [2], `2xxsqrt((K-a)/(0.1))=sqrt((x xxK-a)/(0.1))` or `sqrtx=2`
`:.x=4`
Therefore, if the solution is diluted 4 times, then degree of dissociation of HA in the diluted solution will be 2 times that of the initial solution.
21.

If 720 litres of a gas were collected over water at 25^(@)C and 720 mm then the volume of the dry gas at the same temperature and pressure is (aq. Tension of water at 25^(@)C = 23.8 mm)

Answer»

696.2 litres
360 litres
743.8 litres
1440 litres

Answer :A
22.

If 75% quantity of a radioactive isotope disintegrates in 2 hour , its half-life would be :

Answer»

1 hour
45 MINUTES
30 minutes
15 minutes

SOLUTION :`100% overset(1hr) (to) 50% overset(1 hr) 25% , t = 2` HRS
23.

If 7% of a first order reaction was completed in 60% of the same reaction under the same conditions would be completed in.........

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20 minutes
30 minutes
35 minutes
75 minutes

Solution :`t_(75%)=2t_(50%)`
`t_(50%=((t_(75%))/2))=(60/2)= 30` MIN
24.

If 6gm of CH_(3)COOH is dissoved in benzene to make 1 litre at 300K. Osmotic pressure of solution is found to be 1.64 atm. If it is known that CH_(3)COOH in benzene forms a dimer. Calculate degree of association of acetic acid in benzene?

Answer»

SOLUTION :
`i=1+((1)/(n)-1)beta`
`1.64=0.0821xx300xx[1+((1)/(n)-1)beta]xx0.1`
`RARR 1.64=0.0821xx300[1-(beta)/(2)]0.1`
`(1.64)/(00.821xx30)=(2-beta)/(2)`
`(1.64)/(2.46)=(2-beta)/(2) 4=6.3beta`
`beta=(2)/(3)`
25.

If 6.3 g of NaHCO_(3) are added to 15.0 g of CH_(3)COOH solution, the residue is found to weigh 18.0 g. The mass of CO_(2) released in the reaction is

Answer»

9.3 g
39.3 g
3.3 g
None of these

Solution :`CH_(3)COOH+NaHCO_(3) to CH_(3)COONa+H_(2)O+CO_(2) UARR`
Mass of reactants = Mass of PRODUCTS
`6.3 g + 15.0 g= 18.0 g+` Mass of `CO_(2)`
Mass of `CO_(2) = 6.3g + 15.0g - 18.0 g = 3.3 g`
26.

If 61.25 gm of K ClO_(3) reacts with excess of red phosphorus, what mass of tetraphosphorous dioxide (P_(4) O_(10)) would be produced. K ClO_(3) (s) + P_(4) (s) rarr P_(4) O_(10) (s) + K Cl (s)

Answer»

`142 gm`
`426 gm`
`14.2 gm`
`32.6 gm`

Solution :`10 KClO_(3) (s) + 3P_(4) (s) rarr 3 P_(4)O_(10) (s) + 10 KCl (s)`
`61.25 gm`
`N = (61.25)/(122.5)`
`{:(n = (1)/(2) " mole" ,n = (3)/(10) xx (1)/(2),),(,n = (3)/(20) " mole",):}`
mass of `P_(4)O_(10) = (3)/(20) (3 1 xx 4 + 10 xx 16)`
`= (3)/(20) (284)`
`= 14.2 xx 3`
`= 32.6 gm`
27.

If 6.3 g of NaHCO_(3) are added to 15.0 g of CH_(3)COOH solution, the residue is found to weigh 18.0 g. What is the mass of CO_(2) released in the reaction?

Answer»


ANSWER :3.3 G
28.

If 6.022xx10^(23) atoms are present in 100 mL of urea solution then find out the molarity of urea solution.

Answer»


ANSWER :0.01 M
29.

If 6.022xx10^(20) molecules are present in 100 mL urea solution, then find out the concentration of urea solution.

Answer»

0.01 M
0.001 M
0.2 M
0.1 M

Solution :Mole `= 10^(-3)`
`6.022xx10^(23)` molecule urea = 1 mole urea
`6.022xx10^(20)` molecule urea = (?)
`THEREFORE M=(10^(-3))/(0.1("liter"))=0.01M`
30.

If 60% of a first order reaction was completed in 60 minutes 50% of the same reaction would be completed in approximately :

Answer»

45 minutes
60 minutes
40 minutes
50 minutes .

Solution :(A) `k = (2.303)/(t) "log"([A]_(0))/([A])`
`:. K = (2.303)/(60) "log" ([A]_(0))/(0.4[A]_(0))`
`= (2.303)/(60) "log" (10)/(4)`
`= (2.303)/(60) ` (log 10 -log 4)
Now `t_(1//2) = (0.693)/(k) = (0.693xx60)/(2.303xx0.4)`
`= 45.14 ~~45 ` min .
31.

If 6 litre of H_(2) of Cl_(2) are mixed and exploded in an eudiometer, the volume of HCl formed is:

Answer»

6.0 litre
5.6litre
11.2 litre
11.6 litre

Answer :C
32.

If 60% of a first order reaction was completed in 60 minutes , 50% of the same reaction would be completed in approximately

Answer»

50 MINUTES
45 minutes
60 minutes
40 minutes

Solution :`K = (2.303)/(60) "LOG" (100)/(40) implies 0.152`
`k = (0.693)/(T_(1//2)) T_(1//2) = (0.693)/(0.152) = 45.37` min .
33.

If 60 cm^(3) of ethyl alcohol is mixed in water to form 360 cm^(3) of dilute ethyl alcohol solution, then the percentage by volume of ethyl alcohol in water is _________.

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`16.67%`
`17.05%`
`17.86%`
`18.96%`

ANSWER :A
34.

If 60% of a first order reaction was completed in 60 min, 50% of the same reaction would be completed in approximately

Answer»

`45` MIN
`60` min
`40` min
`50` min

Answer :A
35.

If 5g of a radioactive substance has a t_((1)/(2))=14h, 20g of the same substance will have a t_((1)/(2)) equal to

Answer»

56h
3.5h
14h
28h

Answer :C
36.

If 5.85 gms of NaCl are dissolved in 90 gms of water, the mole fraction of NaCl is

Answer»

`0.1`
`0.2`
`0.3`
`0.0196`

Solution :`5.85 G NaCl=(5.85)/(58.5)` mole = 0.1 mol
`90 g H_(2)O=(90)/(18)` MOLES = 5 moles
mole fraction of NaCl = `(0.1)/(5+0.1)~~0.0196`.
37.

If 5.85 gm of NaCl is dissolved in 90 gm of water, then find out the mole fraction of NaCl.

Answer»


ANSWER :`0.0196`
38.

If 54 g of silver is deposited during an electrolysis reaction, how much aluminium will be deposited by the same amount of electric current?

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(a) 2.7 G
(b) 4.5 g
(C) 27 g
(d) 5.4 g

Solution :1 F DEPOSITS 108 g of Ag `(Ag^(+) + e^(-) to Ag )`
54 g of Ag will be deposited by `1/108xx54=1/2F`
3 F deposite 27 g of Al (`Al^(3+) + 3E^(-) to Al`)
`1/2F` will deposit `27/3xx1/2`=4.5 g of Al
39.

If 500 mL of a 5 M solution is diluted to 1500 mL, what will be the molarity the solution obtained?

Answer»

`1.5M`
`1.66M`
`0.017M`
`1.59M`

Solution :`{:(""M_(1)V_(1),=,""M_(2)V_(2),therefore5xx500=M_(2)xx1500orM_(2)=(5)/(3)=1.66M),("(Before DILUTION)",,"(After dilution)",):}`
40.

If 50% of a radioactive substance has t_(1//2) = 14 hr , 2 g of the same substance will have a t_(1//2) equal to

Answer»

56 HR
3.5 hr
14 hr
28 hr

SOLUTION :`t_((1)/(2))` = contain for a NUCLEIDE
41.

If 50 % of the reactant is converted into a product in first order reaction in 25 minutes how much of it would react in 100 minutes ?

Answer»

93. 75 %
`87.5%`
`75%`
`100 % `

Solution :`t_(1//2) = 25` MIN
Number of half - lives in 100 min =`(100)/(25)=4`
`:.` FRACTION of substance remaining after 100 min `= (1)/(2^(4)) = (1)/(16)`
Thefraction of substance REACTED in 100 min .
`= 1- (1)/(16)=(15)/(16)=0.9375`
% of substance reacted = 93.75 %
42.

If 50 ml of A_(2)B_(3) reacts completely with 200 ml of C_(2) in closed vessel according to the equation , 2A_(2)B_(3)(g)+5C_(2)(g) to 3C_(3)B_2(g)+CA_(4)(g) . The composition of the mixtureof gasesis :

Answer»

10 ml ` C_(2)` , 25 ml `C_(3)B_(2)`, 100 ml `CA_(4)`
25 ml `C_(2)`, 75 ml ` C_(3),B_2,` 25 ml ` CA_(4)`
75 ml ` C_(2)`, 75 ml ` C_(3)B_(2)`, 25 ml ` CA_(4)`
100 ml `C_(2)`, 50 ml `C_(3)B_(2)`, 50 ml `CA_(4)`

SOLUTION :The MOLE ratio is`UNDERSET("Reaction")(2:5)::underset("Product")(3:1)`
43.

If 50 ml of each 1 M NaOH are mixed then correct option is

Answer»

I,I, R
l,ii,R
IV,iii,S
I,ii,Q

Solution :`pH=(pK_(a_1)+pK_(a_2))/2=(5+7)/2=6`
44.

If 50 of 1 M H_3X mixed with 150 mL of 1 M NaOH then correct option is

Answer»

II, I,P
IV,ii,S
IV,III,S
II,I,P

Solution :`[Na_3X]=50/200=0.25 M`
`pH=7+1/2pK_(a_3)+1/2logC`
=11.2
45.

If 50% of the reactant is converted into a product in a first order reaction in 25 minutes, how much of it would react in 100 minutes?

Answer»

0.9375
0.875
0.75
1

Solution :Half-life is 25 mins
`100 OVERSET(25"min")RARR 50 overset(25"min")rarr 25 overset(25"min")rarr 12.5 overset(25 "min")rarr 6.25`
6.25% remaining i.e., 93.75% REACTED.
46.

If 50 ml of 1 M H_3X and 25 mL of 1 M NaOH mixed then correct option is-

Answer»

I,IV,Q
II,ii,P
IV,iii,S
I,iii,Q

Solution :`pH=5+log(25/25)=5`
47.

If 50 ml of 0.2 M KOH is added to 40 ml of 0.5 M HCOOH, the pH of the resulting solution is (K_(a) = 1.8 xx 10^(-4))

Answer»

3.4
7.5
5.6
3.75

Answer :A
48.

If 50 mL of 0.2 M KOH is added to 40 mL of 0.5 M HCOOH. Find the resulting solution. (K_c = 1.8 xx 10^-4) :

Answer»

3.75
5.6
7.5
3.4

Answer :A
49.

If 50 milli ampere of current is passed through copper coulometer for 60 min, calculate the amount of copper deposited.

Answer»

Solution :ELECTRICAL charge input `=Ixx"t coulombs"`
`=50xx10^(-3)Axx60xx60sec`
`="180 coulombs."`
The chemical reaction is, `Cu^(2+)+2E rarr Cu_((s))`
1 gm atom of COPPER requires 2F CURRENT
`therefore" AMOUNT of copper deposted "=(63.5g.mol^(-1)xx180C)/(2xx96500C)`
`=0.0592gm.`
50.

If 50 gm oleum sample rated as 118% is mixed with 18 gm water, then the correct option is

Answer»

The resulting solution contains 18 gm of water and 118 gm `H_(2)SO_(4)`
The resulting solution contains 9 gm water and 59 gm `H_(2)SO_(4)`
The resulting solution contains only 118 gm pure `H_(2)SO_(4)`
The resulting solution contains 68 gm of pure `H_(2)SO_(4)`

Answer :B