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.

How many molecules of hydrogen is required to produce 4 moles of ammonia?

Answer»

15 moles
20 moles
6 moles
4 moles

Solution :`3H_(2)+N_(2) to 2NH_(3)`
To get 2 moles of AMMONIA, 3 mole of `H_(2)` is required.
`therefore` To get 4 moles of ammonia = `(3)/(cancel2)xxcancel(4)^(2) = 6` moles of `H_(2)` is required.
2.

How many molecules of CO_(2)will needed to obtain 1.8 g of glucose according to given reaction. Reaction : 6CO_(2) + 6H_(2) O rarr C_(6) H_(12) O_(6) + 6O_(2) [M. wt C_(6)H_(12)O_(6) = 180 g mol^(-1)] (C=12, H=1, O=16)

Answer»

`0.6xx6.022xx10^(23)`
`6xx6.022xx10^(23)`
`0.06xx6.022xx10^(23)`
`60xx6.022xx10^(23)`

Solution :mole of glucose `=("mass ")/("molecular mass") = (1.8)/(180)`
` =0.01 ` mole glucose ...(1)
Reaction : `6CO_(2) + 6H_(2) O RARR C_(6)H_(12)O_(6) + 6O_(2)`
`:.` 6 mole `CO_(2)` is needed to GET 1 mole glucose
`= (0.01xx6)/(1)`
`= 0.06` mole `CO_(2)` is needed ...(ii)
`=0.06xx6.022xx10^(23)` molecular of `CO_(2)`. ...(iii)
3.

How many moles of hydrogen is required to produce 4 moles of ammonia?

Answer»

15 moles
20 moles
6 moles
4 moles

Solution :`3H_(2) + N_(2) to 2 NH_(3)`
To get 2 moles of AMMONIA, 3 mole of `H_(2)` is required.
`:.` To get 4 moles of ammonia = `3/cancel(2) XX cancel(4^(2))`
= 6 moles of `H_(2)` is required.
4.

How many molecules of CO_(2)is in 44 g CO_(2) ?

Answer»

`6xx10^(23)`
`3xx10^(23)`
`12 xx 10^(23)`
`3xx10^(2)`

ANSWER :A::B::C
5.

How many molecules of an ideal gas are there is 1xx10^(-3)dm^(3)at STP ?

Answer»

SOLUTION :`22.4 DM^(3)` of an IDEAL GAS has `6.023xx10^(23)` molecules at STP ?
`1xx10^(-3) dm^(3)` of an ideal gas contains `=(6*023xx10^(23))/(22.4)xx1xx10^(3)=2.69xx10^(19)` molecules.
6.

How many molecules are there in a 3.46 g sample of hydrogen chloride, HCl? Note:The number of molecules in a sample is related to moles of compound (1 mol HCl = 6.023 xx 10^(23) HCl molecules). Therefore if you first convert grams HCl to moles, then you can convert moles to number of molecules).

Answer»

SOLUTION :`3.46 g "HCl" xx (1 " mol HCl")/(36.5 g "HCl")xx (6.023 xx 10^(23)" HCl molecules")/(1 " mol HCl")`
`= 5.71 xx 10^(22)` HCl molecules
7.

How many molecules are present in 5.6xx10^(-8)cc of Cl_(2) at stP?

Answer»


ANSWER :`1.5xx10^(12)`
8.

How many molecules are present in 32 g of methane?

Answer»

`2xx6.023xx10^(23)`
`6.023xx10^(23)//2`
`6.023xx10^(-23)`
`3.011xx10^(23)`

Solution :METHANE `(CH_(4))` -MOLAR mass=12+4=16g
16 g contains `6.023 xx 10^(23)` molecules.
32 g of methane will contain =`(6.022xx10^(23))/CANCEL(16)xxcancel(32^(2))=2xx6.023xx10^(23)`
9.

How many molecules are present in 2.34 moles of methane?

Answer»


ANSWER :`1.049xx10^(24)`
10.

How many molecules and atoms of phosphorus are present in 0.1 mole of P_(4) molecules ?

Answer»


Answer :`6.022 XX 10^(22)` molecules (ii) `2.409 xx 10^(23)` ATOMS
11.

How many molecules are present in 1 equivalent of hydrogen?

Answer»

Solution :1 EQUIVALENT of H CONTAINS 1 MOLE of H.
`THEREFORE` 1 equivalent of H contains 1/2 mole of `H_2`.
12.

How many mole of FeCl_(3) can be prepared by the reaction of 10g pf KMnO_(4)10.07mole of FeCl_(2) and 500mL of 3M HCl follwing the reaction: 5FeCl_(2) + KMnO_(4) + 8HCI rarr 5FeCl_(3) + KCl + MnCl_(2) + 4H_(2)O

Answer»


ANSWER :`0.315 MOLE`
13.

Find the number of moles of the solute present in 600ml of 0.05M solution.

Answer»


ANSWER :0.03
14.

How many mole of electrons are involved in the reduction of one mole of MnO_(4)^(-)ion in alkaline medium to MnO_(3)^(-)?

Answer»

2
1
3
4

Solution :`{:("MnO_4^(-),MnO_3^(-)),(+2H_2O,+H_2O),(+2E^(-),+2OH^(-)):}`
15.

How many mL of 1 M H_(2)SO_(4) acid solution is required to neutralise 10 mL of 1 M NaOH ?

Answer»

5 mL
2.5 mL
10 mL
20 mL

Solution :N//A
16.

How many mL of 0.125 M Cr^(3+)must be rected with 12.00 mL of 0.200 M MnO_(4)^(-) if theredox products are Cr_(2)O_(7)^(2-) andMn^(2+) ?

Answer»

8 ML
16 mL
24 mL
32 mL

Solution :Reduction equation
OXIDATION equation
`2Cr^(3+)+7_(2)OrarrCr_(2)O_(7)^(2-)+14H^(+)+6e^(-)xx5`
balanced redox equation
applying molarity equation
`therefore (0.125xxV)/(10)=(0.2000xx12.00)/(6)`
or `V_(1)=(0.200xx12.00)/(6)xx(10)/(0.125)=32 mL`
17.

How many mL is one m^(3) ?

Answer»

SOLUTION :`1m^(3) = (10dm)^(3) = 10^(3) DM^(3) = 10^(3)L , 1L = (10cm)^(3) = 10^(3)cm^(3) = 10^(3)mL`
THEREFORE `1m^(3) = (100cm)^(3) = 10^(6)cm^(3) = 10^(6)mL`
One `m^(3)` of GAS is `10^(6)`mL.
18.

How many millilitres of 0.5 M H_(2)SO_(4) are needed to dissolve 0.5 g of copper (II) carbonate ?

Answer»


Solution :`N_(1)V_(1)=N_(2)V_(2)`
`N_(1)= "Normality of " H_(2)SO_(4)=0.5xx2=1N`
`V_(1)= " VOL of " H_(2)SO_(4)`
`N_(2)`= Normality of copper (II) carbonate `=(0.5xx2)/(123.5)N`
`V_(2)`=VOLUME of copper (II) carbonate =1000 ML
Thus, `1.0xxV_(1)=(0.5xx2)/(123.5)xx1000`
or `V_(1)=8.09 mL`
19.

How many milliliters of 0.125 M KMnO_4 are required to react completely with 25.0 mL of 0.250 MFeSO_4 solution in the acidic medium ?

Answer»

Solution :The balanced chemical EQUATION is :
`MnO_4^(-) +5Fe^(2+) +8H^(+) rarr MN^(2+) +5Fe^(3+) +4H_2O`
From the balanced equation it is clear
`{:(1 mol KMnO_4 =, 5 mol FeSO_4),(158,5xx152):}`
Moles of `FeSO_4` present in 25 mL of 0.250 M solution :
`=(0.250)/1000xx25=0.00625` mol
Let us determine the number of moles of `KMnO_4` that must react
5 mol of `FeSO_4` react with `KMnO_4`
= 1mol
0.00625 mol of `FeSO_4` will react with `KMnO_4`
`=1/5xx0.00625=0.00125 mol`
Now, we are the calculate the volume of 1.25 M
`KMnO_4` solution which contain 0.00125 mol .
According to definition of morality , 0.125 M solution means that
`0.125 mol KMnO_4` is present in
= 1000 mL
0.00125 mol of `KMnO_4` is present in
`=1000/(0.125)xx0.00125=10 mL`
This can also be CALCULATED as :
Morality `=("Moles"xx1000)/(Volume)`
`0.125=(0.00125xx1000)/(volume)`
`:. Volume = (0.00125 xx1000)/(0.125) = 10 ` mL
20.

How many milliliters fo a 0.05 M KMnO_4 solution are required to oxidize 2.0 g FeSO_4 in a dilute acid solution ?

Answer»

` 32. 56 mL`
`62.53 mL`
` 25.36mL`
` 52.63 mL`

Solution :Normality `=n_("FACTOR")xx`Molarity
`overset(+7)(M)nO_(4)^(-) rarroverset(+2)(M)n^(2+), n_("factor")=5`
`Fe^(2+) rarr Fe^(3+), n_("factor")=1`
Normality of `KMnO_(4)=(5)(0.05)`
`=0.25 N`
Volume of `KMnO_(4)=V` milliliters
Thus, milliequivalents of `KMnO_(4)=NxxV`
`=0.25 V`
Equivalents of `FeSO_(4)=("Mass"_(FeSO_(4)))/("Gram EQUIVALENT mass"_(FeSO_(4)))`
`["Note that eq. wt. of "FeSO_(4)=("Formula weight")/("Change in O.N.")=152/1]`
Milliequivalent of `FeSO_(4)=2/152xx1000`
ACCORDING to the law of equivalence,
`"Milliequivalents"_(KMnO_(4))="Miliequivalents"_(FeSO_(4))`
`0.25 V= 2/(152) xx 1000`
` V= (2xx 1000)/(152 xx 0. 25) `
` = 52. 63 mL`.
21.

How many milligrams of pure sulphuric acid in 250mL aqueous solution has a pH value 3 ?

Answer»

SOLUTION :`12.25 MG`
22.

How many milliliters of 0.025 M K_2Cr_2O_7 are required to react completely with 25.0 mK of 0.20 M solution of FeSO_4 ?

Answer»

Solution :The BALANCED chemical equation is
`3Fe^(2+)+Cr_2O_7^(2-)+14H^(+)rarr6Fe^(3+)+2Cr^(3+)+7H_2O`
1 mol of `K_2Cr_2O_7 =6 " mol of "FeSO_4`
Moles of `FeSO_4` PRESENT in 25.0 mL of 0.20 M solution
`(0.20)/1000xx25.0 =0.005 mol`
Now 6 mol of `FeSO_4` REQUIRE `K_2Cr_2O_7`
= 1mol
`=1/6xx0.005 = 0.000833 ` mol
Let US calculate the volume of `K_2Cr_2O_7` solution containing 0.000833mol
0.025 mol of `K_2Cr_2O_7` are present in
=100 mL
0.000833 mol of `K_2Cr_2O_7` are present in
`=(1000xx0.000833)/(0.025)=33.3mL`
23.

How many milligrams of bromine are present in a bottle containing 1/10 kg ?

Answer»


SOLUTION :`1 kg = 10^(6) mg :. "CONVERSION FACTOR" = ((10^(6) mg)/((1 kg))`
`1//10kg=0.1kg=((10^(6)mg))/((1kg))xx(0.1kg)=10^(5)`mg
24.

How many milligrams of calcium should be added to decrease the concentration of 200ml of 1.02N HCl to 1.0N HCl?

Answer»


ANSWER :80mg
25.

How many milli-gram of iron (Fe^(2+))are equal to 1mL of 0.1055N K_(2) Cr_(2)O_7equivalent ?

Answer»

5.9 mg
0.59 mg
59 mg
`59 XX 10^(-3) mg`

SOLUTION :`w/56 xx100=1xx0.1055`
`w = (0.1055xx56)/1000 = 5.9` mg
26.

How many methyl groups present in neo pantane.

Answer»


ANSWER :2
27.

How many methyl groups present in neopentance.

Answer»


SOLUTION :Methyl groups present in NEOPENTANE is four.
28.

How many methods can be used for the preparation of the isopropyl benzene? (v). Benzene +H_(3)C-overset(CH_(3))overset(|)(C)=CH_(3)overset(H_(2)SO_(4))to

Answer»


SOLUTION :I,II,III,IV
29.

How many methyl groups are present in 2, 5-Dimethyl-4-ethylnonane ?

Answer»

2
3
4
5

Solution :`{:(""CH_(3)),("|"),(OVERSET(1)(C)H_(3)overset(2)(C)Hoverset(3)(C)H_(2)overset(4)(C)overset(5)(HC)Hoverset(6)(C)H_(2)overset(7)(C)H_(2)overset(8)(C)H_(2)overset(9)(C)H_(3)),("|""|"),(""CH_(3)""CH_(2)),("|"),(""CH_(3)):}`
The no. of `CH_(3)` groups is 4.
30.

How many methods can be used for the preparation of iodobenzene? (i). C_(6)H_(6)+I_(2) (ii). C_(6)H_(6)+I_(2)overset(HNO_(3))to

Answer»

`C_(6)H_(5)N_(2)+CL^(-)+KI`
`C_(6)H_(6)+Kito`

Solution :Only 2, (II) & (iii)
31.

How many “methyl groups” are present in 2,3-dimethyl-4-ethyl heptane

Answer»

2
8
4
5

Answer :D
32.

How many metamers of 3-pentanone are possible? Write their structures and IUPAC names. Can these be regarded as position isomers as well ?

Answer»

Solution :Two metamers (I and II) of 3-pentanone are POSSIBLE.
`underset("3-Pentanone")(CH_(3)CH_(2)-OVERSET(O)overset(||)(C)-CH_(2)CH_(3)) "" underset("2-Pentanone (I)")(CH_(3)-overset(O)overset(||)(C)-CH_(2)CH_(2)CH_(3))``{:(""O""CH_(3)),("||""|"),(CH_(3)-C-CH-CH_(3)),("3-Methyl-2-butanone (II)"):}`
Both these can ALSO be regarded as position isomers of 3-pentanone because they differ in the position of the KETO group.
33.

How many metal cations get precipitated, when (NH_(4))_(2)S solution is added into the aqueous solutions of the following cations? Ag^(+),Cu^(2+),Cr^(3+),Mn^(2+),Zn^(2+),Bi^(3+)

Answer»


SOLUTION :`AG^(+),Cu^(2+),ZN^(2+),BI^(3+),Cr^(3+),Mn^(2+)`
34.

How manymaximumnumberofelectrons in 3porbitalhavespinquantumnumbers= +(1)/(2)?

Answer»

0
6
3
2

Answer :C
35.

Howmanymaximum electroncan be filled in d andf orbitals ?

Answer»

SOLUTION :10 and 14
36.

How many main states of matter ?

Answer»

Solution :(i) Solid (ii) LIQUID and (iii) Gas. In additional care PLASMA and BEC (BOSE Einstein Condense) is KNOWN as states of MATTER.
37.

How many litres of water must be added to 1 litre of an aqueous solution of HCl with pH of 1 to create an aqueous solution of with pH of 2 ?

Answer»

9.0 L
0.1 L
0.9 L
2.0 L

Solution : pH = 1 means `[H^(+)]=10^(-1) M = 0.1 M`
pH = 2 means `[H^(+)]= 10^(-2)M = 0.01 M`
`{:("Applying ",MxxV_(1),=,M_(2)xxV_(2)),(,"(Before dilution)",,"(After dilution)"):}`
`0.1xx1=0.01xxV_(2) "or " V_(2)=10L`
`:.` Volume of WATER to be added = 10 - 1 = 9 L
38.

How many litres of water must be added to 1 litre of an aqueous solution of HCl with a pH of 1 to create an aqueous solution with pH of 2?

Answer»

SOLUTION :`0.1 XX 1 = (1 + V) (0.01) RARR v = 9 L `
39.

How many litres of SO_(2) taken at NTP have to be passed through a solution of HCIO_(3) to reduce 16.9 g of it to HCI?

Answer»

SOLUTION :13.44 LITRES
40.

How many litres of liquid C Cl_(4) (d=1.5 g/cc) must be measured out to contain 1 xx 10^(25) C Cl_(4)molecules?

Answer»


ANSWER :`1.61 L`
41.

How many litres of CO_(2) at STP wil be formed when 100 of 0.1 M H_(2)SO_(4) reacts with excess of Na_(2)CO_(3)?

Answer»

22.4
2.24
0.224
5.6

Answer :C
42.

How many liters of helium gas at NTP has weight same as 3.0115xx10^(23) molecules of methane?

Answer»


ANSWER :44.8
43.

How many layer are adsorbed in chemical adsorption?

Answer»

ONE
two
many
zero.

Solution :Chemiadsorprtion INVOLVES unimolecular layer FORMATION of the ADSORBATE.
44.

How many lattice points are there in one unit cell of each of the following lattices ? (i)Hexagonal close packing and cubic packing ltbr

Answer»

Solution :LATTICE points in face centred cubic or face centred tetragonasl8 ( at corconers) + 6 ( at face centres)
= 14.However, particles per UNIT cell = ` 8 xx 1/8 + 6 xx 1/2= 4`
Lattice points in body centreed cube =8 ( at corners) + 1 ( at body centre) = 9
However, PATICLES , per unit cell ` = 8 xx 1/8 + 1=2`
45.

How many lattice points are there in one unit cell of each of the following lattices ? (i) face centred cubic ,(ii) face centred tetragonal ,(iii) body centred cubic.

Answer»

Solution :Lattice points in face centred cubic or face centred TETRAGONAL =8 (at corners) + 6 ( at face centres ) =14. However , PARTICLES PER unit cell =`8xx1/8+6xx1/2=4`
Lattice points in body centred cube =8 (at corners ) + 1 ( at body centre) =9
However, particles per unit cell =`8xx1/8+1=2`
46.

How many isotopes of Hydrogen ? Give its Names.

Answer»

Solution :HYDROGEN has THREE ISOTOPES :
(i) protium `(._1^1H)` , (ii) deuterium (`._1^2H` and D ) and (III) tritium (`._1^3H` or T)
47.

How many joules of translational kinetic energy is associated with 4 grams of methane at 27^@C?

Answer»

Solution :Number of MOLES of methane ` = ("weight of methane")/("GMW") = (4)/(16) = (1)/(4)`
Kinetic energy is GIVEN as (K.E.)` = (3)/(2)NRT = (3)/(2) xx(1)/(4) xx 8.314 xx 300 = 935` joule
48.

How many isomertic pentanoic acid undergo ? Hence Vohard zelinsky reaction on treatment with Br_(2)//"red" P

Answer»


ANSWER :3
49.

How many isomers with melocular formula C_6H_(12) gives Acetaldehyde in reductive ozonolysis asone of product

Answer»


SOLUTION :HEXANE ISOMERS
50.

How many isomers of C_8 H_(10) when react with hot alkaline KMnO_4 give di-carboxylic acid as a product ?

Answer»


SOLUTION :