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.

The number of nucleons present in chlorine -37 is

Answer»

17
20
54
37

Answer :D
Discussion
2.

The number of nucleons in the isotope of an atom ""_(z)X^(m) are

Answer»

m
Z
m+Z
m-Z

Solution :No. of NUCLEONS = m
Discussion
3.

The number of nucleons in chlorine - 37 is

Answer»

17
20
54
37

Answer :D
Discussion
4.

The number of nodes in s orbital of any energy level is equal to

Answer»

`N`
`2n^2`
`n - 1`
`n - 2`

ANSWER :C
Discussion
5.

The number of nodes and nodal planes in 4p orbital are respectively

Answer»

2, 1
1, 2
2, 3
3, 2

SOLUTION :nodel planes = `L`
NODES = n - l - 1
for 4p orbital, n = 4 , l=1
Discussion
6.

The number of nodes in radial distribution curve of 4s sublevel is

Answer»

1
2
3
4

Answer :C
Discussion
7.

The number of nodal planes in p_(x)-obital is:

Answer»

one
two
three
zero

Answer :A
Discussion
8.

The number of nodal planes for P_(x) orbital is

Answer»

1
2
3
0

Answer :A
Discussion
9.

The number of nodal planes .d. orbital has

Answer»

Zero
one
two
three

Answer :C
Discussion
10.

The number of nodal planes '5d' orbital has, is

Answer»

zero
one
two
three.

Solution :No. of NODAL PLANES = VALUE of L
Discussion
11.

The number of nitrogen molecules present in lc.c of gas at NTP is

Answer»

`2.67 xx 10^(22)`
`2.67 xx 10^(21)`
`2.67 xx 10^(20)`
`2.67 xx 10^(19)`

ANSWER :D
Discussion
12.

The number of neutrons in the dipositive zinc ion (Mass no. of Zn = 65)

Answer»

35
33
65
67

Answer :A
Discussion
13.

The number of neutron in 540 gm water

Answer»

`240xx N_(A)`
`30xx N_(A)`
`540 XX N_(A)`
`18 xx N_(A)`

ANSWER :A::B::D
Discussion
14.

The number of neutorn(s) present in deuterium is

Answer»

0
1
2
3

Answer :B
Discussion
15.

The number of moles of sulphate ions present in the general formula of 1 mole of alum ?

Answer»


SOLUTION :`X_(2)SO_(4).Y_(2)(SO_(4))_(3).24H_(2)O`
Discussion
16.

The number of moles of oxygen required to prepare 1 mole of water is

Answer»

1 MOLE
0.5 mole
2 mole
0.4 mole

Solution :`underset("1 mole")(H_(2)) + underset(0.5 "mole")(1//2O_(2)) to underset("1 mole")(H_(2)O)`
0.5 mole of OXYGEN is required to PREPARE 1 mole of `H_(2)O`
Discussion
17.

The number of moles of oxygen in 1L of air which contains 21% oxygen by volume under standard conditions is :

Answer»

0.186 MOL
0.21 mol
2.10 mol
0.0093 mol

Solution :22.4 L of `O_(2)=1` mol
`0.21 L` of `O_(2) = (0.21)/(22.4)= 0.0093` mol
Discussion
18.

The number of moles of MnO_(4)^(-) and Cr_(2)O_(7)^(-2) separately required to oxidise 1 mole of FeC_(2)O_(4) each in acidic medium respectively

Answer»

0.5 , 0.6
0.6 , 0.4
0.4 , 0.5
0.6 , 0.5

Answer :D
Discussion
19.

The number of moles of lead nitrate needed to coagulate 2 mol of colloidal [AgI]I^(-) is

Answer»

2
1
`1//2`
`2//3`

Solution :`2[AGI]I^(-)+Pb^(2+) rarr PbI_(2)+2AgI`
THUS 2 moles of `[AgI]I^(-)` are coagulated by 1 mole of `Pb^(+)` i.e., 1 mole of Pb `(NO_(3))_(2)`
Discussion
20.

The number of moles of KNnO_(4) that will be needed to react with one mole of sulphite ion in acidic solution is :

Answer»

`4//5`
`2//5`
1
`3//5`

Solution :The ionic equation for the reaction is :
`2MnO_(4)^(-)+6H^(+)+5SO_(3)^(2-)rarr2Mn^(2+)+5SO_(4)^(2-)+3H_(2)O`
5 mole of `SO_(3)^(2-)` IONS are oxidised by `MnO_(4)^(-)` ions = 2 MOL
1 mole of `SO_(3)^(2-)` ions is oxidised by `MnO_(4)^(-)` ions = 2/5 mol.
Discussion
21.

The number of moles of KMnO_(4) that will need to react completely with one mole of ferrous oxalate in acidic solution is :

Answer»

`2//5`
`3//5`
`4//5`
1

Solution :N//A
Discussion
22.

The number of moles of KMnO_4 that will be needed to react completely with one mole of ferrous oxalate in acidic solution is

Answer»

2/5
3/5
4/5
1

Solution :`MnO_(4)^(-)+SO_(3)^(-2) RARR Mn^(+2)+SO_(4)^(2-)`
change in oxidation of Mn = 5
change in oxidation of S= 2
`2MnO_(4)^(-)+5SO_(3)rarr2Mn^(+2)+5SO_(4)^(2-)`
`:.`1 mol of `SO_(3)^(-2) `require `2/5` MOLES `KMnO_4`
Discussion
23.

The number of moles of KMnO_(4) that will be needed to react with one mole of sulphite ion in acidic solution is :

Answer»

`(4)/(5)`
`(2)/(5)`
`1`
`(3)/(5)`

SOLUTION :The redox reaction is :
`2MnO_(4)^(-)+5SO_(3)^(2-)+6H^(+) to 2Mn^(2+) + 5SO_(4)^(2-)+3H_(2)O`
1 mole `SO_(3)^(2-)` will be oxidised by `(2)/(5)` mol of `MnO_(4)^(-)`
Discussion
24.

The number of moles of KMnO_(4) needed to react with one mole of SO_(3)^(2-) in acidic solution is

Answer»

`4//5`
`2//5`
1
`3//5`

Solution :REDUCTION half reactoin
`MnO_(4)^(-)+8H^(+)+5e^(-)rarrMn^(2+)+4H_(2)O`
OXDIATION half reaction
`SO_(3)^(2-=)+H_(2)OrarrSO_(4)^(2-)+2H^(+)+2e^(-)` sinc 1 mole of `SO_(3)^(2-)` gives `2e^(-)` butreduction of 1 mole of `KmnO_(4)` k requires `5 e^(-)` therefore no of MLES of `KMnO_(4)` krequired to react with 1 mole of `SO_(3)^(2-)=2//5`
Discussion
25.

The number of moles of KMnO_(4) reduced by one mole of KI in neutral medium is ("Hint "KI to IO_(3)^(-))

Answer»

ONE
Two
Five
One-fifth

Answer :B
Discussion
26.

The number of moles of KI required to produce 0.4 mole K_(2)HgI_(4) is

Answer»

`1.6`
`0.8`
`3.2`
`0.4`

ANSWER :D
Discussion
27.

Number of moles of KI (potassium iodide) required to produce 0.1 mole of K_2Hg I_(4) is

Answer»

1.6
0.8
3.2
0.4

Answer :D
Discussion
28.

The number of moles of H_(2)O_(2) needed to reduce 2 mole of KMnO_(4) in acidic medium is.

Answer»

2
2.5
5
3

Answer :B
Discussion
29.

The number of moles of H_(2) in 2.24 litre of hydrogen gas at STP

Answer»

0.1 MOLE
0.01 mole
0.001 mole
1 mole

Solution :22.4 litre = 1 MOLAR volume = 1 mole
2.24 litre =`1/22.4 XX 2.24` = 0.1 mole
Discussion
30.

The number of moles of H_(2) in 0.224 litre of hydrogen gas at STP is

Answer»


SOLUTION :22.4 litre of hydrogen GAS at STP CONTAINS 1 MOLE.
`:.` 0.224 litre of hydrogen gas at STP will contain `1/22.4xx0.224 = 0.01`
Discussion
31.

The number of moles of ethanol chloride will produce 10.2 g of ethanoic anhydride on reaction with sodium acetate

Answer»

0.51 mol
1.02 mol
0.10 mol
10.2 mol

Solution :`underset(1 mol)(CH_(3)COCL) + NaOCOCH_(3) rarrunderset(102 G)((CH_(3)CO)_(2) O + NaCl)`
10.2 g of ethanoic anhydride will be OBTAINED from ACETYL chloride `= (1)/(102) xx 10.2 = 0.10 mol`
Discussion
32.

The number of moles of Fe_(2)O_(3) formed when 5.6 lit of O_(2) reacts with 5.6g of Fe?

Answer»

0.125
0.01
0.05
0.1

Answer :A
Discussion
33.

The number of moles of electrons involved in the manufacture of 1 mole of H_2O_2from 50% H_2SO_4

Answer»

2
3
1
4

Answer :A
Discussion
34.

The number of moles of CO_(2) produced when 3 moles of HCl react with excess of CaCO_(3) is

Answer»

1
1.5
2
2.5

Answer :B
Discussion
35.

The number of moles of an acid or base added to one litre of the buffer solution so as to change its pH by one unit is called............of the buffer.

Answer»


ANSWER :BUFFER CAPACITY or buffer INDEX
Discussion
36.

The number of molesof an ideal gas that should be taken in a closed vessel of30Lcapacity at a temperature of 27^(@)C so that the pressure exertedby the gas on the walls of the container is 4.1 atmosphere is

Answer»


ANSWER :5
Discussion
37.

The number of moles of acidified KMnO_(4) required to oxidise one mole of ferrous oxalate (FeC_(2)O_(4)) is :

Answer»

5
3
0.6
1.5

Solution :
`therefore 5 " MOLE" FeC_(2)O_(4)-=3 " mole" KMnO_(4)`
1 mole `FeC_(2)O_(4)-= 0.6 " mole" KMnO_(4)`
Discussion
38.

The number of moles in 0.44 g of CO_2 is ( C = 12, O = 16)

Answer»

100
10
0.1
0.01

Solution :MOLES of `CO_2` `= ("MASS")/("MOLECULAR WEIGHT")= (0.44)/(44)= 0.01`
Discussion
39.

The number of molecules of CO_(2) liberated by complete combustion of 0.1 g of graphite in air is

Answer»

`3.01xx10^(22)`
`6.02xx10^(23)`
`6.02xx10^(22)`
`3.01xx10^(23)`

ANSWER :C
Discussion
40.

The number of molecules in one litre of water is (density of water = 1g/mL)

Answer»

`6XX10^(23)//22.4`
`3.33xx10^(25)`
`3.33xx10^(23)`
`3.33xx10^(23)`

ANSWER :B
Discussion
41.

The number of molecules in a drop of water (0.0018 ml) at room temperature is

Answer»


SOLUTION :0.0018 ml - drop of water = 0.0018 g H2O = molecular mass = 18 g. Number of molecules in 18 g = `6.023 xx 10^(23)`
Number of molecules in 0.0018 g = `(6.023xx10^(23))/18 xx 0.0018 = 6.023xx10^(23)xx10^(-5)`
`= 6.023xx10^(19)` molecules.
[OR]
DENSITY of water at `25^@` C=997.0479 g/L
Mass of 0.0018 ml (or) `0.0018xx10^(-3)L`
`=DXXV`
`=997.05xx0.0018xx10^(-3)`
`=1.795xx10^(-3)g`
Molar mass of water = 18 g
Mole=`(1.795xx10^(-3))/18=9.971xx10^(-5)` mole
`:.`Number of molecules in 0.0018 ml= moles `xx` Avogadro number
`9.971xx10^(-5)xx6.023xx10^(23)`
`6xx10^(19)` molecules
Discussion
42.

The number of molecules in 40 g of sodium hydroxide is ……..

Answer»


SOLUTION :SODIUM hydroxide = NAOH = 23 + 16 + 1 =40 g
40 g= 1 mole = `6.023 xx 10^(23)`
Discussion
43.

The number of molecules in 16g of methane is :

Answer»

`3.0 xx 10^(23)`
`6.022 xx 10^(23)`
`16//6.022 xx 10^(23)`
`16//3.0 xx 10^(23)`

SOLUTION :16G of methane (1.0 mole) contain AVOGADRO's number of molecules.
Discussion
44.

The number of molecules in 16g of methane is .............

Answer»


Solution :Methane: `CH_(4)`
Molecular MASS = 12 + 4=16
16 g of methane CONTAINS Avogadro number of molecules = `6.023 xx 10^(23)` molecules.
Discussion
45.

The number of molecules in 11 g of carbon dioxide approximately (C = 12, O = 16)

Answer»

`0.5 xx10^(23)`
`1.5 xx10^(23)`
`2.5 xx10^(23)`
`3.5 xx10^(23)`

SOLUTION :NUMBER of MOLECULES in 11 G of `CO_2`
`=(6xx10^(23)xx11)/(44) = 1.5 xx10^(23)`
Discussion
46.

The number of molecule in 4.4 gmCO_(2) ..........

Answer»

`6.022xx10^(21)`
`6.022xx10^(22)`
`6.022xx10^(23)`
`6.022xx10^(-17)`

Solution :M.M of `=44 g "mol"^(-1)`
mole of `CO_(2) = (4.4)/(44) = 0.1`
MOLECULS `= 0.1xx6.022xx10^(23)`
`=6.022xx10^(22)`
Discussion
47.

The number of mole of oxalate ions oxidised by one mole of MnO_4^(-) is

Answer»

`1//5`
`2//5`
`5//2`
`5`

Solution :`overset(+7)(MnO_4^(-))overset((5))rarrMn^(2+),overset(+3)(C_2)O_4^(2-)overset((2))rarr2overset(+4)(CO_2)`
`1/5` moles `MnO_(4)^(-)-=1/2` moles `C_(2)O_(4)^(2-)`
1 mol ........ ? `=5/2`
Discussion
48.

The number of moes of KNnO_(4) reduced by one mole of KI in alkaline medium is

Answer»

one
two
five
one FIFTH

SOLUTION :`2KMnO_(4)+H_(2)Orarr2KOH+2MnO_(2)+3[O]KI+3[O]rarrKIO_(3)`
`2KMnO_(4)+H_(2)O+H_(2)O+KIrarr2MnO_(2)+2KOH+KIO`
`therefore 2 KMnO_(4)=KI`
Thus2 MOLES of `KMnO_(4)`are reduced by 1 mole of KI i.e option (b) is CORRECT
Discussion
49.

The number of millimoles of H_2SO_4 present in 5 litres of 0.2N H_2SO_4 solution is

Answer»

500
1000
250
`0.5 XX 10^(-3)`

ANSWER :A
Discussion
50.

The number of lone paris of electrons one the central atoms ofH_(2)O, SnCl_(2), PCl_(3) and XeF_(2) respectively are

Answer»

2,1,1,3
2,2,1,3
3,1,1,2
2,1,2,3

Solution :O is in Group16 . Last shell has 6 electrons . Two
are shared with H-atoms . No of lone
pairs on o = 2
Sn is in Group 14 . Last shell has 4 electrons . Two
areshared . No. of lone pairs on Sn = 1 .
P is in Group 15 . shell has 5 electrons . Three
are shared with Cl atoms . No of
lone pair on P = 1
Xeis in Group 18 . Lastshell has 8 electrons. Two
are shared with two F atoms .
No . of lone pairs on Xe = 3
Discussion