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 moles of MnO_(4)^(-) ions will react with 1 mole of ferrous oxalate in acid medium ?

Answer»

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

SOLUTION :`2KMnO_(4)+3H_(2)SO_(4) to K_(2)SO_(4)+3H_(2)O+5[O]`
`2{:((COO^(-),),(|,Fe^(2+)),(COO^(-),)):}+3H_(2)SO_(4)+3[O]to Fe_(2)(SO_(4))_(3)+4CO_(2)+3H_(2)O`
`therefore6 " mole" KMnO_(4)-= 10 " mole" FeC_(2)O_(4)`
`THEREFORE 1 " mole" FeC_(2)O_(4)` will REACT with `(3)/(5)` mole of `KMnO_(4)`
Discussion
2.

How many moles of Mg_(3)(PO_(4))_(2) consists 0.25 mole oxygen atoms ?

Answer»

`2.55xx10^(-2)`
`0.0225`
`3.125xx10^(-2)`
`1.5xx10^(-2)`

SOLUTION :1 mole `Mg_(3)(PO_(4))_(2)` consists 8 mole .O. atoms
8 mole atoms in 1 mole `Mg_(3)(PO_(4))_(2)`
`:.0.25` mole atoms `= (?)`
`=(0.25xx1)/(8) = 0.03125 = 3.125xx10^(-2)` mole
Discussion
3.

How many moles of Mg can reduce one mole of dil. HNO_3into NH_(4)^(+)ions ?

Answer»


Solution :`4 Mg +10 HNO_3 RARR NH_4NO_3+4Mg(NO_3)_2+3H_2O`
Only one `HNO_3` is reduced to `NH_(4)^(+)`
Discussion
4.

How many moles of methane is required in combustion reaction of Methene to produce 22g CO_(2) ?

Answer»




ANSWER :`0.5 " MOL "CO_(2)`
Discussion
5.

How many moles of methanol are dissolved in 500 mL, 3 M aqueous solution of methanol ?

Answer»

1.66
1.5
15
315

Answer :A::B
Discussion
6.

How many moles of methane are required to produce 22g CO_(2) after combustion?

Answer»

SOLUTION :0.5 MOLES
Discussion
7.

How many moles of methane are required to produce 22 g CO_(2)(g) after combustion ?

Answer»



Solution :ACCORDING to the chemical EQUATION,
`CH_(4(g)) +2O_(2(g)) rarr CO_(2(g)) + 2H_(2)O_((g))`
44 g `CO_(2(g))` is obtained from 16 g `CH_(4(g))`
( `:.` 1 mol `CO_(2(g))` is obtained from 1 mol of `CH_(4(g))` )
Mole of `CO_(2(g)) = 22 g CO_(2(g)) XX ("1 mol" CO_(2(g)))/(44 g CO_(2(g)))`
`= 0.5 "mol" CO_(2(g))`
Hence, 0.5 mol `CO_(2(g))` would be obtained from 0.5 mol `CH_(4(g))` or 0.5 mol of `CH_(4(g))` would be required to PRODUCE 22 g `CO_(2(g))`.
Discussion
8.

How many moles of methane are obtained by the hydrolysis of one mole of aluminium carbide?

Answer»


SOLUTION :`Al_(4)C_(3)+12H_(2)Oto4Al(OH)_(3)+3CH_(4)`
3 MOLES of `CH_(4)` are PRODUCED
Discussion
9.

How many moles of magnesium phosphate, Mg_3(PO_4)_2 will contain 0.25 mole of oxygen atoms ?

Answer»

0.02
`3.125 XX 10^(-2)`
`1.25 xx 10^(-2)`
`2.5 xx 10^(-2)`

ANSWER :B
Discussion
10.

How many moles of magnesium phosphate, Mg_(3)(PO_(4))_(2) will contain 0.25 mole of oxygen atoms ?

Answer»

0.02
`3.125 xx 10^(-2)`
`1.25 xx 10^(-2)`
`2.5 xx 10^(-2)`

Solution :`UNDERSET("1 MOLE")(Mg_(3)(PO_(4))_(2))-=underset("8 mole")(8(O))`
8 mole of oxygen ATOMS are present in `Mg_(3)(PO_(4))_(2)=1` mole
0.25 mole of oxygen atoms are present in `Mg_(3)(PO_(4))_(2)`
`=(1xx0.25)/(8)=3.125xx10^(-2)` mole.
Discussion
11.

How many moles of magnesium phosphate Mg_3 (PO_4)_2 will contain 0.25 mole of oxygen atoms ?

Answer»

(a) `0.02`
(b) `3.125 xx 10^(-2)`
(C) `1.25 xx 10^(-2)`
(d) `2.5 xx 10^(-2)`

Solution :From the formula, `Mg_3(PO_4)_2`, it .s clear that 8 moles of oxygen atoms are present in 1 mole of MAGNESIUM phosphate.
`therefore` 0.25 moles of oxygen atoms will be present in`1/8 xx 0.25 = 3.125 xx 10^(-2)` moles of `Mg_(3)(PO_(4))_(2)`.
Discussion
12.

How many moles of lead (II) chloride are formed from a reaction between 6.5 g of PbO and 3.2 of HCl ?

Answer»

0.011
0.029
0.044
0.333

Solution :`underset("1 MOL")(PbO)+underset("2 mol")(2HCl)rarrunderset("1 mol")(PbCl_(2))+underset("2 mol")(2H_(2)O)`
`underset("0.029 mol")((6.5)/(224))"mol"underset("0.087 mol")((3.2)/(36.5))"mol"`
PbO is limiting REACTION
`:.` No. of MOLES of `PbCl_(2) = 0.029` mol.
Discussion
13.

How many moles of KMnO_(4) required to oxidised acidic medium of 1 mole Fe(C_(2)O_(4)) ?

Answer»

0.6
1.67
0.2
0.4

Solution :`MnO_(4)^(-)+8H^(+)+5e^(-)toMn^(+2)+4H_(2)O`
`FE^(+2)+C_(2)O_(4)^(-2)toFe^(+3)+2CO_(2)+3e^(-)`
`therefore` Moles of `KMnO_(4)` REQUIRED to oxidised 1 mole `Fe(C_(2)O_(4))=3/5=0.6`
Discussion
14.

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

Answer»

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

Solution :`KMnO_(4)` oxidises both `Fe^(2+)` as well as `C_(2)O_(4)^(2-)` ions.
`Fe^(2+) to Fe^(3+) + e^(-)`
`C_(2)O_(4)^(2-) to 2CO_(2) + 2e^(-)`
`MnO_(4)^(-) + 8H^(+) + 5E^(-) to Mn^(2+) + 4H_(2)O`
The OVERALL reaction is:
`5Fe^(2+) + 5C_(2)O_(4)^(2-) + 3MnO_(4)^(-) + 24H^(+) to 3Mn^(2+) + 5Fe^(3+) + 10 CO_(2) + 12H_(2)O`
Obviously, 5 moles of ferrous OXALATE react with 3 moles of `KMnO_4`. Hence, ONE mole of ferrous oxalate will react with `3/5` moles of `KMnO_(4)`.
Discussion
15.

How many moles of KMnO_(4) required to react with (SO_(3)^(-2)) sulphite ion in acidic medium?

Answer»

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

Solution :`2MnO_(4)^(-)+6H^(+)+5SO_(3)^(-2)to2Mn^(+2)+3H_(2)O+5SO_(4)^(-2)`
For 1 MOLE `SO_(3)^(-2),2/5` MOLES `KMnO_(4)` required.
Discussion
16.

How many moles of KMnO_4are needed to oxidised a mixture of 1 mole of each FeSO_4" & "FeC_(2)O_4in acidic medium

Answer»

`4//5 `
`5//4`
` 3//4`
`5//3`

Solution :`2KMnO_(4)+ 10FeSO_(4) +8H_(2)SO_(4) RARR K_(2) SO_(4) +2MnSO_(4)+5Fe_(2)(SO_4)_(3)+8H_2O`
10MOLES g `FesO_4` requires 2 MOLES of `KMnO_4`
1 MOLE of `FeC_(2)O_4` require = 1/5 moles of `KMnO_4`
`6KMnO(4) +10FeC_(2)O_(4)+24H_(2)SO_(4) rarr 3K_2SO_(4) + 6MnSO_(4) +5Fe_(2)(SO_4)_(3)+20CO_(2)+24H_(2)O`
10 mole of `FeC_(2)O_(4)` require = 6 mole of `KMnO_4`
1 mole of `FeC_(2)O_4 ` requires `=6/10 =3/5`
Total moles of `KMnO_(4) ` require `=1/5+3/5 =4/5`
Discussion
17.

How many moles of K_(2)Cr_(2)O_(7) is reduced by 1 mole Sn^(+2) ?

Answer»

`1/6`
`1/3`
`2/3`
1

Solution :
Discussion
18.

How many moles of iodine are liberated when 1 mole of potassium dichromate react with excess of potassium iodide in the presence of concentrated sulphuric acid ?

Answer»

1
2
3
4

Solution :`K_(2)Cr_(2)O_(7)+6Kl+7H_(2)SO_(4)rarr4K_(2)SO_(4)+Cr_(2)(SO_(4))_(3)+7H_(2)O+3l_(2)`
Discussion
19.

How many moles of iodin are liberated when 2 moles of potassium permanganate rect with potassium iodide?

Answer»


Solution :Balanced redox RECTION is
`2MnO_(4)^(-)+10I^(-)+16H^(+)+5E^(2)+8H_(2)O` Thus 2 moles of `KMnO_(4)` react with KI to libreate 5 Moles of `I_(2)`
Discussion
20.

How many moles of hydrogen is required to produced 20 moles of ammonia ?

Answer»

Solution :`3H_(2)+N_(2) to 2NH_(3)`
As per stoichiometric equation, No. of moles of hydrogen required for 2 moles of AMMONIA = 3 moles
No. of moles of hydrogen required for 20 moles of ammonia = `(3)/(2)xx20 = 30` moles.
Discussion
21.

How many moles of Hydrogen atoms are present in 1 mole of C_(2)H_(6) ?

Answer»

18 MOLES
6 moles
3 moles
1 mole

Solution :`C_(2)H_(6)` CONTAINS 6 H ATOMS. 6 moles.
Discussion
22.

How many moles of hydrogen is required to produce 10 moles of ammonia ?

Answer»

Solution :The balanced stoichiometric equation for the formation of ammonia is
`N_(2)` (g) + `3H_(2)` (g)` rarr 2NH_(3)` (g)
As PER the stoichiometric equation, to produce 2 moles of ammonia, 3 moles of hydrogen are required.
`:.` to produce 10 moles of ammonia,
`(3 "moles of"H_(2))/(cancel(2 "moles of" NH_(3))) XX cancel(10 "Moles" of NH_(3))5`
=15 moles of hydrogen are required
Discussion
23.

How many moles of hydrogen is required to produce 20 moles of ammonia?

Answer»

Solution :`3H_(2) + N_(2) to 2NH_(3)`
As per stoichiometric equation,
No. of moles of hydrogen required for 2 moles of AMMONIA = 3 moles
No. of moles of hydrogen required for 20 moles of ammonia = `3/2 xx 20` = 30 moles.
Discussion
24.

How many moles of hydrochloric acid react with one mole of borax to convert all boranes to boric acid.

Answer»


Solution :`underset("1 mole")(Na_(2)B_(4)O_(7))+underset("2 mole")(2HCl)+5H_(2)O RARR 2NaCl+4H_(3)BO_(3)`
Discussion
25.

How many moles of H_(2)SO_(4) can be reduced to SO_(2) by 2 moles of Aluminium?

Answer»


Solution :EQ. AL = Eq. `H_(2)SO_(4)`
`2xx3=nxx2impliesn=3`
Discussion
26.

How may moles of H_(2)O_(2) must be present in 2L of its solution, such that 100 ml of the solution can liberate 3.2 grams of oxygen at 273^(@)C and 0.5 atm pressure?

Answer»


Solution :`PV=((W)/(M))RT`
`(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2)),(0.5xx8960)/(546)=(1xxV_(2))/(273)`
`0.5xxV=(3.2)/(32)xx0.0821xx546`
`V_(2)=2240ml`
`V=8.96` litres = 8960 ML
`100ml H_(2)O_(2)_____2240 ml` oxygen at STP
`1mlH_(2)O_(2)______22.4ml` oxygen at STP
volume strength of `H_(2)O_(2)=22.4`
1 volume `H_(2)O_(2)=3.03g//L=(3.03" moles")/(34)//L`
22.4 volume `H_(2)O_(2)=(22.4xx3.03" moles")/(34)//L`
= 1.99 moles/L `~~` = 2moles/L
For 2 litres = 4 moles
Discussion
27.

How many moles of H_2O form when 25.0 mL of 0.10 M HNO_3 solution is completely neutralised by NaOH?

Answer»


ANSWER :`2.5 XX 10^(-3)` MOLE
Discussion
28.

How many moles of glucose are there in a) 540 gm glucose b) 900 gm glucose

Answer»


ANSWER :3,5
Discussion
29.

How many moles of glucose are present in 720 g of glucose ?

Answer»

SOLUTION :GLUCOSE = `C_(6)H_(12)O_(6)`
Molecular MASS of Glucose =`(12xx6)+(1xx12)+(16xx6)`
72+12+96=180
Number of MOLES of Glucose =` ("Mass of Glucose")/(" Molecular mass of Glucose")`
`720/180 = 4 `moles
Discussion
30.

How many moles of gold are present in 49.25 g of gold rod ? (Atomic mass of gold = 197)

Answer»


SOLUTION :Gram atomic MASS of gold `= 197.0 G`
197.0 g of gold have mass = 1 gramk mol
49.25 g of gold have mass `= ((49.25g))/((197.0g))XX("1 gram mol")=0.25` gram mol = 0.25 mol
Discussion
31.

How many moles of ethane is equaired to produce 44 g of CO_(2(g)) after combustion.

Answer»

Solution :`UNDERSET("ETHANE")(C_(2)H_(6)) + 3xx1/2 O_(2) to underset("Carbon DIOXIDE")(2CO_(2))+3H_(2)O`
1 mole of Ethane `overset("COMBUSTION")(to)` 2 moles of `CO_(2)`
`:.` 1 mole of `CO_(2)`
2 moles of `CO_(2)` is produced by 1 mole of ethane.
1 mole of `CO_(2)` will be produced by = ?
`:.`To produce 1 mole of `CO_(2)` , the required mole of ethane is =`1/2 xx 1` = 0.5 mole of ethane.
Discussion
32.

How many moles of electrons weigh one kilogram?

Answer»

`6.022 XX 10^(23)`
`1/(9.108) xx 10^(31)`
`6.022/9.108 xx 10^(54)`
`1/(9.108 xx 6.022) xx 10^(8)`

ANSWER :D
Discussion
33.

How many moles of electrons are involved in the conversion of 1 mole of Cr_(7)O_(7)^(2-) ions to Cr^(3+) ions?

Answer»


Solution :The balanced equation is
`Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-)rarr2Cr^(3+)+7H_(2)O`
THUS 6 moles of ELECTRONS are needed in the abvoe conversion
Discussion
34.

How many moles of dioxygen are present in 64g of dioxygen? (Molecular mass of dioxygen is 32)

Answer»

Solution :Number of MOLES =`"Mass"/"Molar mass"` Number of moles of dioxygen in 64 G of dioxygen = `(64G)/(32gmol^(-1))=2mol`
Discussion
35.

How many moles of CH_(4) will produce 12.o ethane according to the reaction: CH_(4) + Cl_(2) overset(80%to CH_(3)Cl + HCl 2CH_(3)Cl + 2Na overset(50%)to CH_(3)CH_(3) + 2NaCl

Answer»

2
0.8
0.32
1

Answer :A
Discussion
36.

How many moles of ammonia are produced, on hydrolysis of five moles of Li_(3)N?

Answer»


SOLUTION :`Li_(3)N+3H_(2)Orarr3Li(OH)+NH_(3)` 5 molesof `Lui_(3)` on HYDROLYSIS will produce 5 molesof `NH_(3)`
Discussion
37.

How many moles of AgBr (K_(sp)=5xx10^(-13)"mol"^(2)L^(-2)) will dissolve in 0.01 M NaBr solution ?

Answer»


Solution :SUPPOSE solubilityof AgBr in 0.01 M NABR = s mol `L^(-1)`. Then as `AgBr rarr Ag^(+) + Br^(-)`,
`[Ag^(+)]=s"mol " L^(-1) and "Tota" [Br^(-)]=0.01 + s ~~ 0.01 M`
`K_(sp) = [ Ag^(+)][Br^(-)], i.e., 5xx10^(-13)= s xx0.01 or s = 5 xx 10^(-11) "mol" L^(-1)`
Discussion
38.

How many moles of AgCI will be formed if 10 g each of KCI and NaCI react with excess of silver nitrate?

Answer»


Solution :The CORRESPONDING chemical equations are:
`underset("one mole")(KCl) + AgNO_(3) to underset("one mole")(AgCl) darr + KNO_(3)`
`underset("one mole")(NaCl) + AgNO_(3) to underset("one mole")(AgCl) darr + NaNO_(3)`
The gram molecular mass of KCl `=39.01 + 35.45 = 74.55 g`
`therefore` Number of moles of KCl in 10 g of it `=10/(74.55) = 0.1341`
SINCE, one mole of KCI forms one mole of AGCI, the moles of AgCI formed by 0.1341 moles of KCI = 0.1341.
Similarly, the number of moles of NaCI in 10 g of it `=10/(22.99 + 35.45) = 0.1711`
and the number of moles of AgCI formed by 0.1711 moles of NaCI = 0.1711`therefore` The total moles of AgCI formed =0.1341 + 0.1711 =0.3052
Discussion
39.

How many moles of acidified permanganate are required to oxidise one mole of ferrous oxalate?

Answer»

Solution :Ferrous oxalate `FeC_(2)O_(4)` is a DUAL REDUCTANT
`Fe^(2+)overset(1E^(-))rarr Fe^(3+)" "C_(2)O_(4)^(2-)overset(2e^(-))rarr2CO_(2)`
ONE mole of `FeC_(2)O_(4)` is involved in 3 electron change
`MnO_(4)^(2-) overset(5e^(-))rarr Mn^(2+)`
One mole of `MnO_(4)^(-)` in acid medium is involved in 5 electron change
5 moles of `FeC_(2)O_(4) = 3` moles of `MnO_(4)^(-)`
1 mole of `FeC_(2)O_(4) = ?`
Number of moles of permanganate that can be oxidised by one mole of ferrous oxalate `=(3)/(5)xx1=0.6`
Discussion
40.

How many moles of acidified FeSO_(4) can be completely oxidised by one mole of KMnO_(4)~ ?

Answer»

10
5
6
2

Solution :`2KMnO_(4)+3H_(2)SO_(4) to K_(2)SO_(4) +2MnSO_(4)+3H_(2)O +5[O]`
`([2FeSO_(4)+H_(2)SO_(4)+[O]toFe_(2)(SO_(4))_(3)+H_(2)O]XX5)/ul(2KMnO_(4)+10FeSO_(4)+8H_(2)SO_(4)to K_(2)SO_(4)+2KnSO_(4)+5Fe_(2)(SO_(4))_(3)+8H_(2)O)`
`2" mole" KMnO_(4)-=10 " mole" FeSO_(4)`
`1 "mole" KMnO_(4)-=5 " mole" FeSO_(4)`
Discussion
41.

How many moles of acidified FeSO_(4) can be completely Oxidised by one mole of KMnO_(4)

Answer»

10
5
6
2

Answer :B
Discussion
42.

How many moles H_(2) obtained from 54 gm Al. Give its reaction.

Answer»

Solution :`underset("2 mole = 54 gm")(2Al_((s))+2NaOH_((AQ)))+H_(2)O_((l))to 2NaAlO_(2(aq))+ underset("3 mole")(3H_(2(g)))`
..3 moles `H_(2)` gas obtained from 54 gm AL.
Discussion
43.

How manymoles CO_(2) will be produced on theremal decomposition of 25 gram ? CaCO_(3) ? [C=12, O=16, Ca=40]

Answer»

1
2
1.5
0.25

Solution :Molecular mass of `CaCO_(3) = 40 + 12 + 3(16)`
`= 100 g "MOL"^(-1)`
`"Mole" = ("Mass of compound in gram")/("Molecular mass of the compound in g mol"^(-1))`
`= (25 "gram")/(100 "g. mol"^(-1)) = 0.25` mole of `CaCO_(3)`
Equation of thermal decomposition of `CaCO_(3)`
`CaCO_(3(s)) rarr CaO_((s)) + CO_(2(g))`
According to STOICHIOMETRY of the balanced reaction :
1 mole `CaCO_(3(S))` produces 1 mole `CO_(2)` GAS
`:. 025` mole `CaCO_(3(S))` produces 0.25 mole `CO_(2)` gas
Discussion
44.

How many moles are present in (a) 5.08g of sodium bicarbonate (b) 16.3g of rhombic sulphur ( c) 6.46 g og helium and(d) 23.3g of zine

Answer»


ANSWER :(a) 0.06, (B) 0.06 (C ) 1.6 and (d) 0.36
Discussion
45.

How many moles are present in 54 grams of glucose?

Answer»

SOLUTION :GMW of glucose =180G
180g of glucose =1mole
54 g of glucose=?
Number of MOLES `=("weight")/("GMW")=(54)/(180)="0.3mole"`
Discussion
46.

How many moles are present in 1500mL of semi molar sucrose solutions? What is the weight of solute in the solution?

Answer»

Solution :Number of mole of solute present in V LIT solution is given as
VM, where V=1.5lit and M =0.5mol `L^(-1)`
Number of MOLES of solute `= 1.5xx0.5=0.75`
GRAM molecular weight (GMW) of sucrose = 342g
Weight of solute = number of molesGMW `=0.75xx342=256.5g`
Discussion
47.

How many moles are present in 108 grams of glucose?

Answer»

Solution :Mass of ONE electron `=9.1095 xx10^(-31)`g
Mass of one MOLE electrons = Mass of electronsx Av. Number `=9.1095xx10^(-31)xx6.022xx10&(-23)`
`5.5 xx10^(-7)` kg
Mole of electrons has 0.55 mg mass
Discussion
48.

How many moles and how many grams of sodium chloride are present in 250 mL of 0.5 M NaCl solution ?

Answer»


SOLUTION :MOLARITY of solution (M) `= ("Moles of NACL (N)")/("Volume of solution in mL"/(1000))`
`("0.5 mol L"^(-1))=(n)/(0.250L)`
`n =("0.5 mol L"^(-1)) xx (0.250 L) = 0.125` mol
`:.` No. of moles of NaCl = 0.125 mol
No. of GRAMS of `NaCl=("0.125 mol")xx("58.5 g mol"^(-1))=7.3g`.
Discussion
49.

How many molecules of water of crystallisationarepresentin1.648 gof copper sulphate (CuSO_(4).5H_(2)O) ?

Answer»

Solution :Gram molecular mass of
`CuSO_(4).5H_(2)O = 63.55 + 32.06 + (4 xx 16.0) + 5 xx (2 xx 1.008 + 16.0) = 249.69 g`
This mass contains `6.022 xx 10^(23)`molecules of `CuSO_(4).5H_(2)O`
`therefore` Total number of molecules of `CuSO_(4).5H_(2)O` present in 1.648 g `= (6.022 xx 10^(23))/(249.69)xx 1.648`
`=3.975 xx 10^(21)`
Since, each molecule of `CuSO_4.5H_2O` contains 5 molecules of water of crystallisation, therefore, total number of water molecules present in the given sample
`3.975 xx 10^(21) xx 5 = 1.987 xx 10^(22)`
Hence, 1.648 g of copper sulphate contain `1.987 xx 10^22` molecules of water of crystallisation.
Discussion
50.

How many molecules of water are present in 0.9 g of water ?

Answer»


Solution :No. of moles of 0.9 G of water `= (("1 mol"))/(("18.0 g"))=(0.9g)=0.05` mol
No. of `H_(2)O` molecules present `= ((6.022 xx10^(23)"molecules"))/(("1.0 mol"))XX(0.01mol)`
`= 3.01 xx 10^(22)` molecules.
Discussion