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 O-atoms are in 2.7xx10^(25) molecules of CO_(2) (N_(A)=6xx10^(23))

Answer»


ANSWER :`(##ALN_NC_CHM_MC_E01_013_A01##)`
90
Discussion
2.

How many moles of NH, must be added to 1 litre of 0.75 M AgNO_3in order to reduce the Ag^+concentration to 5 xx 10^(-8)M? K_b [Ag(NH_3)_7^+] = 1xx 10^(-8)

Answer»

SOLUTION :1.9 MOLE
Discussion
3.

How many moles of NaOH would be required for complete neutralization of following compounds ?

Answer»


ANSWER :D
Discussion
4.

How many moles of NaOH would be required for complete Neutralisation of following compounds

Answer»


SOLUTION :
Discussion
5.

How many moles of NaOH can be added to one litre of a solution of 0.1 M in NH_3 and 0.1 M in NH_4Clwithout changing pOH more than one unit? Assume no change in volume, pK_b = 4.75.

Answer»

SOLUTION :0.082 MOLE
Discussion
6.

How many moles of NaOH are consumed in the following Hofmann Bromamide reaction?

Answer»

<BR>

SOLUTION :`R-OVERSET(O)overset(||)C-NH_(2)+HNaOH overset(Br_(2))toR-NH_(2)+2NaBr+NO_(2)CO_(3)`
Discussion
7.

How many moles of NaOH are present in 27mL of 0.015 M NaOH ?

Answer»

`4.05 xx 10^(-3)`
`4.05 xx 10^(-4)`
`4.05 `
`0.0405 `

Solution :MOLES of `NAOH= (0.015)/(1000) xx 27 ml `
` = 4.05 xx 10^(-4) ` mol
Discussion
8.

How many moles of Na^(+) ions are in 20mL of 0.4 M Na_(3)PO_(4)?

Answer»

0.008
0.024
0.05
0.2

Solution :No. of moles of `Na_(3)PO_(4)=(MV)/(1000)=(0.4xx20)/(1000)`
=0.008
Number of moles of `NA^(+)=3XX` Number of moles of `Na_(3)PO_(4)`.
`=3xx0.008=0.024`.
Discussion
9.

How many moles of mild oxidizing agent are needed to oxidize the products formed ?

Answer»

SOLUTION :`CH_(3)COCHO,OHCCH_(2)CH_(2)CHO`
Discussion
10.

How many moles of methyl iodide are required to convert ethylamine, diethylamineand triethylamineinto quaternaryammonium salt, respectively ?

Answer»

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

Answer :A::B::C::D
Discussion
11.

How many moles of methyl alcohol is obtained when 1 mole of hydrated ferric chloride (i.e. FeCl_3 . 6H_2O) is made anhydrous by reacting with 6 moles of 2,2 dimethoxypropane.

Answer»


SOLUTION :`FeCl_3. 6H_2O+6CH_3-UNDERSET(OCH_3)underset(|)oversetoverset(OCH_3)(|)C-CH_3toFeCl_3+12CH_3OH+6CH_3COCH_3`
Discussion
12.

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

Answer»

Solution :The balanced chemical EQUATION for the combustion of methane is :
`underset("1 MOLE")(CH_(4))+2O_(2)rarr underset(=12+2xx16=44g)underset("1 mole")(CO_(2))+2H_(2)O`
Thus, to produce 44 g `CO_(2), CH_(4)` required = 1 mole
`therefore"To produce 22 g of "CO_(2),CH_(4)` required `=(1)/(44)xx22"mole"="0.5 mole"`
Discussion
13.

How many moles of mercury will be produced by electrolysing 1.0 M Mg(NO_(3))_(2) solution with a current of 2.00 A for 3 hours? [Hg(NO_(3))_(2)=200.6" g "mol^(-1)]

Answer»


Solution :`Hg^(2+)+2E^(-)toHg,Q=2Axx(3xx3600s)=21600C`
`2F=2xx96500C` produce `Hg=1` MOLE
`therefore21600`C will produce Hg `=(1)/(2xx96500)xx21600`mole=0.112 mole
Discussion
14.

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

Answer»

0.02
`3.125xx10^(-2)`
`1.25xx10^(-2)`
`2.5xx10^(-2)`

Solution :`because` 8 MOLE oxygen atoms are PRESENT in 1 mole `Mg_(3)(PO_(4))_(2)`
`therefore` 0.25 mole oxygen atoms will be present in `(1)/(8)xx0.25` mole
`Mg_(3)(PO_(4))_(2)`, i.e.,`3.125xx10^(-2)` mole `Mg_(3)(PO_(4))_(2)`.
Discussion
15.

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

Answer»

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

Solution :From the molecular formula of `Mg_3(PO_4)_2` .it is CLEAR that 1 mole of `Mg_3(PO_4)_2` CONTAINS 8 mole of O atoms .
8 mole of O atoms = 1 mole of `Mg_3 (PO_4)_2 ` 0.25 mole of O atoms` = 1/8 xx 0.25 `mole of `Mg_3 (PO_4)_2 = 3.125 xx 10^(-2)`
Discussion
16.

How many moles of lead (II) chloride will beformed from a reaction between 6.5 g of PbO and 32 g of HCI ?

Answer»

0.044
0.033
0.011
0.029

Answer :D
Discussion
17.

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

Answer»

0.044
0.333
0.011
0.029

Solution :`PBO+2HCl rarr PbCl_(2)+H_(2)O`
Molar MASS of PbO `=207+16=223" g mol"^(-1)`
`therefore"6.5 g PbO"=(6.5)/(223)" mole"=0.029" mole"`
Molar mass of HCl = 36.5 g `mol^(-1)`
`therefore"3.2 g HCl"=(3.2)/(36.5)" mole = 0.0877 mole"`
1 mole of PbO reacts with 2 MOLES of HCl
Thus, PbO is the LIMITING REACTANT.
1 mole of PbO produces 1 mole of `PbCl_(2)`
`therefore 0.029` mole of PbO will produce `PbCl_(2)`
= 0.029 mole.
Discussion
18.

How many moles of KMnO_4 are required to oxidise 10 moles of ferrous oxalate in acidic medium

Answer»


Solution :
`3MnO_(4)^(-) 5FeC_2O_4 rarr 5Fe^(+3) + 10 CO_2 + 3Mn^(+2)` , 5 mol of `FeC_2O_4` reacts with 3 mol of `KMnO_4`
Discussion
19.

How many moles of KMnO_(4) are required to liberate one gram mole of oxygen from H_(2)O_(2) in acid medium ?

Answer»


ANSWER :2.5
Discussion
20.

How many moles of KI are required to produce 0.4 moles 0.4 K_2HgI_4 ?

Answer»

0.4
0.88
0.32
1.6

Answer :B
Discussion
21.

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

Answer»


SOLUTION :`Cr_(2)O_(7)^(2-) + 14H^(+) + 6e^(-) to 2Cr^(3+) + 7H_(2)O`
`(Sn^(2+) to Sn^(4+) + 2e^(-))xx3`
`Cr_(2)O_(7)^(2-) + 14H^(+) + 3Sn^(2+) to 3Sn^(4+) + 2Cr^(3+) + 7H_(2)O`
`implies 1mol Cr_(2)O_(7)^(2-) equiv 3MOL of Sn^(2+)`
It is CLEAR from this EQUATION that 3 moles of `Sn^(2+)`reduce one MOLE of `Cr_(2)O_(7)^(2-),`hence 1 mol. of `Sn^(2+)`will reduce `1/3`moles of `Cr_(2)O_(7)^(2-)`.
Discussion
22.

How many moles of ions (cation and anion) are present in 1 mole of Diamminediaquadipyridineiron(II)sulphate

Answer»


SOLUTION :FORMULA of COMPOUND is
`[Fe(H_(2)O)_(2)(en)_(2)(py)_(2)]_(2)(SO_(4))_(3)`
Hence total moles of ions =5
Discussion
23.

How many moles of iodine are liberated when 1 mole of potassium dichromate reacts with potassium iodide?

Answer»

1
2
3
4

Solution :`K_(2)Cr_(2)O_(7) + 4H_(2)SO_(4) rarrK_(2)SO_(4) + Cr_(2)(SO_(4))_(3) + 4H_(2)O+ 3(O)`
`ul2KI + H_(2)SO_(4) + (O) rarrK_(2)SO_(4) + I_(2) + H_(2) O ] xx3)`
`K_(2)Cr_(2)O_(7) + 7H_(2)SO_(4)+ 6KI rarr4K_(2)SO_(4) + Cr_(2)(SO_(4))_(3) + 7GH_(2)O + 3I_(2)`
Discussion
24.

How many moles of iodine are liberated when 1 mole of potassium dichromate reacts with potassium iodide

Answer»

1
2
3
4

Solution :Iodine is LIBERATED from POTASSIUM iodide.
`{:(K_(2)Cr_(2)O_(7)+4H_(2)SO_(4)rarrK_(2)SO_(4)+Cr(SO_(4))_(3)+4H_(2)O+3[O]),([2KI+H_(2)SO_(4)+[O]rarrK_(2)SO_(4)+I_(2)+H_(2)O]xx3),(ulbar(K_(2)Cr_(2)O_(7)+6KI+7H_(2)SO_(4)rarr4K_(2)SO_(4)+Cr_(2)(SO_(4))_(3)+7H_(2)O+3I_(3)...)):}`
Discussion
25.

How many moles of I_(2), are liberated when I mole of potassium dichromate react with potassium iodide?

Answer»

1
2
3
4

Solution :`K_(2)Cr_(2)O_(7)+ 6KI+ 7H_(2)SO_(4) to 4K_(2)SO_(4)+ Cr_(2)(SO_(4))_(3)+ 7H_(2)O+ 3I_(2)`
Discussion
26.

Howmanymolesof I_(2) areliberatedwhen1 moleof potassiumdichromatereact withpotassiumiodide?

Answer»

1
2
3
4

Answer :C
Discussion
27.

How many moles of hydrogen molecules are there in one mole of hydrogen peroxide?

Answer»


ANSWER :1 MOLE
Discussion
28.

How many moles of hydrogen molecule are there in one mole of sulphuric acid ?

Answer»


ANSWER :1 MOLE
Discussion
29.

How many moles of helium gas occupy 22.4L at O^@Cand 1 atm pressure ?

Answer»

0.11
0.9
1
1.11

Solution :1 MOLE of all GASES at N.T.P. OCCUPIES 22.4 L
Discussion
30.

How many moles of HCl must be removed from 1 litre of aqueous HCl solution to change its pH from

Answer»

1
0.02
0.009
0.01

Answer :C
Discussion
31.

How many moles of He gas occupy 22.4 litres at 30^@Cand one atmospheric pressure

Answer»

0.9
1.11
0.11
1

Answer :A
Discussion
32.

How many moles of HIO_4 is required to break down the following molecule ?

Answer»

1
2
3
4

Solution :Only 1 MOLE `HIO_4` is NEEDED
Discussion
33.

How many moles of Grignard readent can react with one mole of following compound?

Answer»


SOLUTION :N//A
Discussion
34.

How many moles of ethyl bromide is required to convert ethanamine to N, N-diethyl ethanamine?

Answer»

Solution :TWO MOLES of ETHYL bromide is required to convert ETHANAMINE to N, N-diethyl ethanamine.
Discussion
35.

How many moles of electrons will weigh one kilogram?

Answer»

`6.023 xx 10^(23)`
`1/(9.108) xx 10^(21)`
`(6.023)/(9.108) xx 10^(54)`
`1/(9.108 xx 6.023) xx 10^(8)`

Solution :Mass of an ELECTRON `=9.108 xx 10^(-31)` kg
Mass of one MOLE of electron `=9.108 xx 10^(-31) xx 6.023 xx 10^(23) = 9.108 xx 6.023 xx 10^(-8)`
`therefore` In 1 kg, no. of moles of electrons.
`=1/(9.108 xx 6.023 xx 10^(-8)) =10^(8)/(9.108 xx 6.023)`
Discussion
36.

How many moles of electrons will together contribute to the charge of 289500 coulomb?

Answer»


ANSWER :3
Discussion
37.

Howmanymolesof electronsarerequiredfor thereductionof (i)3 molesof Zn^(2+)to Zn (ii) 1 mol of Cr^(3+)to Cr? HowmanyFaradaysof electricitywill berequired in each case ?

Answer»


Solution :Given :forreductionof 3 mol `Zn^(3+)` to Zn
Numberof molesof electronsrequired = ?
Reductionhalfreaction.
`Zn^(2+)+ 2e^(-)to Zn`
`:. 1 ` moleof `Zn^(2+) ` requires 2molesof electrons
`:. `3 moles of `Zn^(2+) `willrequires`3xx2= 6`molesof electrons
`:' 1` moleof electrons= 1 F
`:. 6`mol ofelectrons = 6 F
(II)Given :Reductionof 1 molof `Cr^(3+) ` to Cr
REDUCTION HALFREACTION
`Cr^(3+) + 3e^(-)" to " Cr`

Hence1 moleof `Cr^(3+)`will require3 moleof electrons
`:' 1` moleof electrons`=1 F"" :. 3 ` Moleof electrons=3 F `
Discussion
38.

How many moles of electrons are involved in balancing the following redox equations?H_2S + NO_3^(-) to S + NOMn (OH)_2 + H_2O_2 to MnO_2

Answer»


ANSWER :6,2
Discussion
39.

How many moles of electron weigh one kilogram?

Answer»

`6.023xx10^(23)`
`(1)/(9.108)xx10^(31)`
`(6.023xx10^(54))/(9.108)`
`(1)/(9.108xx6.023)xx10^(8)`

ANSWER :D
Discussion
40.

How many moles of CO_(2) will released when following compound treated with heat ?

Answer»


ANSWER :D
Discussion
41.

How many moles of CO_(2) per mole will release on heating of following compounds ? (b_CH_(2)=CH-CH_(2)-underset(O)underset(|)C-OH (c) Me-underset(O)underset(||)C-CH_(2)-SO_(2)H Hint : Give your answer as four digit number for example if compound abcdliberates 1,2,3 and 4 moles of CO_(2) respectively on heating fill 1234 in OMR sheet.

Answer»


ANSWER :3100
Discussion
42.

How many moles of CO_(2) will releasd when following compound is heated?

Answer»


ANSWER :D
Discussion
43.

How many moles of CO_2 will released when following compound treated with heat

Answer»


SOLUTION :
Discussion
44.

How many moles of CO_(2) will realsed when following compound is heated

Answer»


SOLUTION :3
Discussion
45.

How many moles of chlorine are obtained when one mole of K_(2)Cr_(2)O_(7) reacts with excess hydrochloric acid ?

Answer»


ANSWER :3
Discussion
46.

How many moles of AgCl will be precipitated when an excess of AgNO_(3) is added to a molar solution of [CrCl(H_(2)O)_(5)]Cl_(2) ?

Answer»

SOLUTION :TWO MOLES of AGCL will be PRECIPITATED.
Discussion
47.

How many moles of AgCl would be obtained, when 100 ml of 0.1 M Co(NH_(3))_(5)Cl_(3) is treated with excess of AgNO_(3) to give two moles of AgNo_(3) ?

Answer»

0.001
0.02
0.03
None of these

Solution :`underset(10" m"" mol")[[CO(NH_(3))_(5)Cl]]Cl_(2)OVERSET(AgNO_(3))tounderset(20" m"" mol")(2AgCldarr)`
Discussion
48.

How many moles of AgCl would be obtained, when Co(NH_(3) )_(5)Cl_(3) is treated with excess of AgNO_(3)?

Answer»

1 mol
2 mol
3 mol
No ppt is formed

Solution :No. of MOLE.s `=0.1 XX 0.1 = 0.01`
`[Co(NH_(3))_(5)Cl]Cl_(2) to [Co(NH_(3))_(5)Cl]^(+2) +2CL^(-)`
`underset(("EXCESS"))overset(0.01" mole")(Ag^(+))underset((0.02" mole"))overset(0.01" mole")(+Cl^(-)""to) underset((0.02" mole")) overset(0.02" mole" to)(AGCL)`
Discussion
49.

How many moles of AgCl are precipiated on the reaction of one mole of CoCl_(3).5NH_(3) with AgNO_(3) ?

Answer»

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

Answer :B::C
Discussion
50.

How many moles of AgCl will be precipitated when an excess of AgNO_(3) solution is added to one molar solution of [CrCl(H_(2)O)_(5)]Cl_(2)?

Answer»

SOLUTION :TWO MOLES of `AGCL` is PRECIPITATED.
Discussion