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 manymoles of acidifiedKMn04requiredto oxidise one moleof oxalicacid ?

Answer»

5
1.5
0.6
0.4

Solution :0.6
Discussion
2.

How many moles of acetic anhydride will be required to form glucose pertaacetate from 2M of glucose?

Answer»

2
5
10
`2.5`

ANSWER :C
Discussion
3.

How many moles of acetic anhydride will be required to form glucose pentaacetate from 2 M of glucose ?

Answer»

2
5
10
`2*5`

Solution :When ecetylated with acetic ANHYDRIDE `(-O-overset(O)overset(||)(C)-CH_(3))`, glucose forms a pentaacetate. The reaction is as follows :

In the GIVEN reaction, 1 M of glucose is producing glucose pentaacetate i.e., it contains 5 OAC group. HENCE, 5 MOLES of acetic anhydride will be required to form from glucose pentaacetate from 2 M of glucose.
Discussion
4.

How many moles of acetic anhydride will be required to form glucose pentaacetate from 1 mole of glucose?

Answer»

SOLUTION :10 MOLES of ACETIC ANHYDRIDE.
Discussion
5.

How manymoleseachof Ag^(+) ionCu^(2+)ionand Fe^(3+)ionswould bedepositedby passeageof same quantityof electrictythrough solutionsof their salts?

Answer»

samenumberof moleeach
`1:1//2: 1//3`
`1//3 : 1//2 : 1` mole
`1:2:3`

Solution :Ratio DEPENDS on reciprocal of number of ELECTRONS accepted BT CATION.
Discussion
6.

How many moles electron weigh one kilogram

Answer»

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

Solution :1 MOLE of electron WEIGHT `9.108 xx 10^(-31) xx 6.023 xx 10^(23)kg`
So number of moles of electron in 1 kg is `(1)/(9.108 xx 10^(-31) xx 6.023 xx 10^(23)) = (1)/(9.108xx 6.023) xx10^(8)`
Discussion
7.

How many moles chlorine (available ) are released when 254g bleaching powder is allowed to react with CO_(2)"_____________"

Answer»


SOLUTION :`CaOCl_(2) + CO_(2) rarr CaCO_(3) + Cl_(2) , 127 rarr 1 `
`254 rarr ( 254)/( 127) =2 `2 MOLES of `Cl_(2)` are released
Discussion
8.

How many moles are there in 8.8 grams of CO_2.

Answer»


ANSWER :0.2 MOLE
Discussion
9.

How many moles are there in 1 m^(3) of any gas at NTP?

Answer»


ANSWER :44.6moles
Discussion
10.

How many moles and how many grams of sodium chloride (NaCl) are present in 250 mL of a 0.50 M NaCl solution?

Answer»


ANSWER :0.125 MOL, 7.32 G
Discussion
11.

How many moles and how many grams of sodium chloride (NaCl) are present in 250 cm^(3) of a 0.500 M NaCl solution ?

Answer»


ANSWER :0.125 MOLE, 7.312 G
Discussion
12.

How many molecules will be present in one gram molecular mass of hydrogen gas ?

Answer»


ANSWER :`6.02 XX 10^23`
Discussion
13.

How many molecules of water and oxygen atoms are present in 0.9g of water?

Answer»

SOLUTION :Gram-molecular mass of water =18g
Number of moles in 0.9g of water `=(0.9)/(18)=0.05`
Number of water molecular in one mole of water =18g
Number of moles in 0.9g of water `=(0.9)/(18)=0.05`
Number of water MOLECULES in one mole of water
`=6.02xx10^(23)`
Number of molecules of water in 0.05 moles.
`=0.05xx6.02xx10^(23)`
`=3.010xx10^(22)`
As one molecular of water contains one oxygen atom.
So, number of oxygen ATOMS in `3.010xx10^(22)` molecule of water `=3.010xx10^(22)`
Discussion
14.

How many molecules of sulphur are present in 12.8 g of sulphur (atomic mass of S = 32) ?

Answer»

`3.01 XX 10^22`
`2.408 xx 10^23`
`6.02 xx 10^23`
`0.4`

Solution :No. of atoms of sulphur in 12.8 g
` = (6.02 xx 10^23)/(32) xx 12.8`
One of mulecules of sulphur contains 8 atoms ` = (S_8)`
` THEREFORE ` No. of `S_8` molecules
` = (6.02 xx 10^23 xx 12.8)/(32 xx 8) = 3.01 xx 10^22`
Discussion
15.

How many molecules of phenyl hydrazine are used up per molecule of glucose?

Answer»

1
2
3
6

Solution :Three molecules of phenyl hydrazine are used up by molecules of glucose for the formation of glycosazone as explaied below
`{:(""CHO),("|"),(""CHOH),("|"),(" "(CHOH)_(3)),("|"),(UNDERSET(" Glucose ")(""CH_(2)OH)):} OVERSET(C_(6)H_(5)NHNH_(2))underset(-H_(2)O)to {:(""CH=N NHC_(6)H_(5)),("|"),(""CHOH),("|"),(""(CHOH)_(3)),("|"),(underset("Glucose phenyl hydrazine")(CH_(2)OH)):} overset(2C_(6)H_(5)NHNH_(2))underset(-C_(6)H_(5)NH_(2),-NH_(3),-H_(2)O)to {:(" "CH=N NHC_(6)H_(5)),("|"),(""C=N NHC_(6)H_(5)),("|"),(" "(CHOH)_(3)),("|"),(underset("Glucosazone")(""CH_(2)OH)):}`
Discussion
16.

How manymolecules of CH_(3)lwill reactwithCH_(3)NH_(2)to fromtetra methylammonium iodide?

Answer»

2
3
4
5

Answer :B
Discussion
17.

How many molecules are present in 1 kg mole of a substance?

Answer»

SOLUTION :`6.022xx10^(26)`
Discussion
18.

How many molecules are present in 1 equivalent each of hydrogen, oxygen and nitrogen?

Answer»

SOLUTION :1 equivalent of H contains 1 MOLE of H.
`therefore` 1 equivalent of H contains 1/2 mole of `H_2`.
Discussion
19.

How many molecules are present in 2.24 ml of a gas at STP ?

Answer»


ANSWER :`6.022 XX 10^21` MOLECULES
Discussion
20.

How many molecules approximately do you expect to be present in (i) a small sugar crystal which weighs 10 mg (ii) one drop of water with 0.05 cc volume ?

Answer»


Solution :(i) `"10 mg SUGAR"(C_(12)H_(22)O_(11))=0.01//342" mole"=2.92xx10^(-5)"mole"`
(ii) 0.05 cc water = 0.05 G = 0.05/18 mole = `2.78xx10^(-3)" mole"`
Discussion
21.

How many molecuels and atoms of oxygen are present in 5.6 litres of oxygen (O_(2)) at NTP.

Answer»

Solution :We KNOW that 22.4 litres of oxygen at NTP CONTAIN `6.02xx10^(23)` molecules of oxygen.
So, 5.6 litres of oxygen at NTP contain.
`=(5.6)/(22.4)xx6.02xx10^(23)` molecuels ltbr. `=1.505xx10^(23)` molecules
1 molecule of oxygen contain=2 ATOMS of oxygen.
So, `1.505xx10^(23)` molecules of oxygen contain.
`=2xx1.505xx10^(23)` atoms ltbr. `=3.01xx10^(23)` atoms.
Discussion
22.

How many mole of oxygen is required for complete combustion of 1 mole of Alkenen.

Answer»

Solution :VOLUME of HYDROCARBON `=8c.c.`: Volume of`O_(2)=40c.c`
FORMULA No. `1, (8)/(40)=(2)/(3n+)` (For alkane)
`(1)/(5)=(2)/(3n+1)` or `3n+1=10` or `3n=10-1=9 , n=3`
The value of `n` comes in whole number from `1st` formula it MEANS hydrocarbon
Discussion
23.

How many mole of HCl are required to prepare one litre of buffer solution (conataining NaCN+HCl) of ph 8.5 using 0.01 g formula weight of NaCN(K_(HCN)=4.1xx10^(-10)?

Answer»

`8.85xx10^(-3)`
`8.75xx10^(-2)`
`8.85xx10^(-4)`
`8.85xx10^(-2)`

ANSWER :A
Discussion
24.

How many mL of each of two hydrochloric acids of strengths 12 N and 3 N are to be mixed to make one litre of 6 N solution ?

Answer»

SOLUTION :`333.33 "ML " 666.67 mL`
Discussion
25.

How many mole of acetic anhydride will consumed when it reacts with compound (A)

Answer»

3
4
5
6

Answer :C
Discussion
26.

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

Answer :A
Discussion
27.

How many mL of a 0.1 M HCl are required to react completely with 1 g mixture of Na_(2)CO_(3) and NaHCO_(3) containing equimolar amounts of the two?

Answer»

Solution :Step 1. To calculate the number of moles of the components in the mixture.
Suppose `Na_(2)CO_(3)` present in the mixture = xg`""therefore""NaHCO_(3)" present in the mixture "=(1-x)g`
`"Molar mass of "Na_(2)CO_(3)=2xx23+12+3xx16="106 g mol"^(-1)`
`"Molar mass of "NaHCO_(3)=23+1+12+3xx16="84 g mol"^(-1)`
`therefore"Moles of "Na_(2)CO_(3)" in x g "=(x)/(106),"Moles of "NaHCO_(3)" in "(1-x)g=(1-x)/(84)`
As mixture contains equimolar AMOUNTS of the two.
`(x)/(106)=(1-x)/(84)"or"106-106x=84x or x=(106)/(190)g=0.558g`
Thus,`"moles of "Na_(2)CO_(3)=(0.558)/(106)=0.00526,"Moles of "NaHCO_(3)=(1-0.558)/(84)=(0.442)/(84)=0.00526`
Step 2. To calculate the moles of HCl required.
`Na_(2)CO_(3)+2HCl rarr 2NaCl+H_(2)O+CO_(2),NaHCO_(3)+HCl rarr NaCl +H_(2)O+CO_(2)`
`"1 mole of "Na_(2)CO_(3)" required HCl =2 moles"`
`therefore" 0.00526 mole of "Na_(2)CO_(3)"REQUIRES HCl "=0.00526 xx"2 moles = 0.01052 mole"`
`"1 mole of "NaHCO_(3)" requires HCl =1 mole"`
`therefore"0.00526 mole of "NaHCO_(3)" required HCl"="0.00526 mole"`
`therefore"Total HCl required "=0.01052+0.00526" moles = 0.01578 moles"`
Step 3. To calculate volume of 0.1 M HCl
`"0.1 mole of 0.1 M HCl are present in 1000 mL of HCl."`
`"0.01578 mole of 0.1 M HCl will be present in HCl "=(1000)/(0.1)xx0.01578="157.8 mL"`
Discussion
28.

How many mL of concentrated suphuric acid of sp. Gr 1.84 containing 98 % H_(2)SO_(4) solution by weight is required to prepare 200 mL of 0.5 N solution ?

Answer»

SOLUTION :`2.71` ML
Discussion
29.

How many mL of 0.5 M H_(2)SO_(4) are needed to dissolve 0.5 of Copper (II) carbonate ?

Answer»

Solution :`8.09 ML ~~ 8 mL`
`underset("1 mol")(CuCO_(3)) + underset("1 mol")(H_(2)SO_(4)) rarr CuSO_(4) + H_(2)O + CO_(2)`
Let of volume of 0.5 M `H_(2)SO_(4)` be V
Moles of `CuCO_(3)-= "Moles of" H_(2) SO_(4)`
`(0.5)/(123.5) = (0.5 xx V)/(1000)`
`therefore""V = 8.09 mL ~~ 8 mL`
Discussion
30.

How many mL of 0.3M K_(2)Cr_(2)O_(7) (acidic) is required for complete oxidation of 5 mL of 0.2 M SnC_(2)O_(4) solution.

Answer»


ANSWER :2.22 ML.
Discussion
31.

How many ml of 0.150M Na_(2)CrO_(4) will be required to oxidize 40 ml of 0.5 M Na_(2) S_(2) O_(3). CrO_(4)^(2-) + S_(2) O_(3)^(2-) rarr Cr(OH)_(4) ^(-) + SO_(4)^(2-)

Answer»

225ml
355ml
455ml
555ml

Answer :B
Discussion
32.

How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na_(2)CO_(3) and NaHCO_(3) containing equimolar amounts of both ?

Answer»

Solution :Let the amount of `Na_(2)CO_(3)` in the MIXTURE be `x_(g)`, Then, the amount of `NaHCO_(3)` in the mixture is `(1-x)g`.
MOLAR mass of `Na_(2)CO_(3)=2xx23+1xx12+3xx16`
`= 106 g mol^(-1)`
`therefore` Number of moles `Na_(2)CO_(3)=(x)/(106)mol`
Molar mass of `NaHCO_(3)=1xx23+1xx12+3xx16`
`= 84 g mol^(-1)`
Number of moles `NaHCO_(3)=(1-x)/(84)` mol
According to the question,
`(x)/(106)=(1-x)/(84)`
`therefore 84x = 106-106 x`
`therefore 190x=106`
`therefore x=0.5579`
Therefore, number of moles of `Na_(2)CO_(3)`
`=(0.5579)/(106)mol`
= 0.053 mol
And, number of moles of `NaHCO_(3)`
`= (1-0.5579)/(84)`
= 0.0053 mol
HCl reacts with `Na_(2)CO_(3)` and `NaHCO_(3)` according to the following equation.
`2HCl+Na_(2)CO_(3)to 2NaCl+H_(2)O+CO_(2)`
`HCl+NaHCO_(3)to NaCl+H_(2)O+CO_(2)`
1 mol of `Na_(2)CO_(3)` reacts with 2 mol of HCl.
Therefore, 0.0053 mol of `Na_(2)CO_(3)` reacts with `2xx0.0053` mol = 0.0106 mol of HCl.
Similarly, 1 mol of `NaHCO_(3)` reacts with 1 mol of HCl.Therefore, 0.0053 mol of `NaHCO_(3)` reacts with 0.0053 mol of HCl.
Total moles of HCl required `= (0.0106+0.0053)` mol
= 0.0159 mol
In 0.1 M of HCl
0.1 mol of HCl is preset in 1000 mL of the solution
Therefore, 0.0159 mol of HCl is present in
`= (1000xx0.0159)/(0.1)` mol
= 159 mL of the solution
Hence, 159 mL of 0.1 M of HCl is required to react completely with 1 g mixture of `Na_(2)CO_(3)` and `NaHCO_(3)` containing equimolar amounts of both.
Discussion
33.

How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na_2CO_3and NaHCO_3containing equimolar amounts of both ?

Answer»

Solution :STEP 1. To calculate the number of moles of the components in the mixture.
Suppose mass of `Na_2CO_3` PRESENT in the mixture = x g
Mass of `NaHCO_3`present in the mixture = (1 – x) g
Molar mass of `Na_2CO_3 = 2 xx 23 + 12 + 3 xx 16 = 106 g "mol"^(-1)`
Molar mass of `NaHCO_3 = 23 + 1 + 12 + 3 xx 16 = 84 g "mol"^(-1)`
Number of moles of `Na_2CO_3` in x g ` = (x)/(106)`
Number of moles of `NaHCO_3` in (1 – x) g= `(1 - x)/(84)`
As mixture CONTAINS EQUIMOLAR amounts of the two,
`(x)/(106) = (1-x)/(84) " or " 106 - 106x = 84x " or " x= (106)/(106+ 84) = 106/190 g = 0.558 g `
Number of moles of `Na_2CO_3 = (0.558)/(106) = 0.00526`
Number of moles of `NaHCO_3 = (1-0.558)/(84) = (0.442)/(84) = 0.00526`
Step 2. To calculate the number of moles of HCl required.
`Na_2CO_3 + 2HCl to 2NaCl + H_2O + CO_2`
`NaHCO_3 + HCl to NaCl + H_2O + CO_2`
`NaHCO_3 + HCl to NaCl + H_2O + CO_2`
1 mole of `Na_2CO_3` requires = 2 moles of HCI
` therefore ` 0.00526 mole of `Na_2CO_3`requires = 0.00526 x 2 moles = 0.01052 mole
1 mole of `NaHCO_3`requires = 1 mole of HCI
` therefore `0.0526 mole of `NaHCO_3` requires = 0.00526 mole
` therefore ` Total HCl required = 0.01052 + 0.00526 moles = 0.01578 moles
Step 3. To calculate volume of 0.1 M HCl.
0.1 mole of 0.1 M HCl are present in 1000 mL of HCl.
0.01578 mole of 0.1 M HCl will be present in = `(1000)/(0.01) xx 0.01578 = 157.8 mL` of HCl
Discussion
34.

How many millilitres of 6.0 M hydrochloricacid should be used to prepare 150 mL of a solution which is 0.30 M in hydrogen ion

Answer»

`3.0`
7.5
9.3
30

SOLUTION :`M_(1) = 6.0 M` of HCL , `V_(1) = ?`
`M_(2) = 0.30 M` is `H^(+)` CONCENTRATION in solution.
`V_(2) = 150 ml` of solution.
`M_(1)V_(1) = M_(2)V_(2), 6.0 xx V_(1) = 0.30 xx 150`
`V_(1) = (0.30 xx 150)/(6) = 7.5 ml`.
Discussion
35.

How manymillilitres of concentrated sulphuric acidof sp. 1.84 containing 98 %H_(2)SO_(4) by weightare required to prepare 200 mLof 0.50 N solution ?

Answer»

Solution :98 % of `H_(2)SO_(4)`by weightmeans100 g `H_(2)SO_(4)`solutioncontains98 gof `H_(2)SO_(4)`
Volume of 100g of `H_(2)SO_(4) = 100/(1.84 ) mL `
i.e `100/(1.84)` mL of `H_(2)SO_(4)` solution contains98.00g of `H_(2)SO_(4)`
Equivalents of `H_(2)SO_(4) = 98/49 = 2 ""...(EQN .4i)`
(eq. WT . of `H_(2)SO_(4)=49`)
m.e of `H_(2)SO_(4) = 2xx1000 = 2000. "" ...(Eqn .3)`
Normality of `H_(2)SO_(4)` solution = `(m.e)/("volume in mL") "" ....(Eqn.1)`
` = (2000)/(100//1.84) = 36.8 N `
Let the volume of `H_(2)SO_(4)`of normality36.8 N = 36.8 V ....(Eqn.1)
and m.e of 200 mL of `H_(2)SO_(4)` of normality `0.5 N = 0.5 xx 200`
= 100
Since both the solutions of `H_(2)SO_(4)`shouldhave the same number of m.e we have ,
` 36.8 v = 100`
` :. "" v = 100/(36.8) = 2.72 mL `
Discussion
36.

How many millilitres of 0.5 M H_(2)SO_(4) are needed to dissolve 0.5 g of copper (II) carbonate ? (At. Mass : H = 1, C = 12, O = 16, S = 32, Cu = 63.5)

Answer»

Solution :`CuCO_(3)+H_(2)SO_(4)rarr CuSO_(4)+CO_(2)+H_(2)O`
`"1 MOLE of "CuCO_(3)=63.5+12+48=123.5" G required "H_(2)SO_(4)="1 mole"`
`THEREFORE 0.5 g CuCO_(3)" will required "H_(2)SO_(4)=(1)/(123.5)xx0.5"mole"=(1)/(247)"mole"`
`"0.5 mole of 0.5 M "H_(2)SO_(4)" are present in 1000 mL"`
`therefore""(1)/(247)" mole of 0.5 M "H_(2)SO_(4)" will be present in "(1000)/(0.5)xx(1)/(247)mL=8.1mL`
Discussion
37.

How many milimoles of nitrogen dissolve when nitrogen is passed in 1 litre water at 293 K temperature ?[The value of K_(H) is 7.648xx10^(4) bar and partial pressure of N_(2) gas is 0.987 bar.]

Answer»


ANSWER :0.72 MILIMOLES
Discussion
38.

How many milligrams is 1 carat of diamond equal to ?( 1 कैरेट मे कितना मिलीग्राम हीरा होगा ?

Answer»

100 MG
150 mg
200 mg
250 mg

Solution :1 carat=200 mg
Discussion
39.

How many milimoles of CO_(2) gas will dissolve when CO_(2) gas is passed in 900 mL water at 298 K temperature ? [The value of K_(H) is 6.02xx10^(-4) bar and partial pressure of CO_(2) gas is 2xx10^(-8) bar.]

Answer»


ANSWER :1.661 MILIMOLES
Discussion
40.

How many milli gram of iron (Fe^(2+)) are equal to 1 mL of 0.1055NK_2Cr_2O_7 equivalent?

Answer»

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

Answer :A
Discussion
41.

How many methylanilines are formed when 3-methylchlorobenzene is treated with sodamide in liquid ammonia?

Answer»


SOLUTION :
Discussion
42.

How many mi Iii litre of 0.5 N SnCl_2 solution will reduce 600 ml of 0.1 N HgCl_2 to Hg_2 Cl_2

Answer»

60 ml
240 ml
120 ml
30 ml

Answer :C
Discussion
43.

How many metamers are possible for molecular formula C_5H_12O

Answer»

4
6
8
10

Answer :B
Discussion
44.

How many metamersare possiblefor molecular formula C_(4),H_(11),N.

Answer»

0
2
3
4

Solution :Theremetamers are pointpossibleforformula`C_(4) H_(11)N`.
1. `CH_(3) - NH - CH_(2)- CH_(2)- CH_(3)`
2. `CH_(3) - NH - CH(CH_(3))_(2)`
3. `C_(2)H_(5) - NH- C_(2)H_(4)`
Discussion
45.

How many metamers are possible for molecular formula C_4H_10O ?

Answer»

3
7
5
2

Answer :A
Discussion
46.

How many metamers are possible for C_4H_10O ?

Answer»

1
2
3
4

Answer :C
Discussion
47.

How many metamers are possible for 2^(0) amines of formula C_(5)H_(13)N?

Answer»

5
6
7
8

Answer :B
Discussion
48.

How many metals are commercially extracted by electrometallurgy from the given metals? Al, Mg, Na, K, A, Hg. Ti, Th, Zt, B

Answer»


Solution :High electroposotive METAL like Al, Mg, Na & K can be commercially extracted from ELECTROMETALLURGY process.
Discussion
49.

How many metals are commercially extracted by hydro metallurgy from the given metals Ag, Mn, In, Cr, Pb, Au

Answer»


SOLUTION :AGE and Cucan be COMMERCIALLY EXTRACTED by hyrometallurgy
Discussion
50.

How many maximum number of atoms are present in single plane of Al(CH_(3))_(3) molecule.

Answer»

7
4
10
6

Solution : When CUT HORIZONTALLY
Discussion