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.

Mn^(2+) compounds are more stable than Fe^(2+) compounds towards oxidation to their +3 state, because

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`Mn^(2+)` is more stable with high 3rd ionisation energy
`Mn^(2+)` is BIGGER in size
`Mn^(2+)` has completely filled d- orbitals
`Mn^(2+)` does not exist

SOLUTION : `Mn^(2+)` has stable HALF - filled electronic configuration thus it has high third ionisation energy while `Fe^(2)` on losing one more electron will acquire this stable electronic configuration.
Thus `Fe^(2+)` is more prone to get oxidised to +3 oxidation state .
2.

Mn^(++) can be converted into Mn^(7+) by reacting with

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`SO_(2)`
`Cl_(2)`
`PbO_(2)`
`SnCl_(2)`

SOLUTION :STRONG OXIDIZING agents such as `PbO_(2)` or sodium bismuthate `(NaBiO_(3))` oxidize `MN^(2+)` to `MnO_(4)^(-)` or `Mn^(7+)`.
3.

Mn belongs to :

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s-block
p-block
d-block
f-block

ANSWER :C
4.

M = molarity of the solution m = molality of the solution d = density of the solution (in g. ml^(-1) )M = gram molecular weight of solute Which of the following relations is correct

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`m=(M)/(1000d-MM)^(1)`
`m=(Mxx1000)/(d+MM)^(1)`
`m=(Mxx1000)/((1000xxd)-MM^(1))`
`M=(mxx1000)/((1000xxd)-MM^(1))`

ANSWER :C
5.

Mixture(s) showing positive deviations from Raoult's law at 35^(@)C is (are)

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carbon TETRACHLORIDE + methanol
carbon DISULPHIDE + acetone
benzene + toluene
phenol + aniline

Solution :`"CCl"_(4)+CH_(3)OH`= POSITIVE DEVIATION from Raoult's law
= Positive deviation from Raoult's law ltbr Benzene + Toluene = Ideal solution
= NEGATIVE deviation from Raoult's law
6.

Mixtures of solution. Calculate the pH of the following solution. K_(1)=7.5 xx 10^(-3)M, K_(2)=6.2 xx 10^(-8)M, K_(3)=1.0 xx 10^(-12)M 40 ml of 0.020 M Na_(3)PO_(4) + 40 ml of 0.040 M HCl,

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ANSWER :4.66
7.

Mixtures of solution. Calculate the pH of the following solution. K_(1)=7.5 xx 10^(-3)M, K_(2)=6.2 xx 10^(-8)M, K_(3)=1.0 xx 10^(-12)M 40 ml of 0.050 M Na_(2)CO_(3) + 50 ml of 0.040 M HCl,

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ANSWER :8.34
8.

Mixtures of solution. Calculate the pH of the following solution. K_(1)=7.5 xx 10^(-3)M, K_(2)=6.2 xx 10^(-8)M, K_(3)=1.0 xx 10^(-12)M 50 ml of 0.10 M Na_(3) PO_(4) + 50 ml of 0.10 M NaH_(2)PO_(4)

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ANSWER :9.6
9.

Mixtures of solution. Calculate the pH of the following solution. K_(1)=7.5 xx 10^(-3)M, K_(2)=6.2 xx 10^(-8)M, K_(3)=1.0 xx 10^(-12)M 40 ml of 0.010 MH_(3)PO_(4) + 40 ml of 0.10 M Na_(3)PO_(4).

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ANSWER :7.2
10.

Mixture X=0.02 mol of [Co(NH_(3))_(5)(SO_(4))]Br and 0.02 mol of [Co(NH_(3))_(5)Br]SO_(4) was prepared in 2 litre of solution 1 litre of mixture X + excess AgNO_(3)rarrY 1 litre of mixture X = excess BaCl_(2)rarrZ What will be the number of moles of Y and Z formed ?

Answer»

Solution :Only 1st complex will FORM a precipitate of AgBr and only 2nd complex will form precipitate of `BaSO_(4)`
`underset(0.02mol("in 2L"))([CO(NH_(3))_(5)SO_(4)])Br+AgNO_(3)rarr[Co(NH_(3))_(5)SO_(4)]NO_(3)+underset(0.02 mol)(AgBr(Y))`
`underset("0.02 mol"("in 2L"))([Co(NH_(3))_(5)Br])SO_(4)+BaCl_(2)rarr[Co(NH_(3))_(5)Br]Cl_(2)+underset("0.02 mol")(BaSO_(4))(Z)`
Hence, with 1 L of the solution 0.01 mol of Y and 0.01 mol of Z will be formed.
11.

Mixture X= 0.02 mole of [Co(NH_(3))_(5) SO_(4)]Br and 0.02 mol of [Co(NH_(3))_(5)Br]SO_(4) was prepared in 2 litre of solution. 1 litre of mixture X + excess AgNO_(3) to Y 1 litre of mixture X + excess of BaCl_(2) to Z Number of moles of Y and Z are

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0.01, 0.01
0.02, 0.01
0.01, 0.02
0.02, 0.02

Solution :Mixture X will CONTAIN 0.02 mol `Br^(-)`IONS and 0.02 mol `SO_(4)^(2-)` iojnsin 2 L solution. Hence 1 L of mixture X will contain 0.01 mol `Br^(-)` and 0.01 mol `SO_(4)^(2-)`ions. With excess of `AgNO_(3)` , 0.01 mol of AgBr i.e. Y is formed. With excess of `BaCl_(2)` , 0.01 mol of `BaSO_(4)`i.e. Z is formed.
12.

Mixture X of 0.02 mole of [Co(NH_(3))_(5)SO_(4)] ad 0.02mole of [Co(NH_(3))_(5)Br]SO_(4) was prepared in 2 litre of solution: 1 litre of mixture X+ excess of AgNO_(3) to Y 1 litre of mixture X+ excess of BaCl_(2) to Z. Number of moles of Y and Z respectively are:

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0.01, 0.02
0.02, 0.01
0.01, 0.01
0.02, 0.02

Answer :C
13.

Mixture 'X' containing 0.02 mole of [Co(NH_(3))_(5)SO_(4)]Br and 0.02 mole of [Co(NH_(3))_(5)Br]SO_(4) was dissolved in water to get 2 lit of solution. If 1 lit of X+ execss AgNO_(3)toY mole of ppt. and 1 lit of X+ excess BaCl_(2)toZ mole of ppt. Number of mole of Y and Z are

Answer»

0.01, 0.01
0.02, 0.01
0.01, 0.02
0.02, 0.02

Solution :1 lit. of solution CONTAINS 0.01 mole of each
`[Co(NH_(3))_(5)SO_(4)]Br and [Co(NH_(3))_(5)Br]SO_(4)`
`[Co(NH_(3))_(5)UNDERSET(0.01)(SO_(4))]Br+HgNO_(3)tounderset(0.01)(AgBr)darr+[Co(NH_(3))_(5)SO_(4)]NO_(3)`
`[Co(NH_(3))_(5)underset(0.01)Br]SO_(4)+BaCl_(2)tounderset(0.01)(BaSO_(4))darr+[Co(NH_(3))_(5)Br]Cl_(2)`
14.

Mixture X = 0.02 mol of [Co (NH_3)_5 SO_4] BrBrand0.02 mol of[Co (NH_2)_5 Br] SO_4 was prepared in 2L of solution.1 L of mixture X + excess of AgNO_3toY 1 L of mixture X + excess of BaCl_2 toZNumber of moles of Y and Z are :

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0.01 , 0.01
0.02 , 0.01
0.01 , 0.02
0.02 , 0.02

Solution :Mixture X will contain 0.02 mol of `Br^(-)` and 0.02 mol of `SO_(4)^(2-)` ions in 2L solution . ` THEREFORE` 1L of mixture X will contain 0.01 mol of `Br^(-)` and 0.01 mol of `SO_(4)^(2-)` ions with EXCESS of `AgNO_3 ,0.01` mol of AgBr i.e., Y is formed
with excess of `BaCl_2` , 0.01 mol of `BaSO_4` i.e. Z is formed .
15.

Mixtureused for the tips of match stick is

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S+K
`Sb_(2)S_(3)`
`K_(2)Cr_(2)O_(7)+S+red P`
`K_(2)Cr_(2)O_(7)+K+S`

Answer :C
16.

Mixture used on tips of matchsticks is:

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S+K
Antimony SULPHIDE
`K_2Cr_2O_7 + S + RED P`
`K_2Cr_2O_7 + K + S`

Answer :C
17.

Mixture (s) showing positive deviation from Raoult's law of 35^(@)C is (are)

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CARBON TETRACHLORIDE `+` methanol
carbon DISULPHIDE `+` acetone
benzene `+` toluene
phenol `+` aniline

Answer :A::B
18.

Mixture of two metals having mass 2 gm (A = 15, B = 30) and are bivalent and dissolve in HCl and evolve 2.24 L H_(2) at STP . What is mass of A present in mixture ?

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1 gm
1.5 gm
0.5 gm
0.75 gm

SOLUTION :`A+2HCl to Acl_(2)+Hz`
mole `=(x)/(15)""(x)/(15)`
`B+2HClto BCl_(2)+H2`
mole `=(2-x)/(30)""(2-x)/(30)`
mole of `H_(2)=(x)/(15)+(2-x)/(30) =(2.24)/(22.4)=1/10`
`(x)/(15)-(x)/(30)=1/10-1/15x=1` gm
19.

A mixture x containing 0.02 mol of [Co(NH_(3))_(5) SO_(4)]Br and 0.02 mol of [Co(NH_(3))_(5)Br]SO_(4) was prepared in 2 L of solution. 1 L of mixture X +excess AgNO_(3) rarr Y 1 L of mixture X + excess BaCl_(2) rarr Z The number of moles of Y and Z are

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`0.01, 0.01`
`0.02, 0.01`
`0.01, 0.02`
`0.02, 0.02`

Solution :Concentration of IONS in 1 litre of the solution after MIXING will be reduced to half i.e. 0.01 mol.
`underset((0.01"mol"))(Br^(-)+AgNO_(3))tounderset((0.01"mol"))(AGBR(Y))`
`underset((0.01"mol"))(SO_(4)^(2-)+BaCI_(2))tounderset((0.01"mol"))(BaSO_(4)(Z))`
20.

Mixture of volatile components A and B has total vapour pressure (in torr) : P=254-119X_(A) Where X_(A) is mole fraction of A in mixture . Hence, P_(A)^(0)& P_(B)^(0) are (in torr):

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`254, 119`
`119,254`
`135,254`
`154,119`

ANSWER :C
21.

Mixture (s) showing positive deviation from Raoult's law at 35^(@)C is (are)

Answer»


SOLUTION :BENZENE + toluene will form ideal solution.
Phenol + aniline will SHOW negative DEVIATION.
22.

Mixture of solutions. Calculate the pH of the following solutions. K_1= 7.5 xx 10^(-3) M, K_2= 6.2 xx 10^(-8)M , K_3 =1.0 xx 10^(-12)M 40 ml of 0.10 M H_3PO_4 + 40 ml of 0.25 M NaOH.

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ANSWER :12
23.

Mixture of solutions. Calculate the pH of the following solutions.K_1= 7.5 xx 10^(-3) M, K_2= 6.2 xx 10^(-8)M , K_3 =1.0 xx 10^(-12)M 50 ml of 0.12 M H_3PO_4 + 40 ml of 0.15 M NaOH ,

Answer»


ANSWER :4.66
24.

Mixture of solutions. Calculate the pH of the following solutions.K_1= 7.5 xx 10^(-3) M, K_2= 6.2 xx 10^(-8)M , K_3 =1.0 xx 10^(-12)M 40 ml of 0.12 M H_3PO_4 + 40 ml of 0.18 M NaOH ,

Answer»


ANSWER :7.2
25.

Mixture of solutions. Calculate the pH of the following solutions.K_1= 7.5 xx 10^(-3) M, K_2= 6.2 xx 10^(-8)M , K_3 =1.0 xx 10^(-12)M 50 ml of 0.12 M H3PO4 + 20 ml of 0.15 M NaOH

Answer»


ANSWER :2.12
26.

(Mixture of Sn^(+2),Cu^(+2),Pb^(+2)) on allowing H_(2)S gas gives sulphide of metal (B) this mixture a solution of on allowing ammonium sulphide gives residue (C ) and Filtrate (D ). Which are is // are correct.

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two METAL sulphide of mixture (B ) black
RESIDUE (C ) contain black COLOURED metal sulphide
Filtrate (D ) contains `( NH_(4))_(2) SnS_(3)`
Filtrate (D) Contain CuS and PbS

Solution :
27.

Mixture of O_2 and N_2O is used as:

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Fuel
Anaesthetic
In welding
Oxidising agent

Answer :B
28.

Mixture of sand and iodine can be separated by which process?

Answer»


ANSWER :C
29.

Mixture of MgCO_(3) &NaHCO_(3) on strong heating gives CO_(2) &H_(2)O in 3 : 1 mole ratio. The weight % of NaHCO_(3) present in the mixture is :

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0.3
0.8
0.4
0.5

Answer :D
30.

Mixture of MgCl_2 and MgO is called :

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PORTLAND cement
Sorel's cement
Double SALT
NONE of these

Answer :B
31.

Mixture of H- CHO and C_(6) H_(5) - CHO is treated with conc. NaOH then self redox reaction involves (1) oxidation of H - CHO (2) oxidation of C_(6) H_(5) - CHO (3)reduction of H - CHO (4) reduction of C_(6)H_(5) - CHO

Answer»

`1, 3`
`1, 2`
`1, 4`
`2, 3`

ANSWER :C
32.

Mixture of methanol and ethanol is obtained from catalytic hydrogenation of

Answer»




ANSWER :D
33.

Mixture of ethanal and propanal is subjected to aldol condensation, theproduct formed are

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1
2
3
4

Answer :D
34.

Mixture of conc. HCl and anhydrous ZnCl_2 is an important reagent which helps to distingulsh between 1^@,2^@ and 3^@ alcohols. Give one example each for 1^@,2^@ and 3^@ alcohols.

Answer»

SOLUTION :
35.

Mixture of conc. HCl and anhydrous ZnCl_2 is an important reagent which helps to distingulsh between 1^@,2^@ and 3^@ alcohols. Explain how the above reagent helps to distinguish above three types of alcohols.

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Solution :On shaking with Lucas reagent`3^@` alcohol gives immediate TURBIDITY DUE to the formation of `3^@` alky 1 halide. SECONDARY alcohol gives turbidity in about five minutes. Primary alcohol does not give turbidity at room TEMPERATURE.
36.

Mixture of chloroxylenol and terpineol acts as

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Antiseptic
ANTIPYRETIC
ANTIBIOTIC
ANALGESIC

Solution :Antiseptic (dettol).
37.

Mixture of carboxylic acids oblained by the oxidation of hexan-3-one does not contain

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Methanoic acid
Ethanoic acid
Propanoic acid
Butanoic acid

Answer :A
38.

Mixture of chloroxylenol and terpineol acts as :

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Analgestic
Antipyretic
Antiseptic
Antibiotic

Answer :C
39.

Mixture of chloroxylenol and terpineol acts as-

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analgesic
antiseptic
antipyretic
antibiotic

Answer :B
40.

Mixture of acetic acid and propionic acid is obtained by oxidation of

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pentan-2-one
butanone
pentanal
propanone

Answer :A
41.

Mixture of acetic acid and propionic acid is obtained from oxidation of

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`CH_3-COCH_3`
`CH_3-COC_2H_5`
`CH_3-CH_2-COCH_2-CH_3`
`CH_3-CH_2-CHO`

ANSWER :C
42.

Mixture of A contains

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`CO_(3)^(2-) , HCO_(3)^(Θ)` anions
`CO_(3)^(2-) , HSO_(3)^(Θ)` anions
`SO_(3)^(2-) , HSO_(3)^(Θ)` anions
None of these

Answer :b,C
43.

Mixture of 10 moles of Fe_(2)S_(3), 20 moles of H_(2)O and 30 moles of O_(2) react with 5% yield in the given reaction : Fe_(2)S_(3) + H_(2)O + O_(2) rarr Fe(OH)_(3) + S Then moles of Fe(OH)_(3) that can be produced is -

Answer»

`(10)/(3)`
`(20)/(3)`
`20`
10

Solution :`{:(Fe_(2)S_(3),+,3H_(2)O,+(3)/(2)O_(2),overset(50%)(rarr),2FE(OH)_(3),+,3S,),(10 " MOLE ",,20 " mole",,30 " moles",,,,):}`H_(2)O` is the limiting reagent.
moles of `FE(OH)_(3) = (2)/(3) xx 20 xx 0.5 = (20)/(3)` moles
44.

Mixture of acetic acid and propionic acid is obtained by oxidation of

Answer»




ANSWER :B
45.

Mixture of A contain:

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`CO_(3)^(2–) HCO_(3)^(–)` ANIONS
`CO_(3)^(2–), HSO_(3)^(–)` anions
`SO_(3)^(2–), HSO_(3)^(–)` anions
None of these

Answer :B::C
46.

Mixture is heated with dil. H_(2)SO_(4) and the lead acetate paper turns black by the evolved gases. The mixture contains

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Sulphite
Sulphide
Sulphate
Thiosulphate

Solution :It MUST be sulphide as
`PB(CH_(3)COO)_(2)+H_(2)S to underset("BLACK ppt.")(PbS)+2CH_(3)COOH`
47.

One gram equimolecular mixture of Na_(2)CO_(3) and NaHCO_(3)is reacted with 0.1 NHCl. The milliliters of 0.1 N HCl required to react completely with the above mixture is :

Answer»

15.78 mL
157.8 mL
198.4 mL
308 mL

Answer :D
48.

Mixing of two different ideal gases under isothermal reversible condition will lead to

Answer»

INCREASE of Gibbs FREE energy of the system
No change of entropy of the system
Increase of entropy of the system
Increase of ENTHALPY of the system

Solution :During MIXING, `DeltaS_(mix)` is always positive.
49.

Mixing of two different ideal gases under isothermal reversiblecondition will lead to

Answer»

INCREASE of gibbs FREE energy of the system
no change of entropy of the system
increase of entropy of the system
increase of enthalpy of the system .

Solution :During mixing , ` DELTA S_(mix)` is always POSITIVE
50.

Mixing of positively charged colloidal solution with negatively charged colloidal solution brings __________. The decreasing order of coagulating power of Na^(+), Ba^(2+) and Al^(3+)for negatively charged colloidal solution is ________ .

Answer»

mutual COAGULATION , `Na^(+) gt Ba^(2+) gt Al^(3+)`
mutual coagulation , `Al^(3+) gt Ba^(2+) gt Na^(+)`
coagulation , `Na^(+) gt Ba^(2+) gt Al^(3+)`
PEPTIZATION , `Al^(3+) gt Ba^(2+) gt Na^(+)`

Solution :According to Hardy Schulze rule, the coagulating POWER of an ion DEPENDS upon its valency . Higher the valency of ion, greater is its coagulating power.