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.

On addition of an inert gas at constant volume to the reaction : N_2 + 3H_2 hArr 2NH_3 at equilibrium:

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The REACTION halts
forward reaction is favoured
The reaction REMAINS unaffected
Backward reaction is favoured

Answer :C
2.

On addition of ammonium chloride to a solution of ammonium hydroxide

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Dissociation of `NH_(4)OH` increases
Concentrationof `OH^(-)` increases
Concentration of `OH^(-)` decreases
Concentration of `NH_(4)^(+)` and `oH^(-)` increases

Solution :DUE to common ION effect.
3.

On addition of AgNO_3 to NaCl, white ppt. occurs

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INSTANTANEOUSLY
with a measurable SPEED
SLOWLY
very SLOW

Solution :Ionic REACTION are instantaneous one.
4.

On addition of ammonium chloride to a solution of ammonium hydroxide ……….

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DISSOCIATION of `NH_4OH` increases
concentration of OH increases
concentration of `OH^-` decreases
concentration of `NH_4^+ and OH^-` increases

Solution :Due to common ion EFFECT
5.

On addition of AgNO_3 to NaCl, white ppt, occurs:

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Instantaneously
With a MEASURABLE speed
Slowly
None

Answer :A
6.

On addition of 1 ml of solution of 10% NaCl to 100ml corporate gold sol in presence of 0.25g of starch, the coagulation is just prevented. The gold number of starch is

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`0.025`
`0.25`
2.5
25

Answer :D
7.

On addition of 1 ml solution of 10% NaCl to 10 mL gold sol in the presence of 0.025 g of starch, the coagulation is just prevented. Starch has the gold number

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0.025
0.25
0.5
25

Solution :By definition, GOLD NUMBER of starch is the amount of starch in mg added to 10 mL STANDARD gold sol which PREVENTS the COAGULATION of gold on adding 1 mL of 10% NaCl solution. The amount of starch is 0.025 g = 25 mg. Hence, gold number is 25.
8.

On addition of 2mL of 5% NaCl to 10 mL gold sol in the presence of 0.025 g starch , the coagulation is just prevented. What is the gold number of starch ?

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Solution : 0.025 g of STARCH just prevent the coagulation of 10mL GOLD sol on the addition of 2mL of 5% NaCl or 0.025 g of starch just prevents the coagulation of 10mL gold sol on the addition of 1ML of 10% NaCl. Milligrams of starch which just prevent the coagulation of 10mL gold sol. on the addition of 1 mL 10% NaCl = 25
`:.` Gold number of starch = 25 .
9.

On additing (A) into a dilute solution of silver nitrate a white precipitate appears which quickly changes into a black coloured compound. (C)

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Answer :The REACTIONS indicate that the COMPOUND (A) is sodium THIOSULPHATE. It is formed in step (i) by passing gas (B) which is either `I_(2)`.
`2AgNO_(3)+ underset("(Whitw ppt)")(Na_(2)S_(2)O_(3) rarr) Ag_(2)S_(2)O_(3)+2NaNO_(3)`
`Ag_(2)S_(2)O_(3)+underset((C)" black")(H_(2)Orarr) Ag_(2)S+H_(2)SO_(4)`
10.

On adding two or three drops of FeCl_(3) into the excess of solution (A) a violet coloured compound (D) is formed. This colour disapperars quickly.

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ANSWER :The reactions indicate that the compound (A) is sodium thiosulphate. It is formed in step (i) by passing gas (B) which is either `I_(2)`.
`S_(2)O_(3)^(2-)+Fe^(3+) underset((D)" violet")(RARR) [Fe^(3+)(S_(2)O_(3))_(2)]^(-)`
`[Fe(S_(2)O_(3))_(2)]^(-)+Fe^(3+) rarr 2 Fe^(2+)+S_(4)O_(6)^(2-)`
11.

On adding which of the following the pH of 20 mL of 0.1 N HCl will not alter ?

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1 mL of 1 N HCl
20 mL of DISTILLED WATER
1 mL of 0.1 N NaOH
500 mL of HCl of pH = 1.

Solution :pH of 20 mL of `0.1 N HCl= -log [H^(+)]`
`=-log (0.1)=1`
The given solution has pH = 1
`therefore` When 500 mL of HCl of pH = 1 is added to the given solution, pH REMAINS same.
12.

On adding which of the following the pH of 20 ml of 0.1 NaOH will not alter

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1 mL of 1N HCl
20 mL of DISTILLED water
1 mL of 0.1 N NaOH
500 mL of HCl of PH = 1

Answer :D
13.

On adding solid pottasium cyanide to water :

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PH will increase
pH will decrease
pH will not change
Electrical CONDUCTANCE will not change

Answer :A
14.

On adding sold potassium cyanide to water

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PH will increase
pH will decrease
pH will not change
Electrical conducting will not change

SOLUTION :`CN^(-) + H_(2)O rarr HCN + OH^(-)`
Because `OH^(-)` concentration is increased.
15.

On adding NaOH to ammonium sulphate, a colourless gas with pungent odour is evolved which forms a blue coloured complex with Cu^(2+) ion. Identify the gas.

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Solution :Ammonium sulphate reacts with NaOH to evolve `NH_(3)` gas which has pungent smell and forms blue coloured complex with `Cu^(2+)` ions as shown below :
`{:(UNDERSET("Amm. sulphate")((NH_(4))_(2)SO_(4)) + 2 NaOH RARR underset("Ammonia")(2NH_(3))uarr + Na_(2)SO_(4) + 2H_(2)O),(underset("Copper (II) ion")(Cu^(2+)(AQ.))+4NH_(3)(aq.) rarr underset(("Blue coloured complex"))underset("Tetraamminecopper (II) ion")([Cu(NH_(3))_(4)]^(2+)(aq.))):}`
16.

On adding NaOH solution to the aqueous solution of K_(2)Cr_(2)O_(7), the colour of the solution changes from :

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orange to yellow
yellow to orange
orange to red
yellow to PINK.

SOLUTION :Colour CHANGE fromorange to yellow.
17.

On adding Na_(2)S to sodium nitro prusside solution

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`Na_(4)[FE(CN)_(5)NOS]` complex is FORMED
`[Fe(CN)_(5)NOS]^(-4)` complex is formed
a violet colour is formed
All of the above

Solution :`Na_(2)S+Na_(2)[Fe(CN)_(3)NO] to Na_(4)underset("Violet")([Fe(CN)_(5)NOS])`
18.

On adding of conc. H_(2)SO_(4)ot a chloride salt colourless fumes are evolved but in case of iodide salt. Violet fumes come out. This is because

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HIis of violet colour
HI gets oxidised to `I_(2)`
HI changes to `HIO_(3)`
`H_(2)SO_(4)` reducesHI to `I_(2)`

Solution :HI gets oxidised to `I_(2).2I^(-)+H_(2)SO_(4) to SO_(4)^(2o+)+2HI,`
`H_(2)SO_(4) to H_(2)O+SO_(2)+(O), 2HI+O to H_(2)O +I_(2)`
19.

On adding Kl solution to a solution of CuSO_4, a white precipitate of ….. Is formed .

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POTASSIUM SULPHATE
Cuprous sulphate
Cuprous IODIDE
CUPRIC iodide

Answer :C
20.

On adding FeCl_(3) to the acidic Lassaigne solution, a blood red colouration is obtained. It indicates the presence of :

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S
N
N and S
N and halogen

Answer :C
21.

On adding excess of ammonium hydroxide to a copper chloride solution

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A DEEP blue solution is obtained
No change is observed
Blue PRECIPITATE of COPPER hydroxide is obtained
Black precipitate of copper oxide is obtained

Solution :On addingexcess of ammonium hydroxide to a copperchloride solution a deep blue solution of `[Cu(NH_(3))_(4)]^(2+)` ion is FORMED.
22.

On adding excess of ammonium hydroxide to a copper chloride solution :

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BLUE precipitate of COPPER hydroxide hydroxide is obtained
BLACK precipitate of copper oxide is obtained
A DEEP blue solution is obtained
No change is observed

Answer :C
23.

On adding conc. H_2SO_4to a salt, brown colouration was observed on the walls of the test tube, On heating, violet coloured vapours were evolved. The brown colouration is due to the formation of

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`l_(2)`(s)
L
`l_(3)^(-)`
Hl

Answer :C
24.

On adding ammonium hydoxide solution to Al_(2)(SO_(4))_(3) (aq):

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A PERCIPITATE is formed which does not DISSOLVE in EXCESS of ammonium hydoxide
A percipitate is formed which DISSOLVES in excess of ammonia solution
No precipitate is formed
None of these

Solution :`Al_(2)(SO_(4))_(3)+6NH_(4)OH to 2Al(OH)_(3)+3(NH_(4))_(2)SO_(4)`
`Al(OH)_(3)+NaOH to Na^(+)[Al(OH)_(4)]^(-)` sulublecomplex it is insoluble in `NH_(4)OH`
25.

On adding ammonium hydroxide solution to aqueous solution of Al_2(SO_4)_3 :

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A precipitate is FORMED which DISSOLVES in EXCESS of ammonia solution
A precipitate is formed which does not dissolve in excess of ammonium hydroxide
No precipitate is formed
None of these

Answer :B
26.

On adding AgNO_(3) solution to a solution of [Pt(NH_(3))_(3)Cl_(3)]Cl. The percentage of total chloride ion precipitated is–

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100
75
50
25

Solution :% of CHLORINE`=("ionised CL")/("TOTAL Cl") XX 100`
27.

On adding AgNO_(3) solution to KI solution, a negatively charged colloidal sol will be formed in which of the following conditions?

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100 mL of 0.1M `AgNO_(3) + 100 mL`of 0.1 M kl
100 MLOF 0.1 M `AgNO_(3)` + 50 mL of 0.2 m KI
100 mL of 0.2 M `AgNO_(3)` + 100 mL of 0.1 M KI
100 mL of 0.1 M `AgNO_(3)` + 100 mL of 0.15 M KI

Answer :D
28.

On adding ammonium chloride to a solution ammonium hydroxide

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DISSOCIATION of `NH_4OH` increases
Concentration of OH increases
Concentration of OH decreases
Concentration of OH REMAINS unchanged

Answer :C
29.

On adding a solution of (A) into the solution of cupric chloride, a white precipitate is first formed which dis- solves on adding excess of (A) forming a compound (E). Identify (A) to (E) and give chemical equations for the reactions at steps (i) to (iv).

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Answer :The REACTIONS indicate that the compound (A) is sodium thiosulphate. It is FORMED in step (i) by passing gas (B) which is either `I_(2)`.
`2Cu^(2+)+underset("white ppt.")(3S_(2)O_(3)^(2-) rarr) Cu_(2)S_(2)O_(3) darr +S_(4)O_(6)^(2-)`
`underset("White ppt.")(3Cu_(2)S_(2)O_(3)) darr +underset("(excess)")(2Na_(2)S_(2)O_(3)) rarr underset("(E) soluble complex")(Na_(4)[Cu_(6)(S_(2)O_(3))_(5)])`
30.

On adding AgNO_(3) solution to KI solution, a negatively charged colloidal sol will be formed in which of the following conditions ?

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`100ML " of " 0.1M AgNO_3 + 100ml " of " 0.1M KI `
`100ml " of " 0.1M AgNO_3 + 100ml " of " 0.2M KI `
`100ml " of " 0.2M AgNO_3 + 100ml " of " 0.1M KI `
`100ml " of " 0.15M AgNO_3 + 100ml " of " 0.15M KI `

SOLUTION :`AgNO_3 +underset(("EXCESS"))(KIto) -ve` SOL
31.

On adding a solute, vapourpressure increases .(true/false)

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SOLUTION :DECREASES
32.

On adding a solute, vapour pressure _____ .

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SOLUTION :DECREASES .
33.

On adding a solute, osmotic pressure _____ .

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SOLUTION :INCREASES .
34.

On increasing concentration of a solution, osmotic pressure _____ .

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SOLUTION :INCREASES .
35.

On adding a solute, freezing point of solution _____ .

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SOLUTION :DECREASES
36.

On adding a solute,boiling point of solution _____ .

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SOLUTION :ELEVATES (i.e.increase)
37.

On adding a few drops of dil HCl to a freshly precipitated ferric hydroxide, a red coloured colloidal sol 9 obtained. This phenomenon is called:

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PEPTISATION
Dialysis
Protective
Dissolution.

Answer :A
38.

On adding 0.1 M solution each of [Ag^(+)], [Ba^(2+)], [Ca^(2+)] in a Na_(2)SO_(4) solution, species first precipitated is [K_(sp) BaSO_(4) = 10^(-11), K_(sp)CaSO_(4) = 10^(-6), K_(sp)Ag_(2)SO_(4) = 10^(-5)] Least solubility is of BaSO_(4), hence it will precipitate first.

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`Ag_(2)SO_(4)`
`BaSO_(4)`
`CaSO_(4)`
All of these

Solution :SOLUBILITY of `BaSO_(4)(x) = sqrt(K_(sp)) = sqrt(10^(-11))`
`= 3.16 xx 10^(-6)` mol/L.
Solubility of `CaSO_(4)(x) = sqrt(K_(sp)) = sqrt(10^(-6))`
`= 1.0 xx 10^(-3)` mol/L.
Solubility of `Ag_(2)SO_(4) = 3sqrt((K_(sp))/(4))`
(`because` for `Ag_(2)SO_(4), 4x^(3) = K_(sp)`) `= 3sqrt((10^(-5))/(4)) = 1 xx 10^(-2)` mol/L
39.

On adding 20 ml of 0.1 N NaOH solution to 10 ml of 0.1 HCl, the resulting solution will

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have no EFFECT on red or blue LITMUS paper
turn blue litmus red
turn methgyl orange red
turn PHENOPHTHALEIN SOLUTION pink

Answer :D
40.

On addiction of 1ml solution of 10% NaCl to 10ml gold sol in the presence of 0.0250gm mof starch, the coagulation is just prevented Starch has the following gold number

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0.025
0.25
2.5
25

Answer :D
41.

On acid hydrolysis, propane nitrile gives

Answer»

propanal
acetic ACID
PROPIONAMIDE
PROPANOIC acid

ANSWER :D
42.

On a week end, Shubham went on a picnic with this his parents. There was a beautiful view of a lake but suddenly, Shubham saw some fish floating on the surface of water of the lake as they had died. Shubham asked his parents why these fish had died. They told him that fish also need oxygen for their survival as we do. Dissolved oxygen in the water gets depleted due to discharge of human sewage and organic wastes of the industries into the lake water. After reading the above paragraph, answer the following questions: (a) What lesson do you learn from the explanation given by Shubham's parents to him?

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Solution :(a) We should not THROW HUMAN sewage and ORGANIC wastes of the INDUSTRIES into lakes.
43.

On absorbing light of wavelength 3800 A, bromine molecule undergoes dissociation and form atoms. The kinetic energy of one bromine atom assuming that one quantum of radiation is absorbed by each molecule would be (Bond energy of Br_2=190kJ/mol)

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`1.04`x `10^(-19) J`
`2.08`x`10^(-19) J`
`1.25` x `10^(-5) J`
`6.25` x`10^(-1) J`

ANSWER :A
44.

On a week end, Shubham went on a picnic with this his parents. There was a beautiful view of a lake but suddenly, Shubham saw some fish floating on the surface of water of the lake as they had died. Shubham asked his parents why these fish had died. They told him that fish also need oxygen for their survival as we do. Dissolved oxygen in the water gets depleted due to discharge of human sewage and organic wastes of the industries into the lake water. After reading the above paragraph, answer the following questions: (b) Hoe dissolved oxygen gets depleted due to presence of the organic waste?

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SOLUTION :(b) MICROORGANISMS which decompose the organic WASTE need oxygen. The AMOUNT of oxygen consumed is proportional t the amount of waste present. THUS, sufficient dissolved oxygen no longer remains available for the survival of the fish .
45.

On a give condition, the equilibrium concentration of HI, H_(2) and I_(2) are 0.80,1.10 mole/litre. The equlibrium constant for the reaction H_(2)+I_(2)hArr 2HI will be

Answer»

64
12
8
`0.8`

SOLUTION :`H_(2)+I_(2)hArr2HI,[HI]=0.08,[H_(2)]=0.10,[I_(2)]=0.10`
`K_(c)=([HI]^(2))/([H_(2)][I_(2)])=(0.08xx0.80)/(0.10xx0.10)=64`
46.

On a certain hill station, pure water is found to boil at 95^(@)C. How many grams of NaCl must be added to 2 kg of water so that it boils at 100^(@)C?

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SOLUTION :`DeltaT_(b)=100-95=5^(@),DeltaT_(b)=iK_(b)m"or"m=(DeltaT_(b))/(iK_(b))=(5)/(2xx0.52)=4.807`
`"1000 G should CONTAIN NaCl = 4.807 moles "=4.807xx58.5g`
`therefore"2000 g should contain NaCl "=2xx4.807xx58.7=562.42g.`
47.

On a certain day the vapour pressure is 14 mm of Hg for water vapour in air at 20^@C . The saturated vapour pressure is 17.5 mm. How many moles of water vapour per litre of air would be required to saturate air at this temperature ?

Answer»

`9.41 XX 10^(-4) `
`9.14 xx 10^(-4)`
`1.914 xx 10^(-4) `
`4.19 xx 10^(-4)`

ANSWER :C
48.

On 100% dissociation of the compound K_(4)Fe(CN)_(6), the total number of ions produced will be

Answer»


SOLUTION :`K_(4)[FE(CN)_(6)] implies4K^(+)+[Fe(CN)_(6)]^(4-)`
49.

ompound (X) of molecular formula C_5H_8Odoes not react appreciably with Lucasreagent at room temperature but gives a precipitate with amm. silver nitrate. With excess of MeMgBr, 0.42 g of (X) gives 224 mL of CH_4at STP. Treatment of (X) with H_2in the presence of Pt catalyst followed by boiling with excess HI, gives n-pentane. Suggest the structure for (X).

Answer»

SOLUTION :`HC -= "CCH"_2 CH_2CH_2OH`
50.

Omeprazole, rabeprazole, iso flurane, ranitidine, cemitidine.

Answer»

Solution : Isoflurane. It is an ANAESTHETIC WHEREAS OTHERS are ANTACIDS.