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.

In an experiment,4 g of M_(2)O_(x) oxide was reduced to 2.8 g of the metal. If the atomic mass of the metal is 56gmol^(-1), the number of O atoms oxide is

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1
2
3
4

Solution :1 mole of `M_(2)O_(x)` on REDUCTION gives 2 g atoms of M
i.e. `(2xx56+16x)g M_(2)O_(x)" gives M"=2xx56g`
`= 112g`
`therefore 4g M_(2)O_(3)" will GIVE"=(112)/(112+16x)xx6g`
`therefore""(112xx4)/(112+16x)=28`
i.e., `448=2.87(112+16x)`
or `112+16x=160"or"x=3.`
2.

In an ________external source of voltage is used to bring above a chemical change.

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SOLUTION :electrolytical CELL
3.

In an experiment to measure the charge on an electron, the following values ofcharge were found on oil droplets: -1.6 xx 10^(-15), -2.4 xx 10^(-19),-4.0 xx 10^(-19)(in coulomb). What values of electronic charge would be indicated by these results?

Answer»

Solution :FIND the largest common factor.
`-0.8 XX 10^(-19)C `
4.

In an experiment, the following four gases were produced. 11.2L of which two gases at STP will weigh 14g?

Answer»


ANSWER :C,d
5.

In an experiment of 0.04F was passed through 400 ml of 1 M solution of NaCl. What would be the pH of the solution after the electrolysis?

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8
10
13
6

SOLUTION :`NACL+H_(2)Ooverset("Electrolysis")toNaOH+underset("At anode")((1)/(2)H_(2))+underset("At cathode")((1)/(2)Cl_(2))`
400 ml of 1 M NaCl solution contains NaCl
`=(1)/(1000)xx100mol=0.4mol`
As `Na^(+)+e^(-)toNa overset(H_(2)O)toNaOH+(1)/(2)H_(2)`
1 F produces 1 MOL of NaOH
`therefore0.04F` produces NaOH=0.04 mol
Thus, 400 ml of the solution now contain 0.04 mol of NaOH
`therefore`Molar concentration of NaOH SOL.
`=(0.04)/(400)xx1000=0.1M`
`[OH^(-)]=0.1M[H^(+)]=10^(-13)MthereforepH=13`
6.

In an experiment during the analysis of a carbon compound , 145 I of H_(2) was collected at 760 mm Hg pressure and 27^(@)C temperature. The mass of H_(2) is nearly

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10g
12g
24g
6g

Solution :`PV = nRT, N=(PV)/(RT)=(1xx145)/(0.082 XX 300)=5.8 ~~6` MOLE
7.

In an experiment addition of 10ml of 0.05 M BaCl_2 to 20ml of As_2S_3 sol causes coagulation in 1 hr. If the coagulation value of Ba^(+2) ion is x^2 then x = ___?

Answer»


Solution :`m = (n xx 1000)/(V) , n = (0.05 xx 10)/(1000) = 5 xx 10^(-4)`
`20 rarr 5 xx 10^(-4) , 1000 to ?`
`0.025 " mole"` = 25 milimole = `X^2 , x = 5`
8.

In an experiment addition of 4.0 mL of 0.005 M BaCl_(2) to 16.0 mL of arsenius sulphide sol just causes complete coagulation in 2 hrs. The flocculating value of the effertive ion is

Answer»

`CI^(-)` 1.0
`Cl^(-)`, 2.0
`Ba^(2+), 1.0`
`Ba^(2+),0.5`

SOLUTION :`As_(2)S_(3)` sol is negatively charged owing to preferential adsorption of `S^(2-)` ions. Cation would be effective in causing COAGULATION.
Flocculating value = minimum millimoles of the effective ion PER litre of sol `= (4 xx 0.005)/((4+16)xx 10^(-3))`
9.

In an experiment, addition of 4.0 mL of 0.0005 MBaCI_(2)to 16.0 mL of arseniussol justcauses complete coagulationin 2hrs. The flocculatingvalueof the effective ion is

Answer»

`CI^(-), 1.0`
`CI^(-).2.0`
`Ba^(2+), 1.0`
`Ba^(2+), 0.5 `

Solution :`As2_(2)S_(3)`SOL is negativelycharged OWING to preferentialadsorptionof ` S^(2-)` ions. Cation would be effective in causingcoagulation.
Flocculating value = minimum millimolesof the effectiveion perlitre of sol `= (4 xx 0.005)/( (4+16) xx 10^(-3)) = 1.0`
10.

In an experiment, 500ml of 0.5M hydrated oxalic acid is shaken with 5g of activated charocal and filtered. The conc. of filtrate is 0.4 M. If the extent of adsorption (x/m) is 1.26 xx 10^(-x) then x = __?

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Solution :`0.5 = (WT)/126 cdot 1000/500 , wt = 31.5 , 0.4 = (wt)/(126) cdot (1000)/(500) = 25.2`
adsorbed `31.5 = 25.2 5 GR to 6.3 1 to ? 1.26`.
11.

In an experiment, 2.4 g of iron oxide on reduction with hydrogen yield 1.68 g of iron. In another experiment. 2.9 g of iron oxide give 2.03 g of iron on reduction with hydrogen show that the above data illustrate the law of constant proportions.

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Solution :In the first experiment
The mass of IRON OXIDE =2.4g
The mass of iron after reduction =1.68g
The mass of oxygen =Mass of iron oxide-Mass of iron
`=(2.4-1.68)=0.72g`
RATIO of oxygen and iron `=0.72:1.68=1:2.33`
In the second experiment
The mass of iron oxide =2.9g
The mass of iron after reduction =2.03 g
The mass of oxygen `=(2.9-2.03)=0.87g`
Ratio of oxygen and iron =`0.87:2.03=1:2.33`
Thus, the DATA illustrate the law of constant proportions, as in both the experiments the ratio of oxygen and iron is the same.
12.

In an experiment, 35 mL of a gas was collected at 23^@ C and 720 mm pressure. Find its volume at NTP.

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ANSWER :240.3 MM
13.

In an experiment, 2.4 g of iron oxide on iron. In another experiment, 2.9 g of iron oxide gave 2.09 g of iron on reduction. Which law is ilustrated from the above data?

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LAW of CONSTANT proportions
Law of MULTIPLE proportions
Law of CONSERVATION of mass
Law of RECIPROCAL proportions

Answer :A
14.

Inan experiment , 1804 gof mannitolweredissolvedin 100 g of water . The vapour pressureof waterwaslowered by 0.309 mm Hgfrom 17.535 mm Hg. Calculate the molarmass ofmannitol .

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Solution :Given : Mass of a solute (MANNITOL) =`W_(2) =18.04 g`
Mass of a solvent(water) `=W_(1) =100g`
Vapourpressureof a solvent (water) `=P_(o)=17.535 mm HG`
Lowering ofvapourpressure `=Delta P= 0.309 g mm Hg`
Molarmass ofof `H_(2) O = M_(1) =18 g mol^(-1)`
Molarmass ofsolute (mannitol) = ?
`(P_(o)-P)/(P_(o)) = (W_(2) XX M_(1))/(W_(1) xx M_(2))`
`:' P_(o)- P= DeltaP`
`(DeltaP)/(P_(o)) = (W_(2) xx M_(1))/(W_(2) xx M_(2))`
`:. M_(2) = (P_(o))/(DeltaP) xx (W_(2) xx M_(1))/(W_(1))`
`= (17. 535 xx 18.04 xx 18)/(0.309 xx 100)`
`= 184 . 3 g "mol"^(-1)`
15.

In an experiment, 10 litre of air at 1 atmospheric pressure and 27^(@) Cwerepassed through an alkaline Kl Solution : At the end , the iodine entrapped in a solution reacted with 1.5 ml of 0.01 NNa_(2)S_(2)O_(3) Solution: Calculate the %volume of O_(3) in the sample .

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SOLUTION :N/A
16.

In an experiment , 1.0g CaCO_(3) on heating evolved 224 mL of CO_(2) at NTP. What mass of CaO (calcium oxide) is formed?

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Solution :Mass of 224 ML of `CO_(2)=(44)/(22400)xx224=0.44g`
17.

In an experiment, 0.04 F was passed through 400 mL of a 1M solution of NaCl. What would be the pH of the solution after the electrolysis ?

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8
10
13
6

Solution :Electolysis of aq. NaCl gives hydrogen gas at cathode and oxygen gas at anode, the electrolyte solution contains NaOH after electrolysis.
NUMBER of equivalents of NaOH formed = 0.04
NORMALITY, `N = (0.04 XX 1000)/(400) = 0.1`
`[HO^(-)] = 0.1 M`
`pOH = "" :. pH = 13`
18.

In an exothermic reaction, high yield of the product is obtained at

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High TEMPERATURE
LOW temperature
Low concentration
NONE of these

Answer :B
19.

In an exothermic reaction ( H_(r) = enthalpy of reactants and H_(p) = enthalpy of products )

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<P>`H_(r) LT H_(p)`
`H_(r) gt H_(p)`
`H_(r) = H_(p)`
`H_(r) ne H_(p)` and `H_(p) = 0`

Answer :B
20.

In an exothermic reaction DeltaH is

Answer»

POSITIVE
Negative
Zero
Both positive or negative

Solution :`DeltaH=-ve` in EXOTHERMIC REACTION.
21.

In an estimation of sulphur by Carius method, 0.160g of the substance gave 0.466g of barium sulphate. The percentage of sulphur in the compound is[At mass of Ba=137, S=32, O=16] :

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`25.0%`
`4.0%`
`54.0%`
`40.0%`

SOLUTION :`underset(233)(BaSO_(4))-=underset(32)(S)`
`233g` of `BaSO_(4)` CONTAIN SULPHUR `=32G`
`0.466g` of `BaSO_(4)` contain sulphur
`=(32)/(233)xx0.466=0.064`
`%` Sulphur `=(0.064)/(0.160)xx100=40%`
22.

In an equilibrium reaction for which DeltaG=0, the equlibrium constant K =

Answer»

<P>0
1
2
10

Solution :`If DeltaG^(@)=0`
`logK_(p)=0""(becauselog1=0)`
`K_(p)=1.`
23.

In an equilibrium reaction if DeltaG^o = 0 the equilibrium constant, K should be equal to:

Answer»

ZERO
1
2
10

Answer :B
24.

In an endothermic reaction, the value of DeltaH is

Answer»

zero
POSITIVE
negative
constant

Solution :For endothermic reactions standard heat of reaction `(DeltaH)` is positive because in these reactions total ENERGY of reactants is lower than that of PRODUCTS , i.e., `E_(R) lt E_(P)`
So,`DeltaH=E_(P)-E_(R)=+ve`
25.

In an electrolytic cell , one litre of a 1 M aqueous solution of MnO_(4)^(-) is reduced at the cathode . The quantity of electricity required so that the final solution of 0.1 MnO_(4)^(2-) will be

Answer»

10 F
1 F
0.1 F
100 F

SOLUTION :It is a fact
26.

In an electrolytic cell the anode and cathode are respectively represented as :

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POSITIVE ELECTRODE , NEGATIVE electrode
Negative electrode ,positive electrode
Positive and negative electrode both
None

Answer :A
27.

In an electrolytic cell of Ag|AgNO_3|AgNO_3|Ag,when current is passed the concetration of AgNO_3

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Increases
Decreases
Remains same
None

Answer :C
28.

In an electrolysis process, 10 electrons could deposit x kg of a univalent metal M. The atomic weight of M is (N is Avogadro constant.)

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`X XXN`
`x xxNxx10^(3)`
`x xxNxx10^(2)`
`x xx10^(3)`

Answer :C
29.

In an electrolytic cell current flows :

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From CATHODE to ANODE in OUTER circuit
From anode to cathode OUTSIDE the cell
From cathode to anode INSIDE the cell
None

Answer :A
30.

In an electrolysis experiment, current was passed for 5 hour through two cells connected in series. The first cell contains a solution of gold and second contains CuSO_(4) solutions. 9.85 g of gold was deposited in the first cell. If the oxidation number of gold is +3, find the amount of Cu deposited on the cathode of the second cell. Aso calculate the magnitude of the current in ampere. (If = 96500 coulomb) (Au = 197, Zn = 65.4)

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SOLUTION :4.765 G, 0.8037 AMP
31.

In an electrolchemical cell

Answer»

POTENTIAL ENERGY changes into KINETIC energy
Kinetic energy changes into potential energy
Chemical energy changes into electrical energy
Electrical energy changes into chemical energy

Solution :In the electrochemical CELL chemical energy changes into electrical energy.
32.

In an electrode reaction , no electrode potential is developed when

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solution pressure is LESS than osmotic pressure
solution pressure is GREATER than osmotic pressure
solution pressure EQUALS osmotic pressure
both 'a' and 'b'

Solution :It is called NULL electrode .
33.

In an electrochemical process, a salt bridge is used

Answer»

to MAINTAIN electroneutrally is each solution
to complete the circuit so that the current can FLOW
as an oxidizing agent
as a COLOUR indicator

ANSWER :A,B
34.

In an electrochemical cell, the wrong statement is

Answer»

electrons MOVE from cathode to anode.
anode is negative charged
cathode is positive charged
chemical ENERGY is CONVERTED to electrical energy.

Answer :A
35.

In an electrochemical cell, if E is the e.m.f. of the cell involving n mole of electrons, then Delta G^(@) is :

Answer»

`Delta G^(THETA) = nFE^(theta)`
`Delta G^(theta) = -nFE^(theta)`
`E^(theta) = nFDelta G^(theta)`
`Delta G^(theta) = (nF)/(E^(theta))`

ANSWER :B
36.

In an electrochemical cell ,the electrons flow:

Answer»

From CATHODE to ANODE
From anode to cathode
From anode to SOLUTION
From solution to cathode

ANSWER :B
37.

In an electrical field , the particles of a colloidal system move towardscathode . The coagulation of the same sol is studied using K_2SO_4 (i) , Na_3PO_4 (ii) , K_4[Fe(CN)_6] (iii) and NaCl (iv) . Their coagulation power should be ...............

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`II GT I gt IV gt III`
`III gt II gt I gt IV `
`I gt II gt III gt IV `
none of these

SOLUTION :`III gt II gt I gt IV `
38.

In an electrical field, the particles of a colloidal system move towards cathode. The coagulation of the same sol is studied using K_2SO_4 (i), Na_3 PO_4 (ii), K_4[Fe(CN)_6] (iii) and NaCl (iv) Their coagulating power should be

Answer»

II GT I gt IV gt III
III gt II gt I gt IV
I gt II gt III gt IV
none of these

Answer :B
39.

In an attempt to prepare ferric hydroxide sol by adding small amount of ferric chloride to water, one person got a precipitate of ferric hydroxide. How can you help him to convert Fe(OH)_3 precipitate to Fe(OH)_3 sol?

Answer»


Answer :By adding ferric chloride SOLUTION to the ferric HYDROXIDE precipitate, sol of `FE(OH)_3` will be OBTAINED.
40.

in an atomic orbital the sign of lobes indicate the

Answer»

SIGN of PROBABILITY distribution
sign of charge
sign of WAVE function
presence or ABSENCE of electron

Answer :C
41.

In an atom,the signs of lobes indicate the

Answer»

SIGN of charges
sign of PROBABILITY disribution
sign of the WAVE function
presence of ABSENCE of electron.

Answer :C
42.

In an atom the total number of electrons having quantum number n=4, |m_(e )|=1 and m_(s)= - (1)/(2) is….

Answer»


ANSWER :6
43.

In an atom, the maximum number of electron having quantum number n-3,-1lem_(l)le1 and m_(s)=+(1)/(2)

Answer»

SOLUTION :
TOTAL orbitals =1+3+3+=6+1=7
Total electron with `m_(s)=+(1)/(2)=6+1=7`
44.

In an atom no two electrons can have the same value for all the quantum numbers. This was proposed by:

Answer»

Hund
Pauli
Dalton
Avogadro

Answer :B
45.

In an atom, an electron is moving with a speed of 6200 m/s which the position of an electron can be locates is (h=6.6xx10^(-34)kgm^(2)s^(-1), mass of electron e_(m)=9.1xx10^(-31)kg)

Answer»

`1.52xx10^(-4)m`
`5.10xx10^(-3)m`
`1.92xx10^(-3)m`
`3.84xx10^(-3)m`

SOLUTION :`Deltav=0005/100xx600=0.03`
`Delta X xx mDeltav=h/(4pi)`
`Deltax=h/(4pi mDeltav)=(6.6xx10^(-34))/(4xx3.14xx9.1xx10^(-31)xx6.03)`
`=1.92xx10^(-3)m`
46.

In an atom, an electron is moving with a speed of 600 m/s with an accuracy of 0.005%. Certainty with which the position of the electron can be located is, (h=6.6xx10^(-34)kgm^(2)s^(-1),m_(e)=9.1xx10^(-31)kg)

Answer»

`1.52xx10^(-4)m`
`5.10xx10^(-3)m`
`1.92xx10^(-3)m`
`3.84xx10^(-3)m`

SOLUTION :`Deltax=h/(4pimDeltav),Deltav=0.005/100xx600`
47.

In an arrangement of type ABABA .... identical atoms of first layer ( A ) and third layer ( A ) are joined by a line passing through their centers . Identify the correct statement .

Answer»

No void is found on the line
Only TETRAHEDRAL voids are found on the line
Only octahedral voids are found on the line
Equal number of tetrahedral and octahedral voids are found on the line

Answer :B
48.

In an aromatic ring, a functional group with lone pair of electron exerts +M effect, some fnuctional groups like, -NO,-NC-CH=CH_(2) etc., can function both as electron donating (+M) or electron withdrawing (-M) groups. More extended conjugation provide more stabilization. The most stable carbocation is :

Answer»




SOLUTION : +IVE CHARGE is delocalised in all RINGS.
49.

In the Arrhenius equation, k = A exp^(-Ea//RT), A may be termed as the rate constant at…………. .

Answer»

SOLUTION :INFINITE TEMP.
50.

In an aqueous solution of volume 500ml when the reaction given below2Ag^(+)+ 2Ag^(+)+Cu hati Cu^(2) +2AgCu^(2+)+2Ag reached at equilibrium the [Cu^(2+] was xM.When 500 ml of wateris further added, at the equilibrium the [Cu^(2+)] will be

Answer»

`x`
`2x`
less than `x/2`
`(3x)/4`

Solution :After ADDING water equilibrium will SHIFT BACKWARD and NEW concentration will be less than `x/2`