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.

The calcium salt used as manure is:

Answer»

`CaC_2`
`CaCN_2`
`CaCO_3`
`CaSO_4`

ANSWER :B
Discussion
2.

The calcium contained in a solution of 1.048 gm of the substance being analysed was precipitated with 25 ml of H_(2)C_(2)O_(4). The excess of C_(2)O_(4)^(2-) in one fourth of the filtrate was back titrated with 5 ml of 0.1025 N KMnO_(4) solution. To determine the concentration of the H_(2)C_(2)O_(4)solution taken , it was diluted four fold , the titration of 25 ml of the dilute solution used up 24.1ml of KMnO_(4) solution . What is percentage of CaO in the substance being analysed .

Answer»


SOLUTION :N/A
Discussion
3.

The calcium contained in a solution of 1.048 gm of the substance being analysed was precipitated with 25ml of H_(2) C_(2) O_(4). The excess of C_(2) O_(4)^(2-) in one fourth of the filtrate was back titrated with 5ml of 0.1025 N KMnO_(4) solution. To determine the concentration of the H_(2) C_(2) O_(4) solution taken, it was diluted four fold, the titration of 25ml of the dilute solution used up 24.1 ml of KMnO_(4) solution. What is percentage of CaO in the substance being analysed.

Answer»

Solution :EQUIVALENTS of `KMnO_(4)` used up with one fourth of the filtrate containing EXCESS `H_(2)C_(2) O_(4) = ( 5 xx 0.1025)/( 1000) = 5.125 xx 10^(-4)`
`:.` Excess EQUIVALENT of `H_(2) C_(2) O_(4) = 5.125 xx 10^(-4) xx 4 = 2.05 xx 10^(-3)`
Equivalent of `KMnO_(4)` reacting in the second titration `= ( 24.1 )/( 1000) xx 0.1025 = 2.47 xx 10^(-3)`
`:.` Equivalent `H_(2) C_(2) O_(4)`consumed by `CaO = 9.88 xx 10^(-3) - 2.05 xx 10^(-3) = 7.83 xx 10^(-3)`
`:.` Equivalents of `CaO = 7.83 xx 10^(-3)`
Moles of `CaO = ( 7.83 xx 10^(-3))/(2)= 3.915 xx 10^(-3)`
Mass of CaO `= 3.915 xx 10^(-3) xx 56 = 0.22 ` gm
% CaO = `( 0.22 )/( 1.048) xx 100 = 20.99 %`
Discussion
4.

The calcium salt of the final oxidation product of ethanol on dry distillation gives :

Answer»

Formaldehyde
Acetaldehyde
Acetone
Formic acid

Answer :C
Discussion
5.

The C^(14) " to " C^(12) ratio in a wooden article is 13% that of the fresh wood. Calculate the age of the wooden article. Given that the half-life of C^(14) is 5770 years

Answer»

16989 years
16858 years
15675 years
17700 years

Solution :`(N)/(N_(0)) = ((1)/(2))^((T)/(t_(1//2))) rArr (13)/(100) = ((1)/(2))^((T)/(5770))`
Taking LOG `rArr "log" (13)/(100) = (T)/(5770) log 1//2 rArr 16989 yrs`
Discussion
6.

The Ca^(2+) and F ions are located in CaF_( 2)crystal, respectively at face centred cubic lattice points and in

Answer»

TETRAHEDRAL VOIDS
HALF of tetrahedral voids
OCTAHEDRAL voids
half of octahedral voids

ANSWER :A
Discussion
7.

The C_4H_10O(alcohols) produces immediateterbidity with Lucas reagent the alcohol is

Answer»

`CH_3-CH_2-CH_2 -CH_2-OH`
`(CH_3)_2CH-CH_2-OH`
`CH_3CHOHC_2H_5`
`(CH_3)_3C-OH`

ANSWER :D
Discussion
8.

The C-X bond energy order for carbon tetrahalides is :

Answer»

`CF_4gtC Cl_4gtCBr_4gtCI_4`
`C Cl_4gtCBr_4gtCI_4gtCF_4`
`CI_4gtCBr_4gtC Cl_4gtCF_4`
NONE of these

Answer :A
Discussion
9.

The C-O-C bond angle in dimethyl ether is

Answer»

`110^(@)`
`112^(@)`
`111.7^(@)`
`110.2^(@)`

ANSWER :A::C
Discussion
10.

The C-O-C angle in ether is about

Answer»

`180^(@)`
`190^(@)-28`'
`110^(@)`
`105^(@)`

SOLUTION :NA
Discussion
11.

The C-O bond length in C -= O ….. (decreases or increases or remains unchanged) when metal carbonyls are formed.

Answer»

SOLUTION :INCREASES
Discussion
12.

The C-Mg bond in CH_(3)CH_(2)MgBr is

Answer»

ionic
non-POLAR COVALENT
polar covalent
hydrogen.

Solution :`C-Mg` BOND is covalent but polar.
Discussion
13.

The C-O bond is much shorter in phenol than in ethanol. Given season.

Answer»

SOLUTION :`sp^(2)`, which RESULT in partial DOUBLE bond character due to conjugation UNSHARED ELECTRON pair of oxygen with aromatic ring. Because C-O in phenol is attached with C=C while in ethanol C-O is attached `(sp^(3))` with C-C.
Discussion
14.

The C-Mg bond in CH_3CH_2MgBr is :

Answer»

IONIC
NON -POLAR covalent
polar covalent
Hydrogen

ANSWER :C
Discussion
15.

The C-H bond length is minimum in the bond formed by

Answer»

sp-s overlapping (as in ALKYNES)
`sp^2` -s overlapping (as in ALKENES)
`sp^3` -s overlapping (as in alkanes)
NONE of these

Answer :A
Discussion
16.

The C-H bond distance is the longest in:

Answer»

`C_2H_2`
`C_2H_4`
`C_2H_6`
`C_2H_2Br_2`

ANSWER :C
Discussion
17.

The C-H bond distance is longest in

Answer»

`C_(2)H_(2)`<BR>`C_(2)H_(4)`
`C_(2)H_(6)`
`C_(2)H_(2)Br_(2)`

SOLUTION :`C_(2)H_(6)` will have LONGEST C-H bond distance.
Discussion
18.

The C-Cl bond length in chlorobenzene is shorter than C-Cl bond length in CH_(3)-Cl.

Answer»

Solution :Chlorobenzene can be regarded as a resonance hybrid of FIVE structure. As a RESULT C-CL bond in chlorobenzene has some double bond CHARACTER and hence shorter (169 pm) than C-Cl bond in `CH_(3)-Cl` (177 pm) which has pure single bond character.
Alternately, in `CH_(3)-Cl,Cl` is attached to a `sp^(3)`-hybridized carbon while in chlorobenzene it is attached to a `sp^(2)`-hybridized carbon. since a `sp^(2)`-hybridized orbital is smaller in size as compared to a `sp^(3)`-hybridized
ltBrgt orbital of carbon, THEREFORE, `C_Cl` bond in chlorobenzene is shorter (169 pm) than C-Cl bond length in `CH_(3)-Cl` (177 pm).
Discussion
19.

The C-H bond and C- C bond in ethane are formed by which of the following types of overlap ?

Answer»

`SP^(3)-s and sp^(3)-sp^(3)`
`sp^(2)-s and sp^(2)-sp^(2)`
sp-s and sp-sp
p-s and p-p

ANSWER :A
Discussion
20.

The C-C bond length of the following molecules is in the order

Answer»

`C_(2) H_(6) gt C_(2)H_(4) gt C_(6) H_(6) gt C_(2) H_(2)`
`C_(2) H_(2) lt C_(2) H_(4) lt C_(6) H_(6) lt C_(2) H_(6)`
`C_(2) H_(6) gt C_(2)H_(2) gt C_(6)H_(6) gt C_(2)H_(4)`
`C_(2) H_(4) gt C_(2) H_(6) gt C_(2) H_(2) gt C_(6) H_(6)`

Solution : C-C bond length increases in the ORDER:
`C_(2) H_(2) lt C_(2) H_(4) lt C_(6) H_(6) lt C_(2) H_(6)`
Discussion
21.

The C-C bond length of the following molecules is in the order:

Answer»

`C_2H_4 GT C_2H_6gtC_2H_2 `
`C_2H_2 gt C_2H_4gt C_2H_6`
`C_2H_6 gt C_2H_4 gt C_2H_2`
NONE`

ANSWER :B
Discussion
22.

The C - Cl bond in Ethyl chloride is formed by ........ overlaping

Answer»

`sp^3-s`
`sp^3-p`
`sp^3d-p`
`sp^2-p `

ANSWER :B
Discussion
23.

The C - Cl bond in Ethyl chloride is formed by ......... overlaping

Answer»

<P>`sp^3-s`
`sp^3 -p`
`sp^3 d-p`
`sp^2 -p`

ANSWER :B
Discussion
24.

The C - C bond energy from the following data will be (i)C(s)rarr C(g), Delta H=170.9 kcal(ii) (1)/(2)H_(2)(g)rarr H(g), Delta H=52.1 kcal(iii) Heat of formation of ethane = -20.3 kcal(iv)C - H bond energy = 99.0 kcal

Answer»

80.7 KCAL
40.9 kcal
88.3 kcal
26.8 kcal

SOLUTION :`2C(s)+3H_(2)(g)RARR C_(2)H_(6)(g), Delta H=-20.3 kcal`
Also, `Delta H = -[1 BE_(C-C)+6BE_(C-H)]+[2{C(s)rarr C(g)}+3BF_(H-H)]`
`-20.3=-[BE_(C-C)+6xx99]+[2xx170.9+3xx(2xx52.1)]`
`therefore BE_(C-C)=-594+341.8+312.6+20.3`
C - C bond energy = 80.7 kcal
Discussion
25.

The % by volume of C_(4)H_(10) in a gaseous mixture of C_(4)H_(10), CH_(4) and CO_(2) is 40. When 200 ml of this mixture is burnt completely in excess of O_(2) and then pass through aq. KOH then volume contraction will be -

Answer»

220 ml
340 ml
440 ml
560 ml

Solution :`ubrace(C_(4)H_(10), CH_(4),CO_(2))_("200 mL")`
`C_(4)H_(10) = 0.4 xx 200 = 80 ml`
`CH_(4) = x ml`
`CO_(2) = (120 - x) ml`
`C_(4)H_(10) + (13)/(2) O_(2) rarr 4CO_(2) + 5H_(2)O`
`{:(80 ml,,"320 ml",),(CH_(4) + 2O_(2),rarr,CO_(2) + 2H_(2)O,),(x ml,,x ml,):}`
Total `CO_(2) = 320 + x + 120 - x`
`= 440 ml`
Discussion
26.

The byproduct of Solvay used in photography is:

Answer»

`CO_2`
`CaCl_2`
`NH_3`
`CaCO_3`

ANSWER :B
Discussion
27.

The by-product obtained in the preparation of fluorobenzene from benzene diazonium chloride using HBF_(4) is ____________.

Answer»


ANSWER :`BF_(3)`
Discussion
28.

The by-product in Dow's process …….

Answer»

Ketone
Aldehyde
ETHER
CARBOXYLIC acid

Solution :Ether
Discussion
29.

The button cell used in watches function as follows Zn(s)+Ag_(2)O(s)+H_(2)O(l) iff 2Ag(s)+Zn^(2+)(aq)+2OH^(-)(aq) the half cell potentials are Ag_(2)O(s)+H_(2)O(l)+2e^(-) rarr 2Ag(s)+2OH^(-)(aq) E^(o)=0.34V. The cell potential will be

Answer»

0.84V
1.34V
1.10V
0.42V

Answer :C
Discussion
30.

The button cell used in watches function as follows Zn_a + Ag_(2)O(s) + H_2O_((l))

Answer»

0.84 V
1.34 V
1.10 V
0.42 V

Solution :Anodic OXIDATION: (Reverse the given reaction)
`(E_("ox")^@) = 0.76V ` cathodic reduction
`:. E_("CELL")^(@) = (E_("ox")^@) + (E_("RED")^(@)) = 0.76 + 0.34=1.1V `.
Discussion
31.

The butanol can be obtained by the reaction of methyl magnesium iodide with..

Answer»



Methanal
Propanal

Answer :B
Discussion
32.

The burning of hydrogen is called

Answer»

REDUCTION
HYDROGENATION
OXIDATION
Hydration

Answer :C
Discussion
33.

The burning of coal represented by the equation, C(s)+O_2(g)rarrCO_2(g) .Therate of this reaction is increased by :

Answer»

DECREASE in the CONCENTRATION of oxygen
Powdering the lumps of coal
Decreasing the temperature
Providing inert ATMOSPHERE for burning

Answer :B
Discussion
34.

The building unit of all proteins are

Answer»

MONOSACCHARIDES 
lipids
amino acids
primary amines

Answer :C
Discussion
35.

The buffer system which helps to maintain the pH of blood between 7.26 to 7.42 is

Answer»

`H_(2)CO_(3)//HCO_(3)^(-)`
`NH_(4)OH//NH_(4)CL`
`CH_(3)COOH//CH_(3)COO^(-)`
`CH_(3)COONH_(4)`

Answer :A
Discussion
36.

The buffer solution contains equal concentration of X^(-) and HX. The K_(b) for X^(-) is 10^(-10). The pH of the buffer solution is :

Answer»

4
7
10
14

Answer :A
Discussion
37.

What is buffer solution ?

Answer»


ANSWER :[(Pk_b+1) to [(Pk_b+1)]
Discussion
38.

The buffer action of acidic buffer is maximum when its pH is equal to:

Answer»

5
7
10
`Pk_a =1`

ANSWER :D
Discussion
39.

The buffer present in human blood is

Answer»

`CH_3 COOH+CH_3 COONA`
`NH_4 OH+NH_4 CI`
`H_2CO_3 + HCO_3^-`
Both (a) and (B)

ANSWER :C
Discussion
40.

The Brownian movement of colloidal particles is because of:

Answer»

Convection currents in the fluid
Unequal bombardments by the molecules of the DISPERSION MEDIUM on colloidal particles.
SETTING of dispersed phase under gravity
Thermal GRADIENT in the medium

Solution :Unequal bombardments of solvent particles on colloidal particles cause zig-zag motion.
Discussion
41.

The Brownian motion is due to

Answer»

temperature fluctuation within the liquid phase
attraction and repulsion between charges on the colloidal particles
impact of MOLECULES of the DISPERSION medium on the colloidal particles
convection currents

Solution :Molecules of dispersion medium COLLIDE with colloidal particles and cause them to MOVE in a zig-zag way.
Discussion
42.

The Brownian motion is due to ,

Answer»

Temperature fluctuation withing the liquid phase
Attraction and repulsion between CHARGE on the colloidal PARTICLES
Impact of moleculars of the dispersion medium on the colloidal particles
Convective current

Solution :Brownian motion is due to impact of MOLECULES of the dispersion medium on the colloidal particles.
Discussion
43.

The brown ring test is performed for the qualitative detection of:

Answer»

BROMIDES
IODIDES
nitrates
nitrite

Answer :C::D
Discussion
44.

The Brownian motion is due to :

Answer»

temperature fluctuation WITHIN the liquid phase
attraction and repulsion between CHARGE on the colloidal particles
impact of MOLECULES of the DISPERSION medium on the colloidal particles
convective current

Answer :C
Discussion
45.

The brown yellow colour often shown by nitric acid can be removed by:

Answer»

BUBBLING AIR through the WARM acid
Boiling the acid
Passing ammonia through acid
Adding a little MG powder

Answer :A
Discussion
46.

The brown ring test for NO_(2)^(-) and NO_(3)^(-) is due to the formation of complex ion with formula :

Answer»

`[Fe(H_2O)_6]^(2+)`
`[Fe(NO)(CN)_5]^(2+)`
`[Fe(H_2O)_5NO]^(2+)`
`[Fe(H_2O)(NO)_5]^(2+)`

SOLUTION :REACTION of brown ring test
Discussion
47.

The brown ring test for nitrites and nitrates is due to the formation of complex of iron with the formula.

Answer»

`[Fe(H_(2)O]_(5) NO]^(2+)`
`[Fe(H_(2)O]^(2+)`
`[Fe(H_(2)O) (NO)_(5)]^(2-)`
`[Fe(NO) (CN)_(5)]^(2+)`

ANSWER :A
Discussion
48.

The brown ring rest test for nitrates depends upon :

Answer»

REDUCTION of NITRATE to nitric acid
Oxidation of nitric oxide to nitrogen dioxide
reduction of FERROUS SULPHATE ion
oxidising ACTION of sulphuric acid

Answer :D
Discussion
49.

The brown ring test for nitrates depends on

Answer»

the reduction of nitrate to nitric oxide
oxidation of nitric oxide to nitrogen dioxide
reduction of ferrous sulphate to iron
oxidising ACTION of sulphuric acid

Solution :The brown ring is appeared due to the formation of a brown complex between nitric oxide (FORMED as a result of reduction of the nitrate ION by the `Fe^(2+)` IONS) and `Fe^(2+)` ions.
`NO_(3(aq))^(-)+3Fe_((aq))^(2+)+4H_((aq))^(+)to NO_((g))+3F_((aq))^(3+)+2H_(2)O_((l))`
`underset("pentaaquanitrosonium iron (I) (Brown complex)")(Fe_((aq))^(2+)+NO_((g))+5H_(2)O_((l))to [Fe(H_(2)O)_(5)NO^(+)]^(2_((aq))^(+))`
Discussion
50.

The brown ring obtained in the ring test for nitrate is due to the formation of:

Answer»

`Fe_2SO_4`
`FeSO_4`
`FeSO_4cdotNO`
`Fe_2SO_4cdot2NO_3`

ANSWER :C
Discussion