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 reaction of alcohols with alkali metal which of the following alcohols will react fastest

Answer»

secondary
tertiary
primary
all equal

Answer :C
2.

In reaction of alcohols with alkali metal, acid etc. which of the following alcohol will react fastest

Answer»

Secondary
Tertiary
primary
All equal

Solution :Order of REACTIVITY with alkali METAL (e.g. -Sodium) follows the order `1^(@) gt 2^(@) gt 3^(@)`.
3.

In reaction N_(2(g))+3H_(2(g))hArr2NH_(3(g)),DeltaH=-93.6kJ, the yield of ammonia does not increase when

Answer»

PRESSURE is INCREASED
TEMPERATUREE is lowered
Pressure is lowered
Volume of the reaction vessel is decreased

ANSWER :C
4.

In reaction of alcohols with alkali metal, acid etc. which of the following alcohol will react fastest ?

Answer»

Secondary
Tertiary
Primary
All equal

Answer :C
5.

In reaction H_2(g) + Cl_2(g) = 2HCl (g) relation between K_p and K_c is _____.

Answer»


ANSWER :`K_P`=`K_c`
6.

Name the reaction where Lewis acid is used as catalyst for aromatic alkylation.

Answer»


ANSWER :FRIEDEL - CRAFT
7.

In reaction CaF_(2)+H_(2)SO_(4)rarrCaSO_(4)+2HF 6kg of CaF_(2) are treated with an excess of H_(2)SO_(4) and yield 2.0kg of HF. Percentage yield of reaction is:

Answer»

0.5
0.39
0.65
0.79

Solution :`n_(HF)=0.1`
`n_(CaF_(2))"USED"=(0.1)/(2)`
`w_(CaF_(2))"used"=3.9kg`
`% "yiled"=(3.9)/(6)xx100=65%`
8.

In reaction, CO(g) +2H_2(g) hArrCH_3OH(g)DeltaH=-92kJ//mol^(-1) Concentrations of hydrogen, carbon monoxide and methanol become constantat equilibrium . What will happen if (a) volume of the reaction vessel in which reactants and products are contained is suddenly reduced to half ? (b) partial pressure of hydrogen is suddenly doubled ? (c ) an inert gas is added to the system at constant pressure ? (d) the temperature is increased ?

Answer»

Solution :For the equilibrium,
`CO(g) +2H_2 (g) hArr CH_3 OH(g)`
`K_c=([CH_3OH])/([CO][H_2]^2)`
`K_p=(P_(CH_3OH))/(P_(CO)xxP_(H_2)^2)`
(a) When the volume of the vesselis suddenly reduced to half , the PARTIAL pressures of VARIOUS species gets doubled . Therefore, `Q_p=(2P_(CH_3OH))/(2P_(CO)xx(2P_(H_2))^2)=1/4K_p`
SINCE `Q_P` is less than `K_p` , the equilibrium shifts in the forward direction producing more `CH_3OH`.
(b) When partial pressure of hydrogen is suddenly doubled, `Q_p` changes and is no longer equal to `K_p`.
`Q_p=(2P_(CH_3OH))/(2P_(CO)xx(2P_(H_2))^2)=1/4K_p`
Equilibrium will shift from left to RIGHT.
When and inert gas is added to the system at constant pressure, equilibrium shift from lower number of moles to higher number of moles(in backward direction).
(d) By increasing the temperature, `K_p` will decrease and equilibrium will shift from right to left.
9.

In reaction A(g) to B(g) + C(g) if initial pressure fo A(g)is 100 m.m. of Hg half-life of reaction is found to be 10 minute but if initial pressure of A is 200 mm of Hg half-life of reaction is found to be 2.5 minute then what will be order of reaction ?

Answer»


ANSWER :3
10.

In reaction 8BF_(3)+6LiH overset(Delta) to X+LIBF_(4)

Answer»

`B_(4)H_(10)`
`B_(2)H_(6)`
`BH_(3)`
`B_(3)H_(8)`

Solution :`8BF_(3)+6LiH overset(DELTA) to B_(2)H_(6)+LIBF_(4)`
11.

In reaction C_6H_5CH_3overset"oxidation"to A overset"NaOH"to B overset"sodalime"toC Then C is

Answer»

`C_6H_6`
`C_6H_5OH`
`C_6H_5COOoverset+Na`
`C_6H_5ONa`

SOLUTION :`C_6H_5CH_3overset"[O]"to underset"[A]"(C_6H_5COOH)OVERSET"NAOH"to underset"[B]"(C_6H_5COONa)overset(NaOH//CaO)to underset"[C]"(C_6H_6)`
12.

In Rast method, Camphor is used as a solvent to determine molecular weight of non-volatile solutes because

Answer»

It is readily available
It is volatile
Its molardepression constant is high
It is best SOLVENT for ORGANIC substances

Answer :C
13.

In ramsay - Raylighs second method, the mixture of N_2 and O_2 can be seperated by dissolving in NaOH in the form of

Answer»

NITRATES
NITRIDES
OXIDES
AMIDES

ANSWER :A
14.

In Raschig method, chlorobenzene is converted to phenol by ________

Answer»

catalytic OXIDATION
catalytic REDUCTION
catalytic HYDROLYSIS
ACIDIFICATION

Answer :C
15.

In Ramsay - Ray leight first method, which of the following are used to remove CO_2, O_2 and N_2 respectively

Answer»

Soda lime + Potash SOLUTION, red hot Cu & HEATED MG
Soda lime, heated Mg& red hot Cu
Soda lime, red hot `Cu_2O` & heated Mg
Soda lime + Potash soln, red hot CUO & heated Mg

Answer :B
16.

In Ramsay and Rayleigh's isolation of noble gases from air,the nitrogen of the air is finally converted into

Answer»

`NaNO_(2)` only
NOand `NO_(2)`
`NaNO_(3)` only
`NaNO_(2)` and `NaNO_(3)`

SOLUTION :NITROGEN is finally converted into `NaNO_(2)` and `NaNO_(3)`,in Ramsay and Rayleigh.s method.
`N_(2)O_(2)to2NO`
`2NO+O_(2)to2NO_(2)`
`2NO_(2)+2NaOHtoNaNO_(2)+NaNO_(3)+H_(2)O`
17.

In R-S configuration -C-=N group can be represented as

Answer»

`-C=N-C`
`-OVERSET(C) overset(|)(C)=N`
` UNDERSET(N)underset(|)(C)=N-C`
`-underset(N)underset(|)overset(N)overset(|)C-underset(C)underset(|)N-C`

ANSWER :D
18.

In radioactive decay which one of the following moves the fastest

Answer»

`alpha`-particle
`beta-`particle
`GAMMA`-RAYS
Positron

Solution :`gamma-`rays are ELECTROMAGNETIC WAVES
19.

In R '-O-R, the R' is a higher alkyl group, it is come from

Answer»

alkane
alcohol
both 'a' and 'b'
not predicted

Answer :A
20.

In quick vinegar process of acetic acid, the temperature of mixture is

Answer»

300 K
427 K
500 K
350 K

Solution :`2CH_(3)CH_(2)OH+O_(2) underset(300K)OVERSET("ACETOBACTER")tounderset(8-10%" acetiacid (VINEGAR)")(CH_(3)COOH+H_(2)O)`
21.

In question no. (23) as given above, what would be the molecularity of elementary reaction [a] and [b] respectively?

Answer»

3,3
2,2
1,1
3,2

Answer :A
22.

In Question pH of the final solution will be

Answer»

12
2
`11.7`
3

Solution :`[OVERSET(C-)(O)H]=0.01=10^(-2)`
`pOH=2,pH=14-2=12`
23.

In question 5 rate of use of H^(+) is x mol L^(-1) S^(-1) than (a) Rate of use of Br^(-1) (b)How much is the average rate of formation of Br_(2)?

Answer»

Solution :`(6)/(5)X mol L^(-1) s^(-1)` (B) 2x mol `L^(-1) s^(-1)`
24.

In question 266 step(2) can be thought of an/a:

Answer»

Neutralisation
Electrophilic ATTACK at the CARBONYL carbon
Nucleophilic attack of N-lone pair at the carbonyl carbon leading to substitution
Nucleophilic addition reaction

Answer :C
25.

In Question above, the number of ampere hours for which the battery is used containing 1L of the acid is 16.08x ampere hour. Calculate the value of x.

Answer»


Solution :From QUESTION above,
Decrease in amount of `H_(2)SO_(4)` as batteryyield current `=` Change in molarity `xxMw_(2)xx`VOLUME of ACID in `L`
`=(3xx98xx1L)g`
Overall change `:`

`2F-=2 mol of H_(20SO_(4)` or `1F-=1 mol of H_(2)SO_(4)=98G`
`98g H_(2)SO_(4)` requires `=1F`
Ampere` //` hour `=(3Fxx96500A-s)/(3600 s h^(-1))`
`=80.4 A-h`
`:. 16.08x=80.4`
`x=(80.4)/(16.08)=5`
26.

When methyl amine is treated with NaNO_2 and HCl, the product obtained is―

Answer»

`NO_2`
`NH_3`
`N_2 +H_2O`
`RCH_2NO_2`

ANSWER :C
27.

Which of the mass extinctions are thought to be manmade.

Answer»

Neutralisation
Electrophilic ATTACK at the carbonyl CARBON
Nucleophilic attack of N-lone pair at the carbonyl carbon LEADING to substitution
Nucleophilic addition reaction

Answer :C
28.

Genes involved in cancer are

Answer»

`R-CH_2CO_2H`
`R-CH_2COONH_4`
`R-CH_2CN`
`R-CH_2-N=C=O`

ANSWER :D
29.

In Question 266 step (4) can be carried out with NaNO_2 + dil,HCL,The other products of the step is:

Answer»

`NO_2`
`NH_3`
`N_2 +H_2O`
`RCH_2NO_2`

ANSWER :C
30.

In Question 266 an intermediate involved in step (3) is:

Answer»

`R-CH_2CO_2H`
`R-CH_2COONH_4`
`R-CH_2CN`
`R-CH_2-N=C=O`

ANSWER :D
31.

In qualitative inorganic analysis of basic radicals ydochioric acid is preferred to nitric acid forpreparing a solution of given substance .This is because :

Answer»

NITRATES are notdecomposed to selphides
NITRIC ACID containnitrogen
Hydrocholoric acid is not an oxidesing agent
Choride areeasly of converted to sulphides

Solution :`HNO_(3)` is as oxidesingugent but `HCI` is not
32.

In qualitative inorganic amalysis phosphan , if present is to be elemenated in the apperopriate grest in order to detect the radical :

Answer»

`PB^(2+)`
`As^(3+)`
`Ca^(2+)`
`CD^(2+)`

SOLUTION :Phosphate interferes in the usual inorganic analysis after group II.
33.

In qualitative analysis when H_(2)S is passed through an aqueous solution ofsalt acidified with dil. HCL, a block precipitate is obtainned .On boiling the precipitate with dil. HNO_(3) it forms a solution of blue colour . Addition of excess ofaqueous solution of ammonia to this solution given ....... .

Answer»

deep blue precipitate of Cu `(OH)_(2)`
deep blue solution of `[Cu(NH_(3))_(4)]^(2+)`
deep blue solution of Cu `(NO_(3))_(2)`
deep blue solution of Cu `(OH)_(2)`. Cu `(ON_(3))_(2)`

Solution :In qualitative analysis when `H_(2)S` is PASSED through an aqueous sloution of salt acidified with DIL . HCl a black ppt. of CuS is obtained .
`CuSO_(4)+H_(2)Soverset(dil. HCl)tounderset("blackppt")( CuS +)H_(2)SO_(4)`
On boiling CuS with dil. `HNO_(3)` it FORMS a blue coloured solution and the following reactions OCCUR
`3CuS+8HNO_(3)to3Cu(NO_(3))_(2)+2NO+3S+4H_(2)O`
`S+2HNO_(3)toH_(2)SO_(4)+NO`
`2Cu^(2+)+SO_(4)^(2-)+2NH_(3)+2H_(2)OtoCu(OH)_(2)*CuSO_(4)+2NH_(4)OH`
`Cu(OH)_(2)*CuSO_(4)+8NH_(3)tounderset("Tetraammine copper"(||)("Deep blue solution"))(2[Cu(NH_(3))_(4)]SO_(4)+2OH^(-)+SO_(4)^(2-)`
34.

In qualitative analysis, when H_2S is passed through an aqueous solution of salt acidified with dil.HCl, a black precipitate is obtained. On boiling the precipitate with dil. HNO_3, it forms a solution of blue colour. Addition of excess of aqueous solution of ammonia to this solution gives

Answer»

deep BLUE PRECIPITATE of`Cu(OH)_2`
deep blue solution of `[Cu(NH_3)_4]^(2+)`
deep blue solution of `Cu(NO_3)_2`
deep blue solution of `Cu(OH)_2.Cu(NO_3)_2`

Solution :A deep blue solution is formed due to the following REACTIONS :
`Cu^(2+) + H_(2)S overset(dil HCl) rarr underset("(Black ppt.)")(CuS) + 2H^(+) , "" CuS + underset("(dil)")(2NHO_3) rarr underset("(Blue solution)")(Cu(NO_3)_2) + H_2S`
`Cu(NO_3)_2 + 4NH_3 rarr underset("(Deep blue solution)")([Cu(NH_3)_4]^(2+)) + 2NO_(3)^(-)`
35.

In qualitative analysis when H_2S is passed through an aqueous solution of salt acidified with dil. HCl, a black precipitate is obtained. On boiling the precipitate with dil. HNO_3, it forms a solution of blue colour. Addition of excess of aqueous solution of ammonia to this solution gives .............

Answer»

Deep blue PRECIPITATE of `CU(OH)_2`
Deep blue solution of `[Cu(NH_3)_4]^(2+)`
Deep blue solution of `Cu(NO_3)_2`
Deep blue solution of `Cu(OH)_2 .Cu(NO_3)_2`

Solution :`H_(2)S_(g) + Cu_(aq)^(2+) to CUS + 2H_(aq)^(+)`
`underset("dilute")(Cus + 10HNO_(3))to underset("Blue")(Cu(NO_(3))_(2) + 8NO_(2) + 4H_(2)O + H_(2)SO_(4) to NH_(3)(aq) [Cu(NH_(3))_(4)]^(2+)` (Deep blue colour)
36.

In qualitative analysis when H_(2)S is passed through an aqueous solution of salt acidified with dil. HCl, a black precipitate is obtained. On boiling the precipitate with dil. HNO_(3), it forms a solution of blue colour. Addition of excess of aqueous solution of ammonia to this solution gives........ .

Answer»

deep blue precipitate of `Cu(OH)_(2)`
deep blue solution of `[Cu(NH_(3))_(4)]^(2+)`
deep blue solution of `Cu(NO_(3))_(2)`
deep blue solution of `Cu(OH)_(2).Cu(NO_(3))_(2)`

Solution :`underset("Salt")(CuSO_(4))+H_(2)S OVERSET("Dil.HCI")rarr underset(("Black ppt."))(CuS)+H_(2)SO_(4)" ,"" "CuS + 2HNO_(3) rarr underset(("Blue solution"))(Cu(NO_(3))_(2))+H_(2)S`
`Cu(NO_(3))_(2) + 4NH_(3) rarr underset(("Deep blue solution"))([Cu(NH_(3))_(4)]^(2+)) + 2NO_(3)^(-)`
37.

In qualitative analysis, the metals of Group I can be separated from other ion by precipitating them as chloride salts. A solution initially contains Ag^(+) and Pb^(2+) at a concentration of 0.10 M. Aqueous HCl is added to this solution until the Cl^(-) concentration is 0.10 M. What will the concentrations of Ag^(+) and Pb^(2+) be at equilibrium (K_(sp) for AgCl = 1.8 xx 10^(-10), K_(sp) fr PbCl_(2) = 1.7 xx 10^(-5))

Answer»

`[AG^(+)] = 1.8 xx 10^(-9) M, [Pb^(2+)] = 1.7 xx 10^(-3)M`
`[Ag^(+)] = 1.8 xx 10^(-11) M, [Pb^(2+)] = 1.7 xx 10^(-4)M`
`[Ag^(+)] = 1.8 xx 10^(-7) M, [Pb^(2+)] = 1.7 xx 10^(-6)M`
`[Ag^(+)] = 1.8 xx 10^(-11) M, [Pb^(2+)] = 1.7 xx 10^(-5)M`

Answer :A
38.

In qualitative analysis, in order to detect second group basic radical, HS gas is passed in the presence of dilute HCI to

Answer»

increase the dissociation of `H_(2)S`
decrease the dissociation of SALT solution
decrease the dissociation of `H_(2)S`
increase the dissociation of salt solution

Solution :The original solution must be FREE from any oxidising agent otherwise`H_(2)S` gets oxidised with precipitation of free sulphur. This is done by heating the mixture with concentrated HC1 for SUFFICIENT time during PREPARATION of original solution.
39.

In Q.No. 11, which of the following reagents can be used to convert (A) to (B)?

Answer»

<P>a.Conc. `H_(2)SO_(4)` at `413 K (140^(@)C)`
b.`DC C`
c.`P_(2)O_(5)`
d.Conc. `H_(2)SO_(4)` at `383 K (110^(@)C)`

Solution :Alcohols are DEHYDRATED to ALKENE with concentrated `H_(2)SO_(4)` at `160-170^(@)C`. At `10^(@)C`, ether is formed. Both `DC C` (dicyclo hexyl

and `P_(2)O_(5)` can be used for the dehydration of alcohols to alkene. So the answers are `(b)` and `(c )`.
40.

In pyrophosphoric acid the number of hydroxy groups present are:

Answer»

4
3
5
7

Answer :A
41.

In pyranose of glucose oxide (ether) linkage is formed between

Answer»

`C_1` and `C_6`
`C_2 and C_6`
`C_1 and C_5`
`C_2 and C_3`

ANSWER :C
42.

In purine nucleoside, C-1 of sugar from glycosidic linkage with which position of purine

Answer»

1
2
9
8

Answer :C
43.

In Pt(C_(2)H_(4))_(2)""(C_(2)F_(4)) molecule, in which the platinum and all six carbon atoms are almost coplanar, choose the correct statement(s).

Answer»

C-C distance is greater in `C_(2)F_(4)" than "C_(2)H_(4)`
`Pt-C(of C_(2)H_(4))` distance is greater than `Pt-C(ofC_(2)F_(4))`
If `C_(2)H_(4)` is REPLACED by `C_(2)(CN)_(4)` in the complex, then C-C distance is greater in `C_(2)(CN)_(4)` than `C_(2)F_(4)`. (double BOND length)
In case of `underline(C)(CN)_(4)` in the complex, above, the UNDERLINED C-hybridisationis `underset(n to 3)lim sp^(n)`.

Solution :.`C_(2)F_(4)` has greater acceptance of electron into `pi^(*)` than `C_(4)H_(4)`.Hence,C-C B.L. of `C_(2)H_(4)uarr`.
.Due to synergic bonding M-C bond (of `C_(4)F_(4)`) becomes stronger.
. `C_(2)(CN)_(4)` also has synergic bonding. Due to back donation from metal to `pi^(*)`orbital ,C=C approaches c-c behaviour `(sp^(3))`
44.

In [Pt(NH_(3))_(2)Cl_(2)], pt-Cl bond length is 2 Å and Cl-Cl distance is 2.88 Å then the compound is:

Answer»

tetrahedral
square pyramidal
cis-square planar
trans-square planar

Answer :C
45.

In public urinals, we observe some pungent smell. This smell is due to :

Answer»

HYDROLYSIS of urea of urine by urease of atmosphere into `NH_3` and `CO_2`
Formation of sulphamic ACID by urea of urine.
Reaction of `CO_2` of atmosphere with urea MONONITRATE in urine
Hydrogen present in air reacts with NITROGEN forming `NH_3`

Answer :A
46.

In proteins , the alpha - amino acids are linked by …………...........

Answer»

Amide LINKAGE
Phosphodiester linkage
Glycosidic linkage
CARBOXYL linkage

SOLUTION :Amide linkage
47.

In proteins, the amino acids are linked covalently by ……….. .

Answer»

SOLUTION :PEPTIDE BONDS
48.

In pseudo - order reactions

Answer»

The actual order of reaction is different from that expected using rate law expression
The CONCENTRATION at LEAST one reactant is taken in large excess.
The concentration of reactant taken in excess MAY be taken as CONSTANT
All of these

Answer :D
49.

In problem 8 as mentioned above the wavelength of light emitted in the visibleregion by He^(+)ions after collisions with H atoms is

Answer»

`6.5 xx10^(-7)` m
`5.6 xx10^(-7)` m
`4.8 xx10^(-7)` m `4.0 XX 10^(-7)` m

ANSWER :C
50.

In previous problem calculate DeltaS_("gas") if process is carried out at constant volume:

Answer»

`5R In 2`
`(3)/(2) R In 2`
`3R In 2`
`-3R In 2`

SOLUTION :N//A