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.

H_(2)COH-CH_(2)OH on heating with periodic acid gives

Answer»

`._(H)^(H)gtC=O`
`2CO_(2)`
`2HCOOCH`
`{:(CHO),("|"),(CHO):}`

Solution :`{:(""CHO),(HOCH_(2)-CH_(2)OH UNDERSET("Periodic acid")overset(Delta, HIO_(4))RARR " |+ " H_(2)O),(""CHO):}`
Discussion
2.

H_2CO_3ionises in two stages as represented below H_2CO_3 +H_2O hArr H_3O^(+) +HCO_3^(-), HCO_3^(-)+H_2O hArr H_3O^(+)CO_3^(2-)the no. of conjugate acid -base pairs in the above reaction are

Answer»

2
3
4
5

Solution :`H_2CO_3, HCO_3^(-) , HCO_3^(-) , CO_3^(2-), H_2O, H_3O^(+) `
Discussion
3.

"Asparticacid + A"overset("Transaminase")rarr" B + Glutamic acid" In the above reaction A and B respectively,

Answer»

2
3
4
5

Solution :`H_2CO_3, HCO_3^(-) , HCO_3^(-) , CO_3^(2-), H_2O, H_3O^(+) `
Discussion
4.

H_(2)C_(2)O_(4).2H_(2)O (Mol wt =126) can be oxidised intoCO_(2) by acidified KMnO_(4). 6.3 gms of oxalic acid can not be oxidised

Answer»

3.16 gms of `KMnO_(4)`
200 ml of 0.1M `KMnO_(4)`
0.1 mole of `KMnO_(4)`
0.02 MOLES of `KMnO_(4)`

Solution :Eq. `H_(2)C_(2)O_(4)` = Eq. `KMnO_(4)`
`(6.3)(126)xx2=Mxx5`
Moles of `KMnO_(4)=0.02=3.16g=0.2xx0.1M`
Discussion
5.

H_(2)C_(2)O_(4) and KHC_(2)O_(4) behave both as acids and reductants. Amongst the following, the true statement(s) is/are

Answer»

Equal volumes of 1 M solution of each is oxidised by equal volumes of 1 M `KMnO_(4)`
Their equivalent masses are equal to respective molecular masses when BEHAVING as reducing agents
V ML of 1 M solution of each is neutralized by equal volumes of 1 M NaOH
V ml of 1 N solution of each (as an ACID) is neutralized by equal volumes of 1 M `Ca(OH)_(2)`.

Solution :In the first titration
Eq. HCl = `(1)/(2)` eq `NC_(2)CO_(3)`
`(x xx0.05)/(1000)=(1)/(2)xx(0.050xx40xx2)/(1000)impliesx=40ml`
In the SECOND titration
Eq of HCl = Eq `NC_(2)CO_(3)+` Eq `NKHCO_(3)`
`(yxx0.05)/(1000)=(40xx0.05xx2)/(1000)+(40xx0.05xx1)/(1000)`
`y=120implies y-x=80`
Discussion
6.

H_(2)C_(2)O_(4) acts as an acid as well as an oxidising agent. The correct statements (s) about H_(2)C_(2)O_(4) is (are):

Answer»

It FORMS two series of salts
Equivalent weight of `H_(2)C_(2)O_(4)` as an acid for completer neutralisation and as oxidant are same
`100 mL` of `0.1M` solution of `KMnO_(4)` (acid) will be completely reduced by `50 mL` of `1M H_(2)C_(2)O_(4)`
`100 mL` of `0.1N` solution of `Ca(OH)_(2)` will be completely neutralised by `50 mL` of `0.2M H_(2)C_(2)O_(4)`

Answer :A::B::D
Discussion
7.

H_(2)C_(2)O_(4)*2H_(2)O its molecular mass ...........

Answer»

90 u
126 u
124 u
136 u

Solution :MOLECULAR mass ` = 6("ATOMIC mass of H") + 2("atomic mass of C") + 6("atomic mass of CO")`
`=6(1)+2(12)+6(16)=126u`
Discussion
8.

H_(2)C = O or CH_(3)CN acts as a nucleophile as well as an electrophile. Explain

Answer»

Solution :SINCE both O and N are more ELECTRONEGATIVE than C, therefore, C carries a small +ve CHARGE and, O and N carry a small -ve charge.
In other words, both act as an ELECTROPHILE due to the presence of a partial +ve charge on C and act as a nucleophile due to the presence of a partial -ve charge on O or the N.
Discussion
9.

H_(2)C = CH- CH_(2) -underset(CH_(2)) underset("||")C-overset(+)CH_(2) Resonance hybrid of the carbocation is

Answer»

`H_(2)overset(delta+)C=CH=overset(delta+)CH_(2)=underset(delta+)underset(CH_(2))underset("||")C= overset(delta+)(CH_(2))`
`H_(2)C=CH-CH_(2)-underset(delta+)underset(CH_(2)) underset("||")C=overset(delta+)(CH_(2))`
`H_(2)overset(delta+)C=CH-CH_(2)-underset(delta+)underset(CH_(2))underset("||")C=overset(delta+)(CH_(2))`
`H_(2)C =CH-CH_(2)-overset(delta+)(CH_(2))=CH=overset(delta+)(CH_(2))`

Answer :B
Discussion
10.

H_2C = CH_2 + CO + H_2 underset(100^(@)C , " high " P)overset(Co_2(CO)_8)to

Answer»




SOLUTION :HYDROFORMYLATION
Discussion
11.

H_2Ais a weak diptotic acid . IF the pH of 0.1 MH_2Asolution is 3 and concentration of A^(2)is 10 ^(-12)"at "25 ^(@) CSelect correct statement (s)

Answer»

`[H^(+) ]_("total")~~[H^(+) ]` from first step ionization of acid `H_2A `
Concentration of `OH^(-) ` in solution is `10 ^(-3) ` M
The value of `K_(a_1) `is NEARLY `10 ^(5)`
`p^(ka_2)- p^(ka_1) =9`

Solution :` [H^(+) ]=sqrt(K_(a_1) xx C ) rArr 10^(-3)=sqrt( K_(a_1)xx 10^(-1)) `
` K_(a_1) =10 ^(-5) rArr [A^(-2) ]=K_(a_2) rArr P^(K_(a2 )) = 12 `
` P^(K_(a2))-P^(K_(a3)) = 7 `
Discussion
12.

What is the concentration of CN^(-)ions in a solution with 0.1M HCland 0.01M HCN where K_a of HCN is 10 ^(-6)?

Answer»

`[H^(+) ]_("total")~~[H^(+) ]` from first step ionization of acid `H_2A `
Concentration of `OH^(-) ` in solution is `10 ^(-3) ` M
The value of `K_(a_1) `is NEARLY `10 ^(5)`
`p^(ka_2)- p^(ka_1) =9`

Solution :` [H^(+) ]=sqrt(K_(a_1) XX C ) rArr 10^(-3)=sqrt( K_(a_1)xx 10^(-1)) `
` K_(a_1) =10 ^(-5) rArr [A^(-2) ]=K_(a_2) rArr P^(K_(A2 )) = 12 `
` P^(K_(a2))-P^(K_(A3)) = 7 `
Discussion
13.

H_(2) will not reduce which of the following oxide ?

Answer»

Aluminium oxide
calcium oxide
Ferrous oxide
None of the above

Solution :`H_(2)` Cannot reduce `Al_(2)O_(3),CaO and FEO.`
Discussion
14.

H_(2) SO_(2) solution (20 mL) reacts quantitatively with a solution of KMnO_(4) (20 mL) acidified is just dilute H_(2) SO_(4). The same volume of the KMnO_(4) solution is just decolourised by 10 mL of MnSO_(4) in neutral medium. simulataneously forming a dark brown precipitate of hydrated MnO_(2). The brown precipitate is dissolved in 10 mL of 0.2 M sodium oxalate under boiling condition in the presence of dilute H_(2) SO_(4). Write the balanced equations involved in the reactions and calculate the molarity of H_(2) O_(2) solution.

Answer»

Solution :Take the balanced equations as follows:
`2MnO_(4)^(ɵ) + 6 H^(o+) + 5 H_(2) O_(2) RARR 2 Mn^(2+) + 8 H_(2) O + 5 O_(2)`
`2 MnO_(4)^(ɵ) + 2 Mn^(2+) rarr 5 MnO_(2)`
`MnO_(2) + C_(2) O_(4)^(2-) + 4 H^(o+) rarr Mn^(2+) + 2 H_(2) O + 2CO_(2)`
To find molarity of `H_(2) O_(2)`
GIVEN:
`20 mL H_(2) O_(2) -= 20 mL KMnO_(4)//H^(o+)`
`20 mL KMnO_(4) -= mL MnSO_(4) rarr MnO_(2)`
`MnO_(2) -= 10 mL (0.2 M) Na_(2) C_(2) O_(4)`
From stoichiometry of above reactions:
1 mmol of `Na_(2) C_(2) O_(4) -= 1 mmol MnO_(2)`
`IMPLIES 0.2 XX 10 mmol of Na_(2) C_(2) O_(4) -= 2 mmol of MnO_(2)`
Also, 5 mmol of `MnO_(2) -= 3 mmol of Mn^(2+)`
`implies 2 mmol of MnO_(2) -= (3)/(5) xx 2`
`-= (6)/(5) mmol of Mn^(2+) (MnSO_(4))`
Also, 3 mmol of `MnSO_(4) -= 2 mmol KMnO_(4)`
`implies (6)/(5) mmol of MnSO_(4) -= (2)/(3) xx (6)/(5) = mmol KMnO_(4)`
Now same amount of `KMnO_(4)` reacts completely with `H_(2) O_(2)`.
`2 mmol of H_(2) SO_(4) -= 5 mmol of H_(2) O_(2)`
`implies 2 mmol of KMnO_(4) -= (5)/(2) xx (4)/(5) = 2 mmol of H_(2) O_(2)`
Now mmol of `H_(2) O_(2) = MV_(mL)`.
`implies 2 = M xx 20 implies M = 0.1`
Discussion
15.

H_(2) + S hArr H_(2) S + energy In this reversible reaction , select the factor which will shift the equilibrium to the right .

Answer»

adding heat
adding `H_(2) S`
blocking HYDROGEN GAS reaction
removing hydrogen sulphide gas

Solution :removing hydrogen sulphide gas
Discussion
16.

p^(I) of H_(3)overset(o+)(N)-CH_(2)-COOK is 5H_(3)overset(o+)(N)CH_(2)COOH hArr H_(2)NCH_(2)COO^(-)+H^(+)K_(a)=10^(-4)H_(2)overset(o+)(N)CH_(2)COO^(-)hArr H_(2)NCH_(2)COO^(-)+H^(+)K_(a)=10^(-X) What is x ?

Answer»

`x+y`
` (x+y)//2`
` (x+y)^(1//2)`
` (x+y)^(1//3)`

SOLUTION :` NaH_2 PO_3 ` is AMPHIPROTIC SALT
`PH= ( pK_1 + pK_2)/( 2 )=(x+y)/(2) `
Discussion
17.

H_2 O+H_3 PO_3 hArr H_3O^(+) +H_2O PO_3^(-),pK_1 = x H_2O+H_2PO_3 hArr H_3 O^(+)HPO_3 ^(-2), pK_2 =y Hence , pH of 0.01 M NaH_2 PO_3 is :

Answer»

`x+y`
` (x+y)//2`
` (x+y)^(1//2)`
` (x+y)^(1//3)`

SOLUTION :` NaH_2 PO_3 ` is AMPHIPROTIC SALT
`PH= ( pK_1 + pK_2)/( 2 )=(x+y)/(2) `
Discussion
18.

Hmlocule has two elections and two nuclei. In which form of hydrogen the spin of electrons and also the spin of nuclei are in opposite directions

Answer»

Orhohydrogen
PARAHYDROGEN
Metahydrogen
B-HYDROGEN

Solution :In para hydrogen spin of PROTONS (NUCLEI) are OPPOSITE each other.
Discussion
19.

H_(2) gas is mixed with air at 25^(@)C under a pressure of 1 atmosphere and exploded in a closed vessel. The heat of the reaction, H_(2(g))+(1)/(2)O_(2(g))rarrH_(2)O_((v)) at constant volume, DeltaU_("298 K")=-"240.60 kJ mol"^(-1) and C_(V) values for GH_(2)O vapour and N_(2) in the temperature range "298 Kand 3200 K are 39.06 JK"^(-1)"mol"^(-1) and "26.40 JK"^(-1)"mol"^(-1) respectively. The explosion temperature under adiabatic conditions is (Given : n_(N_(2))=2)

Answer»

2900 K
`2900^(@)C`
2917 K
`3000^(@)C`

Solution :If the process is CARRIED out ADIABATICALLY and isochorically,
`DeltaU=DeltaU_("heating")+DeltaU_("298 K")=0`
or `DeltaY_("heating")=-DeltaU_(298K)`
`=int_(298K)^(T_(F))n SigmaC_(v)dT=+240.60" kJ mol"^(-1)`
`SigmanC_(v)=n.C_(v(H_(2)O_((v)))+nC_(v(V_(2(g))))`
`=(39.06+2xx26.40)=91.86JK^(-1)mol^(-1)`
by using the value of `SigmanC_(v)` in the above equation
`(91.86)(T_(f)-298)=240600"J mol"^(-1)`
`T_(f)-298=(240600)/(91.86)=2619K`
`T_(f)=2619+298=2917K`
Discussion
20.

H_(2) gas is not liberated at both cathode and anode by electrolysis of which of the following aqueous solution?

Answer»

NaH
HCOONa
NaCl
LiH

Solution :`M^(+)H^(-)+H_(2)O RARR MOH+H_(2)`
Discussion
21.

H_(2) gas is liberated when H_(2)O_(2) reacts with

Answer»

Acidified `KMnO_(4)`
`O_(3)`
Acidified `K_(2)Cr_(2)O_(7)`
HCHO

Solution :`H_(2)O_(3)+2HCHO to 2HCOOH+H_(2)`
Discussion
22.

H_(2)+Cl_(2)to 2HCl,Delta H_(1),N_(2) +3H_(2)to 2NH_(3), DeltaH_(2)" and "NH_(3)+3Cl_(2)to NCl_(3)+3HCl,DeltaH_(3) Then calculate the heat of formation of nitrogen trichloride.

Answer»


ANSWER :`DeltaH_(3)+(DeltaH_(2))/2-(3DeltaH_(1))/2`
Discussion
23.

H_(2(g)) + 1/2O_(2(g)) rarr H_2O_((l)) , DeltaH = -286.2KJ H_2O_((l)) rarr H_((aq))^(+) + OH_((aq))^(-) , DeltaH = +57.3 KJ Enthalpy of ionization OH^(-) in aqueons solution is

Answer»

`-228.5KJ`
`+228.5KJ`
`-343.5KJ`
ZERO

ANSWER :A
Discussion
24.

H_2 + CO + R - CH = CH_2to R -CH_2-CH_2 -CHO This reaction is known as

Answer»

HYDROGENATION
Hydroformylation
Carbonation
DECARBOXYLATION

ANSWER :B
Discussion
25.

H_2 + Cl_2 rarr 2HCl , DeltaH_1, N_2 + 3H_2 rarr 2NH_3, DeltaH_2 and NH_3 + 3Cl_2 rarr NCl_3 + 3HCI, DeltaH_3. Then calculate the heat of formation of nitrogen trichloride.

Answer»


ANSWER :`DeltaH2 + DeltaH_2//2 - 3DeltaH-1//2`
Discussion
26.

N_(2) + 3H_(2) rarr 2NH_(3), Delta H = - 46K. Cals. From the above reaction, heat of formation of ammonia is

Answer»

SOLUTION :`Delta H _(3) + Delta H _(2) // 2-3 Delta H _(1) //2`
Discussion
27.

H_(2)" & "D_(2) do not differ in

Answer»

FREEZING point
Boiling point
Bond length
Bond energy

Solution :Metal REACTS with water to liberate HYDROGEN GAS
Discussion
28.

H-undersetunderset(O)("||")C-CNin IUPAC called :-

Answer»

CYANO methanal
2-Oxo ehtane NITRILE
Cyano ETHANAL
Formonitrile

Answer :B
Discussion
29.

H-underset(H)underset(|)(H)overset((a))(……….)Hoverset((b))(-)underset(H)underset(|)(O)….H-underset(H)underset(|)(O) Here a and b are hydrogen bond and covalent bonds , their lengths are

Answer»

`0.97 A^@ ,0.97 A^@`
`1.73 A^@ ,0.97 A^@`
`1.73 A^@ 1.73A^@`
`0.97 A^@ ,1.73 A^@`

ANSWER :B
Discussion
30.

H^(+)+OH^(-)to H_(2)O,DeltaH^(@)=-57Kj//mol DeltaH_("ionistion")^(@)[HCN]=45KJ//mol If 200mL of (1)/(10)M Ba(OH)_(2)solution is mixed in 500mL Of (1)/(10) M HCN solution ,then ,heat evolved will be :

Answer»

600 Joule
240 Joule
12 Joule
480 Joule

Answer :d
Discussion
31.

H^(+) + OH^(-1) to H_(2) O + 13.7 K Cal calculate the enthalpy of neutrilisation reaction of 1 mole H_(2) SO_(4) with base.

Answer»

`13.7` K. CAL
`27.4` K. Cal
`6.85` K. Cal
`3.425` K. Cal

Answer :B
Discussion
32.

'H', 'M' and 'Q' are the aq chlorides of the element 'X', 'Y' and 'Z' respectively. 'X', 'Y' and 'Z' are in the same period of the periodic table. 'Q' gives a white ppt with NaOH but this white ppt dissolves as more NaOH is added. When NaOH is added to 'M' a white ppt forms which does not dissolve when base is added. H does not gives a ppt with NaOH. Which of the following statement are correct? I. The three element are metals II. The electronegativity values decreases from 'X' to 'Y' to 'Z' III. 'X', 'Y'and 'Z' could be Na, Mg and Al IV. The first ionisation energy increases from 'X' to 'Y' to 'Z'.

Answer»

I, II, III
only `I` and `III`
only `II` and `IV`
only `III` and `IV`

ANSWER :B
Discussion
33.

H^(-) is aLewis base. Give reason.

Answer»


Answer :`H^(-)` can donate a pair of ELECTRON. HENCE it is a lewis base.
Discussion
34.

H^(+) ions in aqueous solutions exist as ..........ions.

Answer»


ANSWER :`H_(3)O^(+)` (HYDRONIUM)
Discussion
35.

H-H, X-X and H-X bond energies are 104 Kcal/mole 60Kcal/mole and 101Kcal/ mole. Assuming the electronegativity of hydrogen to be 2.1 the electronegativity of unkonwn element X is (sqrt(19)= 4.36)

Answer»

3.5
3
4
2.5

Answer :B
Discussion
36.

H-atoms in ground state (13.6 eV) are excited by monochromatic radiation of photon of energy 12.1 eV. Find the number of spectral lines emitted in H-atom.

Answer»

2
3
4
5

Answer :B
Discussion
37.

H_((aq))^(+) + OH_((aq))^(-) rarr H_(2)O_((l)) , Delta H = -ve and Delta G= -ve then the reaction is

Answer»

Spontaneous and instantaneous
Spontaneous and endothermic
Spontaneous and SLOW
Non spontaneous ans slow

ANSWER :A
Discussion
38.

H_(2)SO_(4)(aq)+2NaOH_((aq))to Na_(2)SO_(4)(aq) +2H_(2)O_((l)). Suggest the hest of this reaction.

Answer»

SOLUTION :For the FORMATION of one mole of WATER, heat evolved =` 57.3kJ. `For 2 moles of water formed, heat evolved `= 2 xx 57 .3 KJ = 114 .6 KJ.`
The heat of reaction is `=- 114.6 KJ.`
Discussion
39.

H_(2)SO_(4(aq)) + 2KOH_((aq)) rarr K_(2)SO_(4(aq)) + 2H_(2)O_((l)), Delta H for the above reaction is

Answer»

`-13.7`K.Cal
`+57.3`K.J
`-27.4`K.Cal
`-137`K.J

ANSWER :C
Discussion
40.

[H^+]=1.41xx10^(-3) M in 0.08 M solution of HOCl . Then what is the percentage of dissociation of it ?

Answer»

Solution :`{:(,HOCl HARR, H^(+) + , OCl^(-)),(,"0.08 M" ,,),("At equili.",(0.08 - C alpha),c alpha,c alpha),(,=0.08 M,=1.41xx10^(-3)M,):}`
% of DISSOCIATION = `"dissociated [HOCl]"/"Initial [HOCl]"XX100`
`=(1.41xx10^(-3))/0.08xx100`=1.766%
OR
`c alpha=1.41xx10^(-3)` So, `alpha (1.41xx10^(-3))/c`
`=(1.41xx10^(-3))/0.08`=0.017625
% of dissociation = `100 alpha`= 100 x 0.017625
=1.7662% `approx` 1.77%
Discussion
41.

H_2 + Cl_2 rarr 2HCl + 44 K.Cal. Heat of formation of HCl is

Answer»

`-44 K.Cal`
`+44 K.Cal`
`-22 K.Cal`
`+22 K.Cal`

ANSWER :D
Discussion
42.

Gypsum on heating gives :

Answer»

`CaSO_(4)(1)/(2)H_(2)O`
`CaSO_(4)`
`CaO+SO_(3)`
`CaS+O_(2)`

Solution :`CaSO_(4).2H_(2)Ooverset(120-130^(@)C)rarrunderset("Plaster of paris")(CaSO_(4).(1)/(2)H_(2)O)`
`CaSO_(4).2H_(2)Ooverset(gt200^(@)C)rarrunderset("DEAD BURNT GYPSUM")(CaSO_(4))+2H_(2)O`
Discussion
43.

Gypsum on heating to 390 K gives

Answer»

`CaSO_(4) . 2H_(2)O`
`CaSO_(4)`
`CaSO_(4) . 1//2 H_(2)O`
`SO_(3)` and `CaO`

Answer :c
Discussion
44.

Gypsum on heating to 390 K gives.....

Answer»

`CaSO_(4) 2H_(2)O`
`CaSO_(4)`
`CaSO_(4)*(1)/(2)H_(2)O`
`SO_(3) and CaO`

Answer :C
Discussion
45.

Gypsum on heating at higher than 473 K gives anhydrous CaSO_(4), which is known as …

Answer»

KEEN cement
Cement clinker
Portland cement
Dead BURNT plaster

Answer :D
Discussion
46.

Gypsum on heating beyond 200^(@)C. Gives

Answer»

LIME
DEAD BRUNT gypsum
Ca
plaster of paris

Solution :`CaSO_(4),(1)/(2)H_(2)O(or)2CaSO_(4).H_(2)O`
Discussion
47.

Gypsum is heated to 190^(@)C.The percentage loss in it's weight is

Answer»

26.4%
`15.7%`
`20.9%`
`36%`

ANSWER :C
Discussion
48.

Gypsum is used in ___________

Answer»

PLASTER board
surgical splints
soil additive
all the above

Answer :D
Discussion
49.

Gypsum is a hydrate of calcium sulphate 1.0 g of the sample contains 0.791 g of CaSO_(4). How many of CaSO_(4) are there in the sample ? Assuming that the rest of sample is water, how many moles of water are there in the sample ? Show that the result is in consistent with the formula CaSO_(4).2H_(2)O

Answer»

Solution :Mass of `CaSO_(4)=40 +32 + 64 = 136 "g mol"^(-1)`
Molar mass of `CaSO_(4)=40+32+64=136 "g mol"^(-1)`
Mass of water present in 1.0 g of the sample `= 1.0 - 0.791 = 0.209 g`
Moles of `CaSO_(4)=("Mass of "CaSO_(4))/("Molar mass")=((0.791g))/((136"g mol"^(-1)))=5.816xx10^(-3)mol`
Moles of water `= ("Mass of water")/("Molar mass")=((0.209g))/((18 gmol^(-1)))=11.6xx10^(-3)mol`
Ratio of `CaSO_(4) : H_(2)O = 5.816 xx 10^(-3) : 11.6 xx 10^(-3) = 1:2`
`:.` Molecular formula of HYDRATED salt `= CaSO_(4).2H_(2)O`.
Discussion
50.

Groups from 13 to 18 in the periodic table are called p-block elements. Give reason.

Answer»

SOLUTION :(i) The elements whose last electron enters into the p-orbital of the outermost SHELL are having similar properties and thus form a GROUP. The .np orbital of these elements is being progressively filled. Hence, these elements are named as p-block elements.
(ii) The groups of `13^(th)` to `18^(th)` in the periodic table BELONGS to p-block.
Discussion