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 concentration of commercial H_2O_2is 3.125 M than its grade is ______volume.

Answer»

SOLUTION :35 VOLUME
Discussion
2.

The concentration of Ag_2 Cr_2O_7 is 6.5xx10^(-5)M in concentrated solution of Cr_2O_7^(2-)at a temp. then calculate K_2Cr_2O_7

Answer»

SOLUTION :`K_(sp)=[Ag^+]^2 [Cr_2O_7^(2-)]`
`therefore K_(sp)=4S^3=4(6.5xx10^(-5))^3`
`=1.098xx10^(-12)`
Where S=Solubility of `Ag_2Cr_2O_7=6.5xx10^(-5)` M
Discussion
3.

The concentration of Al^(3+)ion in aqueous solution of Al_(2)(SO_(4))_(3) is 0.28 M Then the concentration of SO_(4)^(2-) ion in this solution will be:

Answer»

0.28 M
0.042 M
0.42 M
0.84 M

Solution :Ionisation of `Al_(2)(SO_(4))_(3)` in its aqua solution
`Al_(2)(SO_(4))_(3) overset(H_(2)O)rarr 2Al_((aq))^(3+) + 3SO_(4(aq))^(2-)`
ACCORDING to STOCHIOMETRY of the REACTION `2AL^(3+)`IONS and `3SO_(4)^(2-)` ions are formed.
`SO_(4)^(2-)` ions `=(0.28xx3)/(2) = 0.42 M`
Discussion
4.

The concentration of acetic acid required to get 3.5 xx 10^(-4) mole/lit of H^(+) ion is [K_(a) = 1.8 xx 10^(-5)]

Answer»

`6.8 XX 10^(-3) mol lit^-1`
`6.8 N`
`1.9 4 N`
`1.94 xx 10^(-2) N`

Answer :A
Discussion
5.

The concentration of 500 ML NaOH solution is 0.02 M. How many grams of FeSO_4 added in this solution for precipitation of Fe(OH)_2 ? The K_(sp)of Fe(OH)_2 is 1.5 xx 10^(-15), Molecular mass of Fe(OH)_2 is 152 g "mol"^(-1))

Answer»

SOLUTION :More than `2.85xx10^(-10)` GM
Discussion
6.

The concentration term is used in the calculation of vapour pressure of solution is ……………..

Answer»

SOLUTION :MOLE FRACTION
Discussion
7.

The compund Yb Ba_(2)Cu_(3)O_(7) which shows superconductivity has copper in oxidation stte…………..Assume that the reare earth element ytterbium isin the usual + 3oxidation state

Answer»


Answer :`7//3 overset(+3)ya overset(+2)Ba_(2)overset(X)Cu_(3)overset(-23)O_(7) therefore 3+4+3x-14 or x=7//3`
Discussion
8.

The compressibilty factor for a real gas at high pressure is:

Answer»

`1 + (RT)/(Pb)`
`1`
`1 + (Pb)/(RT)`
`1 - (Pb)/(RT)`

SOLUTION :For 1 mol gas
`(P+ (a)/(V^(2))) (V - b) = RT`
Pressure correction can be NEGLECTED at high pressure.
THUS, `P (V - b) = RT`
`PV = Pb + RT`
`(PV)/(RT) = 1 + (Pb)/(RT)`
`Z = 1 + (Pb)/(RT)`
Discussion
9.

The compression factor (compressibility factor) for one mole of a van der Waals gas at 0^(@)C and 100 atm pressure is found to be 0.5. Assuming that the volume of a gas molecule is negligible, calculate the van der Waals constant a.

Answer»

Solution :`Z=(PV)/(NRT),i.e., 0.5=(100xxV)/(1xx0.082xx273)"or"V=0.1119" L"`
`(P+(a)/(V^(2)))(V-B)=RT` for 1 MOL
Neglecting b, `(P+(PV)/(V^(2)))V=RT "or"PV+(a)/(V)=RT"or" (PV)/(RT)+(a)/(VRT)=1`
or `"" a=(1-(PV)/(RT))VRT=(1-0.5)0.1119xx0.082xx273=1.252" atm "L^(2)mol^(-2)`
Discussion
10.

The compression factor for one mole of real gas at 0^@C and 100 atm is 0.5. Calculate the van der Waals' constant 'a', if 'b' is zero.

Answer»


ANSWER :`1.253 L^2 ATM MOL^(-2)`
Discussion
11.

The compressibility of a gas is less than unity at STP. Therefore,

Answer»

`V_(m) GT 22.4L`
`V_(m) LT 22.4L`
`V_(m)=22.4L`
`V_(m)=44.8L`

ANSWER :B
Discussion
12.

The compressibility of a gas is less than unity at N.T.P.Therefore.

Answer»

`V_(m) GT 22.4 L`
`V_(m) lt 22.4 L`
`V_(m) = 22.4 L`
`V_(m) = 44.8 L`

Solution :`Z = (PV)/(nRT)`
For ideal gas, `V_(m) = 22.4 L` at N.T.P. At a given TEMPERATURE and pressure if `Z` is LESS than unity, then `V_(m) lt 22.4 L`
Discussion
13.

The compressibility of a gas is less than unity at S.T.P

Answer»

`V_(m)gt22.7` LITRES
`V_(m)gt22.7` litres
`V_(m)=22.7` litres
`V_(m)=45.4` litres

Answer :C
Discussion
14.

The compressibility factor (Z) for one mole ofa gas is more than one under S.T.P. conditions. Therefore

Answer»

`V GT 11.2 L`
`V lt 22.38 L`
`V gt 22.38 L`
`V = 22.38 L`

Solution :`Z = (PV)/(NRT)`, if `Z gt 1`
`(PV)/(nRT) gt 1` or `V gt (nRT)/(P)`
`V gt ((1MOL) XX (0.0821 L atmK^(-1) mol^(-1)) xx 273K)/((1atm)) gt 22.4 L`
Discussion
15.

The compressibilityfactor ( z) for a real gas at its Boyle temperature is "_____________".

Answer»

1
0
`GT 1 `
`LT 1 `

ANSWER :A
Discussion
16.

The compressibility factor Z for an ideal gas will be .........

Answer»

`1.5`
`1.0`
`2.0`
zero

Answer :B
Discussion
17.

The compressibility factor of an ideal gas is

Answer»

1.0 
1.5 
2 
Infinity 

Solution :Z for IDEAL GAS = 1
Discussion
18.

The compression factor (compressibility factor) for one mole of a van der Waals gas at 0°C and 100 atmospheric pressure is found to be 0.5. Assuming that the volume of a gas molecule is negligible, calculate the van der Waals constant a.

Answer»

1.253 ATM LIT`""^2MOL^(-2)`
12.53 atm lit`""^2mol^(-2)`
0.125 atm lit`""^2mol^(-2)`
22.53 atm lit`""^2mol^(-2)`

ANSWER :A
Discussion
19.

The compressibility factor for a real gas at high pressure is :

Answer»

`1+ (RT)/(PB)`
1
`1+(Pb)/(RT)`
`1-(Pb)/(RT)`.

SOLUTION :`1+(Pb)/(RT)`
Discussion
20.

The compressibility factor for a real gas at high pressure is

Answer»

1
`1+Pb//RT`
`1-Pb//RT`
`1+RT//Pb`

SOLUTION :For real gases, `(P+(a)/(V^(2)))(V-b)=RT`
At high pressure, `P GT gt a//V^(2)`
Discussion
21.

The compounds which on heating produce at least one polymeric product, are:

Answer»

`Na(NH_(4))HPO. 4H_(2)O`
`NaH_(2)PO_(4)`
`MgHPO_(4)`
`NaHSO_(4)`

Solution :`NaH_(2)PO_(4) overset(DELTA)RARR (HPO_(3))_(n)//(NaPO_(3))_(n)`
Discussion
22.

The compound(s) which have -O-O- bond (s)is / are

Answer»

`BaO_(2)`
`Na_(2)O_(2)`
`CrO_(5)`
`Fe_(2)O_(3)`

Solution :Only peroxide such as `BaO_(2), Na_(2)O_(2) & CrO_(5)` can form peroxy BONDS (-O-O)
Discussion
23.

The compounds which gives the most stable carboniurn ions on dehydrohalogenation is

Answer»

`CH_3I`
`C_2H_5I`
`(CH_3)_3Cl`
`(CH_3)_2CHI`

Solution :Order of STABILITY: tertiary`gt` secondary`gt` PRIMARY
Discussion
24.

The compound(s) which generate(s) N_(2) gas upon thermal decomposition below 300^(@)C is (are)

Answer»

`NH_(4)NO_(3)`
`(NH_(4))_(2)Cr_(2)O_(7)`
`Ba(N_(3))_(2)`
`Mg_(3)N_(2)`

Solution :`NH_(4)NO_(3)` on HEATING gives `N_(2)O` while `Mg_(3)N_(2)` does not decompose below `300^(@)C`. Thus, `(NH_(4))_(2)Cr_(2)O_(7)andBa[N_(3)]_(2)` on thermal DECOMPOSITION give `N_(2)` i.e., options (b,c) are correct.
Discussion
25.

The compound(s) which contain ionic, covalent and coordinate bonds is (are)

Answer»

`H_(2)SO_(4)`
`NH_(4)CI`
`K_(4)[FE(CN)_(6)]`
`CaC_(2)`

Answer :A::B::C
Discussion
26.

The compounds which are used for denaturation of ethyl alcohol are

Answer»

Paphtha
`CH_(3)OH`
Pyridine
Anhy. `CaCl_(2)`

ANSWER :B::C
Discussion
27.

The compound(s) that undergo slower electrophilic substitution than benzene is/are

Answer»





Solution :`-CH = CH_2`, has overall electron WITHDRAWING effect, deactivate the ring. In both (c) and (d) STRONG electron withdrawing REASONANCE effect deactivat ring for ELECTROPHILIC substitution reaction.
Discussion
28.

The compound(s) that direct incoming electrophile predominantly at meta position is/are

Answer»





Solution :Both `-CF_3 and -CN` are META DIRECTION GROUPS. `-CH_2Cl` is deactivating but ortho/para DIRECTING due to hyperconjugation bye benzylic hydrogens. In (d) electrophilic attack occur at ortho/para positions of the ring directly BONDED to oxygen atom.
Discussion
29.

The compounds P,Q and S were separately to nitration using HNO_3//H_2SO_4 mixture.The major product formed in each case respectively is :

Answer»




ANSWER :C
Discussion
30.

The compounds(s) of alkaline earth metals, which are amphoteric in nature is / are

Answer»

BeO
MgO
`Be(OH)^(2)`
`MG(OH)_(2)`

Solution :The compounds of Be such as BeO and `Be(OH)_(2)` are amphoteric due to HIGH POLARISING power
Discussion
31.

The compounds of alkaline earth metals have the following magnetic nature.

Answer»

Diamagnetic
Paramagnetic
Ferromagnetic
Anti ferromagnetic

Answer :A
Discussion
32.

Fill in the blanks The compounds of Al are predominantly ............ due to the ............ size and ............ charge of A1^(3+) ions.

Answer»

SOLUTION :COVALENT , SMALL , HIGH
Discussion
33.

The compounds formed at anode in the electrolysisof an aqueous solution of potassium acetate are…………….

Answer»

`CH_(4)` NAD `H_(2)`
`CH_(4)` and `CO_(2)`
`C_(2)H_(6)` and `CO_(2)`
`C_(2)H_(4)` and `CI_(2)`

Answer :c
Discussion
34.

The compounds CH_3 CH = CH CH_3 and CH_3 CH_2 CH = CH_2 (a) are tautomers (b) are position isomers (c) contain same number of sp^3 sp^3, sp^3 - sp^2 and sp^2 - sp^2 carbon-carbon bonds (d) exist together in dynamic equilibrium (e) are optical isomers

Answer»


ANSWER :B
Discussion
35.

The compounds C_(2)H_(5)OC_(2)H_(5) and CH_(3)OCH_(2)CH_(2)CH_(3) are

Answer»

CHAIN isomers
geometrical isomers
metamers
CONFORMATIONAL isomers

Solution :They are called metamers because there is different in the NATURE of ALKL group.
Discussion
36.

The compounds A and B are mixed in equimolar proportion to from the products, A+B harr C+D. At equilibrium, one third of A and B are consumed. The equilibrium constant for the reaction is

Answer»

0.5
4
2.5
0.25

Solution :
`K_(C)=((a)/(3) XX (a)/(3))/((2a)/(3) xx (2a)/(3))=(1)/(4)`
Discussion
37.

The compounds A, B, C have R or S configurations respectively

Answer»

R,R,S
R,S,R
R,R,R
S,R,S

Answer :3
Discussion
38.

The compound Y is

Answer»

`BCl_(3)`
`BF_(3)`
`B_(2)H_(6)`
`B_(2)O_(3)`

Answer :C
Discussion
39.

The compound Y in the above sequence is

Answer»

2-Methyl-2-phenyl-1-propanol
2-Phenyl-2-propanol
Acetophenone
2-Methyl-1-Phenyl-2-propanol

Solution :
Discussion
40.

The compound (X) on heating gives a colourless gas and a residue that is dissolved in water to obtain (B). Excess of CO_(2) is bubbled through aqueous solution of B, C is formed. Solid (C) or heating gives back X. (B) is …......

Answer»

`CaCO_(3)`
`Ca(OH)_(2)`
`Na_(2)CO_(3)`
`NaHCO_(3)`

SOLUTION :`CaCO_(3)OVERSET(Delta)toCaO+CO_2`
`CaO+H_2OtoCa(OH)_2 , Ca(OH)_2+CO_2to CaCO_3+H_2O`
Discussion
41.

The compound (X) on heating gives a colourless gas a nd a residue that is dissolved in water to obtain (B). Excess of CO_2 is bubbled through aqueous solution of B, C is formed. Solid (C) on heating gives back X. (B) is

Answer»

`CaCO_(3)`
`CA(OH)_(2)`
 `Na_(2) CO_(3)`
 `NaHCO_(3)`

Solution :`caCO_(3)overset(triangle)toCaO+CO_(2)`
`CaO+H_(2)OtoCa(OH)_(2)`
`Ca(OH)_(2)+CO_(2)toCaCO_(3)+H_(2)O`
Discussion
42.

The compound "X" is used as

Answer»

INSECTICIDE 
SOLVENT 
REFRIGERANT 
MOTOR fuel

Answer :A
Discussion
43.

The compound 'X' is

Answer»




SOLUTION :
Discussion
44.

The compound X is

Answer»

`C Cl_(4)`
`CO_(2)`
`CH_(3)OH`
`CO`

ANSWER :B
Discussion
45.

The compound X is

Answer»

`B_(2)H_(6)`
`[B(OH)_(4)]^(-)`
`B_(2)O_(3)`
`H_(2)B_(4)O_(7)`

Answer :C
Discussion
46.

The compound X in the reaction is :

Answer»




ANSWER :B
Discussion
47.

The compound with the lowest oxidation state of iron is:

Answer»

`Fe_(2)O`
`Cr^(2+)`
`Ti^(4+)`
`MN^(2+)`

Answer :A
Discussion
48.

The compound with the highest boiling point is

Answer»

n-hexane
n-pentane
2,2-dimethylpropane
2-methylbutane.

Answer :A
Discussion
49.

The compound with molecular formula C_8H_10which will give only two isomers on electrophilic substitution with Cl_2//FeCl_3or with HNO_3//H_2SO_4is

Answer»

p-dimethylbenzene
m-dimethylbenzene
o-dimethylbenzene
ethylbenzene

Solution :Since `CH_3` groups are o,p-directing , therefore , o-dimethylbenzene (i.e., o-xylene) gives only two ISOMERIC products on chlorination or NITRATION.
Discussion
50.

The compound with molecular formula, C_(6)H_(14) has two tertiary carbons. Its IUPAC name is

Answer»

n-hexane
2-methylpentane
2, 3-dimethylbutane
2, 3-dimethylpentane

Solution :2, 3-Dimethylbutane as SHOWN in ANS 15 has TWO `3^(@)` carbon atoms.
Discussion