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.

Calorific value is in the order

Answer»

FATS GT CARBOHYDRATES gt PROTEINS
Carbohydrates gt Fats gt Proteins
Proteins gt Carbohydrates gt Fats
Fats gt Proteins gt Carbohydrates

SOLUTION : Fats gt Carbohydrates gt Proteins
Discussion
2.

Calomel reacts with ammonium hydroxide to form :

Answer»

`HG(NH_2)CL`
`H_2N-Hg-Hg-Cl`
`Hg_2O`
Hgo

Answer :A
Discussion
3.

Calomel is the name of :

Answer»

`HgCl_(2)`
`Hg_(2)Cl_(2)`
`HgCl_(2)+Hg`
`Hg_(2)Cl_(2)+Hg`

Solution :`underset("acid")(ZN(OH)_(2))+underset("Base")(2OH^(-)) to underset("Salt")(ZnO_(3)^(2-)+underset("WATER")(2H_(2)O)`
`underset("Base")(Zn(OH)_(2))+underset("Acid")(2H^(+)) to underset("Salt")(Zn^(2+)+underset("Water")(2H_(2)O)`
Discussion
4.

Calomel electrode is reversible to

Answer»

`HG^(+2)` ions
Hg
`Hg_(2)Cl_(2)`
`CL^(-)` ions

Solution :`Hg_2Cl_2+2e^(-) to 2Hg_((s))+2CL^(-)`
Discussion
5.

Calomel electrode is an example of

Answer»

indicator electrode
primary reference electrode
secondary reference electrode
gas electrode

Solution :The EMF of calomel electrode is to be determined FIRST CONNECTING it with order reference electrode.
Discussion
6.

Calomal is :

Answer»

`Hg_2Cl_2` and HG
`HgCl_2`
Hg + `HgCl_2`
`Hg_2Cl_2`

ANSWER :D
Discussion
7.

Can all the disaccharides are reducing sugars ?

Answer»

Solution :Sucrose is a non reducing sugar. This is because glycosidic linkage takes place between anomeric HYDROXYL ofglucose `(C_1)` and anomeric hydroxyl GROUP of FRUCTOSE `((C_1))` and anomeric hydroxyl group of fructose `(C_2)`
Maltose and LACTOSE are examples of reducing sugars.
Discussion
8.

Caliche is:

Answer»

CRUDE saltpetre
Impure nitre
Impure carnallite
Ashes of SEA weeds

Answer :A
Discussion
9.

Calgon(a water softener) is:

Answer»

`Na_2[Na_4(PO_3)_6]`
`Na_2[Na_4(PO_3)_4]`
`Na_2[Na_2(PO_3)_6]`
`Na_4[Na_2(PO_4)_6]`

ANSWER :A
Discussion
10.

Calgon used as a water softener is

Answer»

`Na_(2)[Na_(4)(PO_(4))_(5)]`
`Na_(4)[Na_(2)(PO_(3))_(6)]`
`Na_(2)[Na_(4)(PO_(3))_(6)]`
`Na_(4)[Na_(4)(PO_(4))_(6)]`

SOLUTION :CALGON is `Na_(2)[Na_(4)(PO_(3))_(6)].`
Discussion
11.

Calgon is an industrial name given to :

Answer»

NORMAL sodium phosphate
Sodium METAL aluminate
Sodium hexa MATA phosphate
Hydrated sodium ALUMINIUM silicate

Answer :C
Discussion
12.

Calgon causes the softening of hard water by

Answer»

perecipitating `Ca^(2+)` and `Mg^(2+)` IONS as phosphates
PRECIPITATING `Ca^(2+)` and `Mg^(2+)` ionsas suphates
SEQUESTRATION of `Ca^(2+)` and `Mg^(2+)` ion
sequestration of `CI^(-)` and `SO_(4)^(2-)` ion

ANSWER :D
Discussion
13.

Calculte the number of molecules of molecules of oxalic acid in 100 mL of 0.02 N oxalic acidsoliution.

Answer»


Solution :Step I. Calculation of no. of moles of oxalic acid per litre of the solution.
`"Molarity"=("No. of moles of oxalic acid")/("Volume of solution in litre")`
`(0.01" MOL L"^(-1))=("No. of moles of oxalic acid")/(1L)`
No. of moles = 0.01 mol.
Step II. Calculation of no. of molecules of oxalic acid per 100 ML of solution.
1000 mL of oxalic acid solution contains moles = 0.01
`"100 mL of oxalic acid solution contains moles"=(0.01)/(1000)xx100=0.001 mol`
`therefore "Molecules of oxalic acid"=0.001xxN_(0)=10^(-3)xx6.022xx10^(23)=6.022xx10^(20)" molecules".`
Discussion
14.

Calculation the mole fraction of ethylene glycol (C_(2)H_(6)O_(2)) and water in a solution containing 20% of C_(2)H_(6)O_(2) by mass.

Answer»

Solution :`20%` of `C_(2)H_(6)O_(2)` by MASS means that 20 g of `C_(2)H_(6)O_(2)` are PRESENT in 100 g of the solution, i.e.,
Mass of solute `(C_(2)H_(6)O_(2))=20G",Mass of solvent "(H_(2)O)=100-20g=80g`
`"MOLAR mass of "C_(2)H_(6)O_(2)="62 g mol"^(-1)",Molar mass of "H_(2)O="18 g mol"^(-1)`
`therefore"No. of moles of "C_(2)H_(6)O_(2)=(20)/(62)=0.322",No. of moles of "H_(2)O=(80)/(18)=4.444`
`therefore"Mole fraction of "C_(2)H_(6)O_(2)" in the solution "=(n_(C_(2)H_(6)O_(2)))/(n_(C_(2)H_(6)O_(2))+n_(H_(2)O))=(0.322)/(0.322+4.444)=0.068`
Mole fraction of `H_(2)O`in the solution `=1-0.068=0.932`
Discussion
15.

Calculation the molarity and normality of a solution containing "9.8 g of "H_(2)SO_(4) in "250 cm"^(3) of the solution.

Answer»

Solution :Mass of `H_(2)SO_(4)` dissolved = 9.8 g,`"Volume of the solution = 250 cm"^(3)="0.250 L"`
Calculation of molarity : MOLAR mass of `H_(2)SO_(4)="98 g MOL"^(-1)`
`therefore"No. of moles of "H_(2)SO_(4)=("Mass in g")/("Molar mass")=("9.8 g")/("98 g mol"^(-1))="0.1 mole"`
Calculation of normality : Eq. mass of `H_(2)SO_(4)=("Mol. mass of "H_(2)SO_(4))/("Basicity of "H_(2)SO_(4))=(98)/(2)=49`
`therefore"No. of g equivalent of "H_(2)SO_(4)=("Mass in g")/("Eq. mass")=(9.8)/(2)=0.2`
`"Noramlity "=("No. of g eq of the SOLUTE")/("Volume of solution in LITRES")=("0.2 g eq")/("0.250 L")="0.8 g eq L"^(-1)="0.8 N".`
Discussion
16.

calculateth resonance energy of N_(2)O Delta_(f)H^(-) of N_(2)O = 82 kJ mol ^(-1) bond energy of N=O = 607 kJmol ^(-1) bond energy of O=O = 498 kJ mol ^(-1) nond energy of N = N = 418 kJ mol^(-1) bond energy of N= N = 946 kJ mol ^(-1)

Answer»

`82 kJ mol ^(-1)`
`-88 kJ mol ^(-1)`
`- 82 kJ mol ^(-1)`
`+ 88 kJ mol ^(-1)`

Solution :`NequivN(g) + 1/2 O_(2)(g)to N=N=0`
calculated value of `Delta_(F)H^(Theta)=`
`Delta_(f)H^(-)=[BE(NequivN)+1/2BE(=O)]`
`[BE(N=N)+BE(N=O)]`
`[946+1/2(498)]- [418+607]=170 kJ mol^(-1)`
Respmace energy = observed `Delta_(f)H^(Theta)` - calculated `Delta_(f)H^(Theta)`
= 82 -170 =-88 kJ `mol^(-1)`
Discussion
17.

Calculated the mass urea (H_2NCONH_2)required to dissolved in 2.5 kg of water to form a 0.25 molal aqueous solution.

Answer»

Solution :Molar mass of urea = `60G mol^(-1)`
Molality (m) =(mass of urea)/(Molar mass of urea xxMass of WATER in kg)
`THEREFORE` Mass of urea = (Molalityxx Molar lass of urea xxMass of water in kg)
= `0.25xx 60 XX 2.5 =37.5g`
Discussion
18.

Calculated the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0g of polymer of molar mass 185000 in 450 mL of water at 37^@C.

Answer»

Solution :`pi= (W_2RT)/(M_2V)`
`W_2 = 1.0g1, T=273+37=310K`
`M_2` = 185000 V = 450mL = 0.450 L
R = 0.083 L bar `mol^(-1) K^(-1)`
`THEREFORE pi = (1xx0.083xx310)/(185000xx0.45) = 0.000309`bar
1 bar = `10^5` pascal
OSMOTIC pressure, `pi = 0.000309xx10^5 = 30.9 Pa`
Discussion
19.

Calculated spin only magnetic moment of Cr^(x+) is 4.9 BM. Find the 'x' value.

Answer»

Solution :Magnetic moment `mu= 4.9 BM = sqrt(n(n+2))`
Number of unpaired ELECTRONS `= n = 4`
with FOUR unpaired electrons, the configuration is `3d^(4)4s^(0)`.
`Cr^(2+)` ion posses 4 unpaired electrons, THUS .X. value is 2.
Discussion
20.

Calculate % yeild of the reaction if 200 g KHCO_(3) produces22 g of CO_(2) upon strong heating.

Answer»


ANSWER :`050`
Discussion
21.

Calculate x+y+z for H_(3)PO_(3) acid, where x is no. of lone pairs, y is no. of sigma bonds and z is no. of pi bonds.

Answer»


ANSWER :13
Discussion
22.

Calculate workdone in adiabatic compression of one mole of an ideal gas (monoatomic) from an initial pressure of 1atm to final pressure of 2 atm. Initial temperature = 300 K. (a) If process is carried out reversibly (b) If process is carried out irreversible against 2 atm external pressure. Computer the final volume reached by gas in two cases and describe the work graphically.

Answer»

Solution :(a) `T_(2)=395.8,V_(2)=16.24L,w_("REV")=1194.72J`,
`V_(2)^(')=17.24L,T_(2)^(')=420K,w_("irrev")=1496.52J`
Discussion
23.

Calculate work done for an ideal gas( In 2=0.7)

Answer»


SOLUTION :N//A
Discussion
24.

Calculate wedge_(m)^(0)" for "CaCl_(2) and MgSO_(4). lambda_(Ca^(2+))^(0)=119.0S cm^(2)" mol"^(-1) lambda_(Cl^(-))^(0)=76.3S cm^(2) mol^(-1) lambda_(Mg^(2+))^(0)=106 Scm^(2) mol^(-1) lambda_(SO_(4)^(2-))^(0)=160 Scm^(2) mol^(-1)

Answer»

Solution :`wedge_(m)^(0) (MgSO_(4))=lambda_(Mg)^(0)+lambda_(SO_(4)^(2-))^(0)`
`=106+160=266 S cm^(2) MOL^(-1)`
`wedge_(m)^(0) (CaCl_(2))=lambda_(Ca^(2+))^(0)+2lambda_(Cl^(-))^(0)=119+2 xx 76.3 =271.6Scm^(2) mol^(-1)`
Discussion
25.

Calculatew=?(In 2=0.7)

Answer»


SOLUTION :N//A
Discussion
26.

Calculate wedge_(m)^(@) for CaCl_(2) and MgSO_(4) from the following data: lamda^(@) (S cm^(2)mol^(-1)):Ca^(2+)=119.0,Mg^(2+)=106.0,Cl^(-)=76.3,SO_(4)^(2-)=160.0.

Answer»

Solution :By Kohlrausch law, `wedge_(m)^(@)(CaCl_(2))=lamda_(Ca^(2+))^(@)+2lamda_(Cl^(-))^(@)=119.0" S "cm^(2)MOL^(-1)+2xx76.3" S "cm^(2)mol^(-1)`
`=(119.0+152.6)" S "cm^(2)mol^(-1)=271.6" S "cm^(2)mol^(-1)`
`wedge_(m)^(@)(MgSO_(4))=lamda_(Mg^(2+))^(@)+lamda_(SO_(4)^(2-))^(@)=106.0" S "cm^(2)mol^(-1)+160.0" S "cm^(2)mol^(-1)=266" S "cm^(2)mol^(-1)`.
Discussion
27.

Calculate wavelength of photon emitted when an electron goes from n=3 to n=2 level of hydrogen atom.

Answer»


ANSWER :656 NM
Discussion
28.

Calculate W and DeltaU for the conversion of 1 mole of water into 1 mole of steam at a temperature of 100^(@)C and at a pressure of 1 atmosphere. Latent heat of vaporisation of water is 9720 cal/"mole".

Answer»

SOLUTION :p=1 atm
=76 cm
=76xx13.6xx981 "dynes//cm"^(2)`
=1.013xx10^(6) "dymes//cm"^(2).
`V_(1)`= volume of 1 "mole" of water at `100^(@)C=18` mL.
`V_(2)` = volume of 1 "mole"of steam at `100^(@)C`
`=(373)/(273)xx22400=30605 mL`. (Charles.s law)
Now we have,
`W=-p(V_(2)-V_(1))` ...(Eqn.2)
`=-1.013xx10^(6)XX(30605-18) ergs
`=(-1.013xx10^(6)xx30587)/(4.18xx10^(7))"calories"`
=-741 "calories".
Again we have,
`DeltaU=q+W`
`=9720+(-741)`
`=8979 "calories"`.
Discussion
29.

Calculate w and DeltaU for the conversion of 0.5 mole of water at 100^(@)C to steam at 10 atm pressure. Heat of vaporisation of water at 100^(@)C is 40670Jmol^(-1)

Answer»

Solution :Volume of 0.5 MOLE of steam at 1 atm pressure
`=(nRT)/(P)=(0.5xx0.0821xx373)/(1.0)=15.3L`
Change in volume `=Vol. ` of steam-vol. of water
`=15.3-"negligible"=15.3L`
WORK done by the system
`W=P_(ext)xx`volume change
`=1xx15.3=15.3` litre-atm
`=15.3xx101.3J=1549.89J`
'w' shoud be negative as the work has been done by the system on the surroundings.
`w=-1549.89J`
heat requird to convert 0.5 MOLEOF water at `100^(@)C` to steam
`=0.5xx40670J=20335J`
Accoridng to first LAWOF thermodynamics.
`DeltaU=q+w=20335-1549.89=18785.11J`
Discussion
30.

Calculate volume of O_(2) liberatd when 2.5 A current is passed for 1 hour through aqueous solution of Na_(2)SO_(4) having inert electrode at 1 bar pressure and 300 K temperature. [Volume of 1 mole of O_(2) at STP=22 Litre]

Answer»


ANSWER :`573.93mL`
Discussion
31.

Calculate volume of H_(2) gas kept at STP if it contains as many H atoms as in 98 gm H_(3)PO_(4). [atomic mass of p =31 ]

Answer»


ANSWER :`34.05`L
Discussion
32.

Calculate vapour pressure of a 5% (by weight ) solution of water in glycerol (mol.wt.92.1) at 100^(@)C, assuming Raoult's law to be valid and neglecting the vapour pressure of glycerol.

Answer»

Solution :Since pure water boils at `100^(@)C` and b.p. is defined as that temperture at which the vapour pressure of the liquid is equal to atmospheric pressure , i.e. 760MM, the vapour pressure of pure water at `100^(@)C` is 760mm.
In this problem since glycerol `(92.1)` is the solute and water is the solvent and their weight PER cents are `95%` and `5%` respectively, Raoult.s law cannot be applied as the solution is not at all DILUTE. But according to the question, we apply Raoult.s law
`(p^(0)-p)/(p^(0))=(n_(1))/(n_(1)+n_(2))`............[Eqn.1)
`{:("Here" p^(0)=760mm,,n_(1)="mole of glycerol"=(95)/(92.1)),(,,n_(2)="mole of water"=(5)/(18)):}}"SUPPOSING the weight of the solution as 100g"`
SUBSTITUTING in Eqn. (1), we get `p=161.2mm`
Discussion
33.

Calculate value of (X+Y+Z)/(10), here X is O-N-O bond angle in NO_(3)^(-) Y is O-N-O bond angle in NO_(2)^(+) and Z is F-Xe-F adjacent bond angle in XeF_(4).

Answer»


Solution :`X=120(sp^(2)` hybridization)
Y=180 (sp -hybridization)
`Z=90(sp^(3)d^(2)` hybridization)
`(120+180+90)/(10)=(390)/(10)=39`
Discussion
34.

Calculate value of "x+y" if x is the total number of sigma bonds and y is total number of pi bonds in ligand EDTA and phenanthrolene. (EDTA Ethylene diamine tetraacetate, Phen=1,10-N,N-phenanthrolene)

Answer»


ANSWER :5
Discussion
35.

Calculate value of |x-y|, here x and y are the total number of bonds in benzene and benzyne respectively which are formed by overlapping of hybridized orbitals.

Answer»


SOLUTION :
Discussion
36.

Calculate value of |x-y|, here x and y are the total number of bonds in benzene and benyne respectively which are formed by overlapping of hybridized orbitals .

Answer»


SOLUTION :
x=12, y=11
|x-y|=1
Discussion
37.

Calculate value of -U/100, for AB(s), from following data of Born-Haber's cycle. [where U is lattice energy in kJ/mol]

Answer»


SOLUTION :`-424=108+495+(242)/(2)-348+U`
`-424=+376+U`
`U=-424-376=-800kJ`/mol
`-((-800))/(100)=8`
Discussion
38.

Calculate total number of possible isomers (structrural and stereoisomers) of the compound [Ir(en)_2(NH_3)(NO_2)]^(2+)

Answer»


SOLUTION :
Discussion
39.

Calculate total number of orbitals having (n+l) value=8 and magnetic quantum number a nonzero quantity. [Divide answer by 6]

Answer»

SOLUTION :`(n+l)"VALUE"=8rArr 5f,6d,7p,8s`
Total orbitals=16
Orbital having non-zero magnetic QUANTUM
number =16-4=12
Discussion
40.

Calculate total number of electron present in Cr having n + l + m = 6

Answer»

3
1
5
6

Solution :`Cr_(24) = 1s^(2) 2P^(6) 3s^(2) underset(3 + 1 + 1 = 5)underset(darr)(3p^(6)) "" underset(("ONE 'd' orbital"))underset(3 + 2 + 1 = 6)underset(darr)(3D^(5)4s^(1))`
Number of electrons having `n + L + m = 6` from `3d^(5) = 1`
Total `=1`
Discussion
41.

Calculate to the correct number of significant figures : 4.26 - (15.635/5.0)

Answer»

1.13
1.2
1.1
1.133

Answer :B
Discussion
42.

Calculate thewavelength of carbondioxide molecule whose velocity is 440ms^(-1).

Answer»

SOLUTION :`2XX10^(-11)m`
Discussion
43.

Calculate the % yield of monochloro products when isopentane is chlorinated, relative rates of reasction of reaction is 3^(@)H-5.0,2^(@)H-3.8and1^(@)H-1.0

Answer»

SOLUTION :
Discussion
44.

Calculate the Zn-Cu cell (Daniell) potential.

Answer»

Solution :`(E_(Zn^(2+)|Zn)^(THETA)=-0.76V and E_(Cu^(2+)|Cu)^(Theta)=0.34V)`
According to the convention, anode half-cell has less REDUCTION potential and CATHODE half-cell has high reduction potential so, oxidation reaction on the zinc electrode is available on left side of cell is as follows:
`Zn_((S))toZn_((aq," 1M")^(2+)+2e^(-)`. . . (i)
Reduction reaction on the copper electrode available on right side of cell is as follows:
`Cu_((aq," 1M"))^(2+) +2e^(-) to Cu_((S))` . . . (ii)
Sum of reaction (i) and (ii) gives redox reaction which is as follows:
`Zn_((S))+Cu_((aq," 1M"))^(2+)=Zn_((aq," 1M"))^(2+)+Cu_((S))`
Now emf of this cell `=E_(cell)^(Theta)=E_(R)^(Theta)E_(L)^(Theta)`
`THEREFORE DeltaE_(cell)^(Theta)=-0.34V-(-0.76V)=1.10V`
Discussion
45.

Calculate the work performed when 2 moles of hydrogen expand isothermally and reversibly at 25^(@)C from 15 to 50 litres.

Answer»

SOLUTION :We have,
`W=-2.303"nRT"log.(V_(2))/(V_(1))`…. (EQN. 4b)
`=-2.303xx2xx2xx298xxlog.(50)/(15)`
`=-1436 calories.
Discussion
46.

Calculate the work of an isothermal reversible expansion of three moles of watervapour from 5.0 xx 10^4 to 2.0 xx 10^4 Pa at 330 K.

Answer»

SOLUTION :`W = -2.303 NRT LOG p_1/p_2`
=-7.54kJ
Discussion
47.

Calculate the work of an isobaric reversible expansion of three moles of an idealgas while it is heated from 298 K to 400 K.

Answer»

SOLUTION :`W = -NR(T_2 - T_1)`
` -2.45kJ`
Discussion
48.

Calculate the work done when 1 mol of an ideal gas is compressed reversible from 1 bar to 4 bar at a constant temperature of 300 K.

Answer»

4.01 kJ
`- 8.02 kJ`
18.02 kJ
3.458 kJ

Solution :`w= - 2.303 NRT "log" (p_(1))/(p_(2))`
`= - 2.303 xx 1 xx 8.314 xx 300 "log" (1)/(4)`
`3458.3 J = 3.458 kJ mol^(-1)`
Discussion
49.

Calculate the work done during combustion of 0.138 kg of ethanol, C_(2)H_(5)OH(l) at 300 K. Given : R=8.314 JK^(-1)mol^(-1), molar mass of ethanol =46g mol^(-1)

Answer»

`-7482J`
`7482J`
`-2494J`
`2494J`

SOLUTION :The CONBUSTION of ethanol, involves the following reaction .
`C_(2)H_(5)(L)+3O_(2)(g)to2CO_(2)+3H_(2)O`
Given,
Mass of ethanol `=0.138kg=138g`
Temperature `=300K`
`R=8.314JK^(-1)mol^(-1)`
Molar mass of ethanol `=46gmol^(-1)`
No. of moles of ethanol `=("Mass of ethanol")/("Molar mass of ethanol")`
`=(138)/(46)=3`
WORK done (W) during combustion of 0.138 Kg of `C_(2)H_(5)OH=nRT`
`W=nRT`
`W=3xx8.31JK^(-1)mol^(-1)xx300K`
`W=7482.6J`
`~~7482J`
Discussion
50.

Calculate thework done during combustion of 0.138 kg of ethanol, C_(2)H_(5)OH_(1) at 300 K. Given: R= 8.314 JK^(-1) mol^(-1) molar mass of ethanol = 46 gm mol^(-1)

Answer»

`-7482` J
7482 J
`-2494` J
2494 J

Solution :`underset(0.138 kg) underset(darr)(C_(2)H_(5)OH) +underset(3)underset(darr)(3O_(2)) to underset(2)underset(darr)(2CO_(2)) +underset(3)underset(darr)(3H_(2)O)`
Hence Mole `=138/46 =3 `moles of `C_(2)H_(5)OH`.
`3C_(2)H_(5)OH(l) + 9O_(g) to 6CO_(2)(g) + 9H_(2)O (u)`
`deltan = 6-9 =-3`
WORK =`-deltanRT`
`=-(-3) xx 8.314 xx 300`
`=-7482` J
Discussion