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.

At times NH_(4)OH is added before adding (NH_(4))_(2)CO_(3) to precipitate group V cations explain.

Answer»

SOLUTION :Ammonium hydroxide reats with bicarbonate present as impurity in `(NH_(4))_(2)CO_(3)` as follows:
`NH_(4)HCO_(3)+NH_(4)Ohto(NH_(4))_(2)CO_(3)+H_(2)O` ltbr. The presence of bicarbonate is undesirable as the bicarbonates of Ba, Sr and Ca are soluble in WATER.
Discussion
2.

At times warming is suggested while precipitating group V cation explain.

Answer»

Solution :The IMPURITY of bicarbonate ion in `(NH_(4))_(2)CO_(3)` fomrs decomposes these bircarbonates into normal carbonates which are precipitated
`Ca(HNO_(3))_(2)toCaCO_(3)+H_(2)O+CO_(2)`
the solution MUST not be boiled otherwise insoluble carbonates with ammonium chloride are converted into soluble chlorides.
Discussion
3.

At the top of a mountain a thermometer reads 0^@Cand a barometer reads 710 mmHg. At the bottom of the mountain the temperature is 30^@Cand the pressure is 760 mmHg. Compare the density of the air at the top with that at the bottom

Answer»

SOLUTION :`1.04:1`
Discussion
4.

At the time of Independence, Indian leaders were committed to the aims of Liberty, Equality, Fraternity and_________.

Answer»

SOLUTION :DEMOCRACY
Discussion
5.

At thetemperaturecorrespondingtowhich ofthepointswill bereducedto Febycouplingthereaction 2 FeOto2 Fe +O_2will al ofthefollowingreactions ?(i)C +O_ 2toCO_2 (ii)2 C+ O_ 2to2 COand(iii)2 CO+O_2to2 CO_2

Answer»

POINTA
Point B
Point D
Point E

Solution :At pointsB andE, `Delta_ fG ^(@)""_((C, CO_ 2)) , DELTA _ f G ^(@) ""_((CO,CO_2)),Delta _f G ^(@) ""_((C,CO)) `curvesall liebelow` Delta_fG ^(@)""_((Fe, FeO )) `curve,therefore,at pointsB andE,FeOwill bereducedto Febyallthethree reactions , i.e.,options (b) and(d) arecorrect.
Discussion
6.

Atthetemperatureabove1073Kcokecanbeused toreduceFeOto Fe.Howcan you justify thisreductionwithEllinghamdiagram ?

Answer»

Solution : Attemperaturesabove1073 K,` Delta_f G^(@) ` (C,CO) curveliesbelow`Delta _ f G^(@)(Fe,FeO) `CURVE , i.e.,` Delta_ fG^(@) (C,CO ) ltDelta _fG^(@)(Fe, FeO ) `,THEREFORE, cokecan reduceFeOtoFe.
Discussion
7.

At the temperature corresponding to which of the points in Figure, FeO will be reduced to Fe by coupling the reaction 2FeO to 2Fe + O_2 with all of the following reactions ? (1) C + O to CO_2(2) 2C + O_2 + 2CO and (3) 2CO + O_2 to 2CO_2

Answer»

POINT A
Point B
Point D
Point E 

Solution :At point B and point E, the values of `DeltaG^@` for the given oxidation reactions is LOWER than that of REDUCTION reactions.
Discussion
8.

At the sublimation temperature, for the process CO_(2(s))rarrCO_(2(g))

Answer»

`DeltaH,DELTAS and DeltaG` are all positive
`DeltaH gt 0, DeltaS gt 0 and DeltaG lt 0`
`DeltaH lt 0, DeltaS gt 0 and DeltaG lt 0`
`DeltaHgt0, DeltaSgt0 and DeltaG=0`

SOLUTION :Since the process is at EQUILIBRIUM `DeltaG=0 " for "DeltaG=0`, they should be `DeltaH gt 0, DeltaS gt 0`.
Discussion
9.

At the same temperature which of the following solutions will be isotonic?

Answer»

3.42 GM of sucrose PER liter of water and 0.18 gm glucose per LITRE of water.
3.42 gm of sucrose per litre and 0.18 gm glucose in 10 litre of water.
3.42 gm of sucrose per litre of water and 0.585 gm of sodium chloride per litre of water.
3.42 gm of sucrose per litre of water and 1.17 gm of sodium chloride per litre of water.

Answer :B::C::D
Discussion
10.

At the same temperature, following solution will be isotonic

Answer»

3.24 g of SUCROSE per litre of water and 0.18 gm glucose per litre of water
3.42 gm of sucrose per litre and 0.18 gm glucose in 0.1 litre water
3.24 gm of SURCOSE per litre of water and 0.585 gm of sodium CHLORIDE per litre of water
3.42 gm of sucrose per litre of water and 1.17 gm of sodium chloride per litre of water

ANSWER :B
Discussion
11.

At the same temperature and pressure, which of the following gases will have highest kinetic energy per mole ?

Answer»

HYDROGEN
oxygen
methane
all have same.

ANSWER :D
Discussion
12.

At the same temperature and pressure, which of the following gas will be adsorped in more proportion ?

Answer»

`Cl_(2)`
`N_(2)`
`H_(2)`
`NH_(3)`

SOLUTION :`NH_(3)`
Discussion
13.

At the same tempeature , hydrogen is more soluble in water than helium. Which of them have a higher valuem of K_(H) and why?

Answer»

SOLUTION :As `p_(A)=K_(H)x_(A)`. Thus, at constant temperature, for the same partial pressure of different GASES, `x_(A)=1//K_(H)`. In other words, SOLUBILITY is inversely proportional to Henry's constant of the gas. Higher the value of `K_(H)`, lower is the solubility of the gas.
As `H_(2)` is more soluble than helium, `H_(2)` will have lower value `K_(H)` than thatof helium.
Discussion
14.

At the same external pressure , more the temperature lesser is the solubility of a gas in water.

Answer»
Discussion
15.

At the same constant temparature ,more the external pressure ,more is the solubility of a gas in water.

Answer»
Discussion
16.

At the same conditions of pressure, volume and temperature, work done is maximum for which gas if all gases have equal masses?

Answer»

`NH_(3)`
`N_(2)`
`Cl_(2)`
`H_(2)S`

SOLUTION :When p,V and T are same and mass is also same, WORK done depends only UPON molecular mass.
`W prop (1)/(M)` (where M=molecular mass)
Among the given gases, `NH_(3)` has lowest molecular mass, so work done is MAXIMUM for it.
Discussion
17.

Atthe root mean square (rms) speed of a gas X (molecular weightis equal to the most probable speed of gas Y atThe molecular weight of the gas Y is :

Answer»

<P>

Solution :Given, `u_(rm s) = u_(m p) rArr sqrt((3RT)/(M(X))) = sqrt((2RT)/(M(Y))) rArr (3R xx 400)/(40) = (2R xx 60)/((M_(Y)) rArr M (Y) = 4`
Discussion
18.

At the limiting value of radiusratio r_(+)//r_(-)

Answer»

FORCES of ATTRACTION are larger than the forcesof repulsion
Forces of attraction are smaller than the forces of repulsion
Forces of attraction andrepulsion are just equal
None of these

Solution :At the limiting VALUE of `r^(+)//r^(-)`,forces of attraction and repulsion are equal.
Discussion
19.

At the Nagnal fertilizer plant in Punjab, hydrogen is produced by the electrolysis of water. The hydrogen is used for the production of ammonia and nitric acid (by oxidation of ammonia). If the avrage production of ammonium nitrate is 5000 kg/day, estimate the daily consumption of electricity per day.

Answer»

SOLUTION :`2.8 XX 10^(5)` amp/day
Discussion
20.

At the high pressure Langmuir adsorption isotherm takes the form

Answer»

`(x)/(m) =(AP)/(1+bp)`
`(x)/(m) = (a)/(B)`
`(x)/(m) =ap`
`(m)/(x) = (b)/(a) +(1)/(ap)`

Solution :At HIGH pressure 1 is NEGLECTED in the denominater of langmuir equation `(x)/(m) = (ap)/(1+bp)`
Discussion
21.

At the isoelectric point for amino acid the species present are:

Answer»




ANSWER :D
Discussion
22.

At the given point of intersection of the two curves shown concentration of B is given by ………for, A=3B

Answer»

`4//3A_(0)`
`3//4A_(0)`
`1//3A_(0)`
`1//4A_(0)`

ANSWER :B
Discussion
23.

At the equlibrium position in the process of adsorption……….

Answer»

`DeltaHgt0`
`DeltaH=TDeltaS`
`DeltaHgt T DELTAS`
`DeltaHlt T DeltaS`

Solution :At EQUILIBRIUM, `DeltaG=0.` Putting in `DeltaG=DeltaH-T DELTA S,` we get `DeltaH=T DeltaS.`
Discussion
24.

At the equilibrium position in the process of adsorption _________ .

Answer»

`DELTAH GT 0`
`DeltaH = TDeltaS`
`DeltaH gt TDeltaS`
`DeltaH LT TDeltaS`

SOLUTION :`DeltaG = DeltaH- TDeltaS ""`At equilibrium , `DeltaG = 0, "" :. DeltaH-TDeltaS = 0 or DeltaH = TDeltaS`
Discussion
25.

At the equilibrium position in the process of adsorption

Answer»

`DELTAH GT 0`
`DeltaH = TDeltaS`
`DeltaH gt TDeltaS`
`DeltaH lt TDeltaS`

SOLUTION :`DELTAG = DeltaH- TDeltaS ""`At EQUILIBRIUM , `DeltaG = 0, "" :. DeltaH-TDeltaS = 0 or DeltaH = TDeltaS`
Discussion
26.

At the boiling point of liquid its vapour pressure is greater than atmospheric pressure . Is it true or false?

Answer»

SOLUTION :EQUAL to ATMOSPHERIC PRESSURE .
Discussion
27.

How does the boiling point of a liquid change with decrease in atmospheric pressure ?

Answer»

SOLUTION :EQUAL to ATMOSPHERIC PRESSURE .
Discussion
28.

At temperatures above 1073K, coke can be used toreduce FeO to Fe. How can you justify this reduction with Ellingham diagram?

Answer»

Solution :As per Ellingham diagram at TEMPERATURES greater than 1073 K
`TRIANGLEG (C, CO) lt triangleG (Fe, FeO)`
HENCE coke can reduce FeO to Fe.
Discussion
29.

At temperatures above 1073 K coke can be used to reduce FeO to Fe. How can you justify this reduction with Ellingham diagram?

Answer»

Solution :At temperature above 1073 K,
`Delta_fG_((C","CO))^(Theta) < Delta_fG_((FE","FeO))^(Theta)`. Hence, FeO can be REDUCED by COKE.
Discussion
30.

At the boiling point of a liquid its vapour pressure is equal to _____.

Answer»

SOLUTION :ATMOSPHERIC PRESSURE
Discussion
31.

At temperaturee T. a copound AB_(2(g)) dossociates according to the reaction 2AB_(2(g))hArr2AB_((g))+B_(2(g)) with a degree of dissociation x, which is small compared with unity. The expression for K_(P), in terms of x and the total pressure. F is

Answer»

`(Px^(3))/(2)`
`(Px^(2))/(2)`
`(Px^(3))/(3)`
`(Px^(2))/(2)`

Solution :`2AB_(2(g))hArr2AB_((g))+B_(2(g))`
`{:("Initially",1,0,0),("At EQULIBRIUM",(1-x),x, x//2):}`
Total no. of moles at equlibrium
`=(1-x)+x+x/2=(2+x)/(2)`
Partial pressure = mole fraction `XX` total pressure
APPLYING `K_(p)=(P_(AB)^(2)xxP_(B_(2)))/(P_(AB_(2))^(2))`
`=((((x)/(1+x))/(2)xxP)^(2)xx(((x/2)/(a+x))/(2)xxP))/((((1-x)/(1+x))/(2)+P)^(2))=(Px^(3))/((2+x)(1-x)^(2))`
Since `x ltlt1so(1-x)^(2)` can be neglected and (2+x) and be taken as 2.
`thereforeK_(p)=(Px^(3))/(2)`
Discussion
32.

At temperature T_(1), the equilibrium constant of eaction is K_(1). At a higher temperature T_(2),K_(2) is 10% of K_(1). Predict whether the equilibrium is endothermic or exothermic.

Answer»


ANSWER :EXOTHERMIC
Discussion
33.

At temperature T, a compound AB_2 (g)dissociates according to the reaction2AB_2(g) iff 2AB (g) + B_2(g)with a degree of dissociation, x, which is small compared to unity. Deduce the expression for x in terms of the equilibrium constant, K_p , and the total pressure, p.

Answer»

<P>

SOLUTION :`X = ((2K_p)/(p))^(1//3)`
Discussion
34.

At temperature T, a compound AB_2(g) dissociates according to the reaction 2AB_2(g) hArr 2AB(g) + B_2 (g) withadegree of dissociation x, Whichis small compared with unity. Predict the expression for K_p in termsof x and the total pressure P.

Answer»

`Px^3`/2
`Px^2`/3
`Px^3`/3
`Px^2`/2

Answer :A
Discussion
35.

At temperature T, a compound AB_2(g) dissociates according and the reaction 2AB_2(g) hArr2AB(g) + B_2(g) with a degree of dissociation ‘x’. Which is small compared to unity. Deduce the expression for ‘x’ in terms of the equilibrium constant K_P and the total pressure P.

Answer»

<P>

SOLUTION :`x=root3((2K_p)/P)`
Discussion
36.

At temperature of 298 K , the e.m.f. of the following electrochemical cell Ag_((s))|Ag^(+)(0.1M)||Zn^(2+)(0.1M)|Zn_((s)) will be ___________. (Given E_(cell)^(@) = -1.562 V)

Answer»

`-1.532 V`
`-1.503 V`
`1.532 V`
`-3.06 V`

ANSWER :A
Discussion
37.

At temperature of 298 K the emf of the following electrochemical cell Ag_((s)) |Ag^(+ )(0.1M ) ||Zn^(2+) (0.1 M )|Zn_((s)) willbe( givenE_("cell")^@=-1.562V)

Answer»

`-1.532`V
`-1.503 V`
`1.532 V`
`-3.06 V`

Solution :`2Ag _((s))TO2AG^(+)( 0.1 M )+ 2e^(-)`
`2e^(-)+Zn ^(2+)( 0.1M )toZn_((s))`
` 2Ag _((s)) +Zn^(2+) (0.1 M )toZn_((s))`
` 2Ag_((s)) +Zn^(2+) ( 0.1 M ) to2Ag ^( +) ( 0.1 M )+ Zn_((s))`
` thereforeE_("CELL ") =E_("cell")^(@ )- ( 0.059 1)/( 2 )log_(10)([AG^( +)]^2 )/([Zn^(2+)])`
` thereforeE_("Cell") = -1.562- ( 0.0591 )/(2)log_(10)((0.1 )^2)/( 0.1)`
`= -1.562- 0.03lpog_(10)10^(-1)`
` E_("cell") =-1.562+ 0.03=- 1.532V `
Discussion
38.

At temperature above 85 K, decarboxylation of acetic acid becomes a spontaneous process under standard state conditions. What is the standard entropy change (in J/K-mol) of the reaction. CH_(3)COOH (aq) rarr CH_(4)(g) +CO_(2)(g) {:("Given :",DeltaH_(f)^(@)[CH_(3)COOH (aq)],=-484" kJ/mole"),(,DeltaH_(f)^(@)[CO_(2)(g)],=-392" kJ/mole"),(,DeltaH_(f)^(@)[CH_(4)(g)],=-75" kJ/mole"):}

Answer»


SOLUTION :`DELTAS=(DELTAH)/(T) =(-392-75-(-484))/(85) =200 J//K` mole
Discussion
39.

At temperature327^(@)C and concentration C, the osmotic pressure of a solution is P. The same solution at concentration C/2 and a temperature 427^@C ofshows osmotic pressure of 2 atm. The value of P will be :

Answer»

`(12)/(7)`
`(24)/(7)`
`(6)/(5)`
`(5)/(6)`

Solution :`pi V=CRT`
`(pi_(1))/(pi_(2))=(C_(1)RT_(1))/(C_(2)RT_(2))`
`pi_(1)=P, pi_(2)=2` atm. `C_(1)=C, C_(2)=(C )/(2)`
`T_(1)=600K, T_(2)=700 K`
`(P)/(2)=(2xxC xx R XX600)/(C xx R xx 700)`
`P=(24)/(7)`.
Discussion
40.

At t^@C temperature, the observed vapour density of A is 17.5 for the reaction 2A_((g)) hArrƒ B_((g)) + 2C_((g)). If molecular weight of A is 48, then the percentage dissociation of A will be :-

Answer»

`70.27%`
`74.28%`
`37.14%`
`85.71%`

ANSWER :C
Discussion
41.

At T (K), the vapour pressures of pure liquids A and B are 100 mm and 160 mm respectively. An ideal solution is formed by mixing 2 moles of A and 3 moles of B at the same temperature. The mole fraction of A and B in the vapour state respectively are

Answer»

0.706, 0.294
0.294, 0.706
0.40, 0.60
0.60, 0.40

Solution :Vapour pressure of solution,
`p_("TOTAL") = p_(A) + p_(B), chi_(A)p_(A)^(@) + chi_(B)p_(B)^(@) ""[because p_(A) = chi_(A)p_(A)^(@)]`
Also vapour pressure of component 1, `p_(1) = y_(1) p_("total")`
where y, is the mole fraction of component 1 in vapour phase.
Given,
Vapour pressure of pure liquid A, `p_(A)^(@) = 100 mm`
Vapour pressure ofpure liquid B, `p_(B)^(@) = 160 mm`
`:.` Total vapour pressrure solution `= p_(A) + p_(B)`
`p_("total") = chi_(A)p_(A)^(@) + chi_(B)p_(B)^(@) = (2)/(5) xx 100 + (3)/(5) xx 160`
`= 40 + 96 = 136 mm`
Also `p_(A) = y_(A) p_("total")`
where, `y_(A)` is the mole fraction of A.
Mole fraction of A, `y_(A) = (p_(A))/(p_("total")) = (40)/(136) = 0.294`
`:. y_(B) = 1 - y_(A) = 1 - 0.294 = 0.706`
Discussion
42.

At sun atmosphere which of the following forms is stable :

Answer»

ORTHO H
Para H
Molecular H
None

Solution :At higher TEMP (SUN) molecular HYDROGEN DISSOCIATE in to atomic hydrogen.
Discussion
43.

At STP the volume of nitrogen gas required to cover a sample of silica gel, assuming Langmuir monolayer adsorption, is found to be 1.33cm^3 g^(-1) of the gel. The area occupied by a nitrogen molecule is 0.14nm^2? What is the surface area per gram of silica gel (in m^2) ?

Answer»


Solution :`22400 rarr6 XX 10^(23)`
`1.33 RARR ? ""3.5 xx 10^(-18)` molecules
one molecule AREA `- 0.14 xx 10^(-18)`
` 3.5 xx 10^(19)` molecule `rarr ? "" 5`
Discussion
44.

AT STP the volume of nitrogen gas required to cover a sample of silica gel, assuming Langmuir monolayer adsorption, is found to be 1.3^30^1of the gel. The area occupied by a nitrogen molecule is 0.1 0^2What is the surface area per gram of silica gel? [GivenN_A =6 diamond ^2]

Answer»

`5.5^2 6^1`
`3.4^2 8^1`
`1.6^2 1^7`
None of these

Solution :Number of molecules PER gram of `N_2` in monolayer
=`600^(23)
22,400
Cross-sectional area of a MOLECULE =`10^15`
Area covered by molecules per gram =313113
\ surface area =`5.5^2 6`
Discussion
45.

At STP the volume of nitrogen gas required to covera sample of silica gas, assuming larngmulr monolayer adsorption is found to be 1.30 gcm^(-2) g^(-1) of the gel. The area occupied by a nitrogen molecule is 0.16nm^(2). What is the surface area per gram of silica gel ?

Answer»

`5.568m^(2)G^(-1)`
`3.48m^(2)g^(-1)`
`1.6m^(2)g^(-1)`
None of these

Solution :No. of MOLECULE PER gm of `N_(2)` in MONOLAYER `=(6xx10^(23))/(22400)xx1.30=3.48xx10^(19)`
area of a molecule `=1.6xx10^(19)m^(2)`
`:.""` Area covered by molecule per gram`=3.48xx10^(19)xx1.6xx10^(-19)=5.568m^(2)`
`:. ""`Surface are `=5.568m^(2)g^(-1)`
Discussion
46.

At S.T.P. the volume of 7.5 g of a gas is 5.6 L. The gas is

Answer»

NO
`N_(2)O`
CO
`CO_(2)`

SOLUTION :5.6 L at S.T.P WEIGHS = 7.5 g
`therefore 22.4` L at S.T.P weighs `=(7.5 xx 22.4)/5.6 = 30`
`therefore` Molar MASS = 30 gmol, which is the molar mass of NO.
Discussion
47.

At STP , the order of root mean square velocity of moleculesH_2,N_2,O_2 and HBr is

Answer»

`N_2 GT O_2 gt H_2 gt HBR`
`HBr gt O_2 gt N_2 gt H_2 `
`HBr gt H_2 gt O_2 gt N_2`
`H_2 gt N_2 gt O_2 gt HBr `

ANSWER :D
Discussion
48.

At STP ,the order of root mean square speed of molecules H_2 , N_2 , O_2 and HBr is :

Answer»

`H_2 GT N_2 gt O_2 gt HBR`
`HBr gt O_2 gt N_2 gt H_2`
`HBr gtH_2 gtO_2 gtN_2`
`N_2 gt O_2 gt H_2 gt HBr`

ANSWER :A
Discussion
49.

At S.T.P., the density of nitrogen monoxide is :

Answer»

1. `3.0 G L^(-1)`
2. `30 g L ^(-1)`
3. `1.34 g L ^(-1)`
4. `2.68 g L^(-1)`

SOLUTION :`30 // 22.4 = 1.34 g L^(-1)`
Discussion
50.

At STP 1.12 litre of H_(2) is obtained on flowing current for 965 seconds in a solution . The value of current is

Answer»

10 Amp.
1.0 Amp.
1.5 Amp.
2.0 Amp.

Solution :1.12 LITRE of `H_2` at STP =0.1 g of `H_2`.
`therefore W =Zit`
`therefore I =(W)/(Zt) = (0.1 xx 96,500)/(Exxt) =(0.1xx96,500)/(1xx965) =10A`
Discussion