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 zero kelvin, most of the ionic crystals possess

Answer»

Frenkel DEFECT 
Schottky defect 
Metal EXCESS defect 
No defect 

ANSWER :D
2.

At which temperature, the natural rubber becomes brittle:

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LESS than `10^(@)C`
between `10^(@)C- 60^(@)C`
More than `60^(@)C`
NONE of these

SOLUTION :None of these
3.

At which temperature, P C K K K_P/K_Cvalue will be 1/4 for dissociation equilibrium of ammonia [2 NH_3 ƒ N_2 + 3H_2]?

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`6.09 K`
`12.18 K`
`24.36 K`
`18.27K`

ANSWER :D
4.

At which temperature, ceramic matters are known as super conductor ?

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0K
200K
150K
15K

Solution :150K
5.

At which temperature, natural rubber becomes soft ?

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LESS than `60^(@)C`
More than `60^(@)C`
Less than `10^(@)C`
More than `100^(@)C`

Solution :More than `60^(@)C`
6.

At which temperature both rhombic and monoclinic sulphur are stable ?

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369 K
396 K
`396^(@)` C
`369^(@)` C

ANSWER :A
7.

At which temperature 0.006 % w/V urea solution has osmotic pressure 0.0246 atmosphere ?

Answer»


ANSWER :300 K
8.

At which of the following four conditions, the density of nitrogen will be the largest?

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STP
273 K and 2 atm
546 K and 1 atm
546 K and 2 atm

Answer :B
9.

At which of the following concentration would a solution of an electrolyte show a maximum molar conductivity ?

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0.005 M
0.004 M
0.003 M
0.002 M

Answer :D
10.

At which one of the following condition, a reducing agent is suitable for reducing a metal oxide

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SUM of the `DELTA G`values for oxidation-of metal and oxidation of reductantshould be negative
Sum of the `Delta G`valuesfor oxidation of metal and oxidation ofreductantshould bepositive
Sum of the `Delta G`values for reduction of metal OXIDE and oxidation of reductantshould be negative
Sum of the`Delta G`values for oxidation of metal and reduction of reductant should be negative

Answer :C
11.

At which one of the following condition, a reducing a metal oxide

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Sum of the `DeltaG` VALUES for oxidation of METAL and oxidation of reductant should be negative
Sum of the`DeltaG` values for oxidation of metal and oxidation of reductant should be positive
Sum of the `DeltaG` values for REDUCTION of metal OXIDE and oxidation of reductant should be negative
Sum of the `DeltaG` values for oxidation of metal and reduction of reducant should be negative

Answer :C
12.

At what value of 'n' the formation of six membered of ring take place ?

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ANSWER :C
13.

At what temperature would the volume of a given mass of a gas at constant pressure be twice to its volume at 0^@C

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`100^@C`
`273^@C`
`373^@C`
`446^@C`

ANSWER :B
14.

At what temperature would CO_(2) molecule have an rms speed equal to H_(2) at 27^(@) C.

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`4400 K`
`2200 K`
`6600K`
`1100 K`

Solution :`rms_(CO_(2)) = rms_(H_(2))`
`(sqrt((3RT)/(M)))_(CO_(2)) = (sqrt((3RT)/(M)))_(H_(2))`
`(3 xx R xx T)/(44) = (3 xx R xx 300)/(2)`
`T = 6600 K`
15.

At what temperature will the RMS velocity of SO_(2) be the same as that of O_(2) at 303 K?

Answer»

606K
273K
403K
303K

Solution :RMS velocity, `mu = sqrt(3RT//M)`
Temperature at which ROOT mean square of `SO_(2)` be the same as that of `O_(2)` at 303K.
`sqrt((3RT)/(M))_(O_(2)) = sqrt((3RT)/(M_(2)))_(SO_(2)) or sqrt((303)/(32)) = sqrt((T)/(64))`
`RARR T= (303 xx 64)/(32) = 606K`
16.

At what temperature, will the rms velocity of a gas at 50^(@)C be doubled ?

Answer»

626K
1019K
`200^(@)C`
`1019^(@)C`

Solution :`u_("r.m.s.") = SQRT((3RT)/( M)):. u_("r.m.s.") prop sqrt( T )`
For `u_("r.m.s.") ` to be doubled, temperature must be raised by 4 TIMES.
`T = ( 50 + 273 )K = 323 K`
`:. 4T = 4 xx 323 K = 1292K`
`= ( 1292 - 273) C = 1019^(@)C`
17.

The root mean square velocity of SO_(2) gas becomes the same as that of methane at 27^(@)C when the temperature is :

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`327^(@)C`
`127^(@)C`
`54^(@)C`
`927^(@)C`

ANSWER :D
18.

At what temperature will O_(2) molecules have the same root mean square speed as N_(2) molecules at 227^(@C ?

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Solution :For `O_(2): U_("rms") = sqrt((3RT)/(32)),`
For `N_(2): U_("rms") = sqrt((3R xx500)/(28))`
SINCE, both have same `U_("rms")`
`therefore (3RT)/(32) =(3Rxx500)/(28)`
`therefore T=571.4 K =298.4^(@)C`
19.

At what temperature will the average speed of CH_(4) molecules have the same value as O_(2) has at 300 K

Answer»

1200K
150 K
600 K
300 K

Solution :`(V_(av)CH_(4))/(V_(ab)O_(2))=sqrt((T_(CH_(4)))/(T_(O_(2))).(M_(O_(2)))/(M_(CH_(4))))=1`
`(T_(CH_(4)))/(300).(32)/(16)=1, T_(CH_(4))=150K`
20.

At what temperature will be rate of effusion of N_2 be 1.625 times the rate of effusion of SO_2 at 500^@C :

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273 K
893 K
110 K
173 K

Answer :B
21.

At what temperature will be total kinetic energy (KE) of 0.3 mole He be the same as the total KE of 0.40 mole of Ar at 400 K:

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400K
373K
533 k
300 k

Answer :C
22.

At what temperature will a 5% solution(wt.vol) of glucose develops an osmotic pressure of 7 atm.

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33.94K
306.94K
273K
`33.94^(@)C`

ANSWER :B::D
23.

At what temperature, the sample of neon gas would be heated to double of its pressure, if the initial volume of gas is/are reduced to 15% at 44.4 K

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`319^(@)C`
`592^(@)C`
`128^(@)C`
`90^(@)C`

ANSWER :A
24.

At what temperature, the rate of effusion of N_(2) would be 1.625 times that of SO_(2) at 50^(@)C

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110 K
173 K
373 K
273 K

Solution :`(r_(N_(2)))/(r_(SO_(2)))=(V_("RMS")N_(2))/(V_("rms")SO_(2))=sqrt((T_(N_(2)))/(T_(SO_(2))).(M_(SO_(2)))/(M_(N_(2))))=sqrt((T_(N_(2)))/(323)XX(64)/(28))`
`1.625=sqrt((T_(N_(2)))/(323).(16)/(7))`
`T_(N_(2))=((1.625)^(2)xx323xx7)/(16)=373K`
25.

At what temperature natural rubber becomes brittle ?

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LOWER than `60^(@)C`
Lower than `10^(@)C`
Lower than `0^(@)C`
Higher than `60^(@)C`

SOLUTION :`Lower than 0^(@)C`
26.

At what temperature in the celsius scale, V ( volume ) of a certain mass of gas at 27^(@)C will be doubled keeping the pressure constant ?

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`54^(@)C`
`327^(@)C`
`427^(@)C`
`527^(@)`

Solution :`(V_(1))/( T_(1)) = ( V_(2))/(T_(2))` at constant PRESSURE
`(V)/( 300) = ( 2V)/( T_(2))`
or `T_(2) = 300 xx2 = 600`
or `= 600 - 273 = 327^(@) C`
27.

At what temperature in the celsius scale, V ( volume) of a certain mass of gas at 27^(@)C will be doubled keeping the pressure constant

Answer»

`54^(@)C`
`327^(@)C`
`427^(@)C`
`527^(@)C`

Solution :`(V_(1))/(V_(2))=(T_(1))/(T_(2)):' T_(2)=(T_(1)V_(2))/(V_(1))=300K , (2V)/(V)=600K`
28.

At what temperature does the average translational kinetic energy of a moleculein a gas become equal to the kinetic energy of an electron accelerated from rest through a potential difference of 1 volt?1 eV = 1.602 xx 10^(-12) erg.

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SOLUTION :`7.73 XX 10^3 K`
29.

At what temperature does an aqueous solution containing 3xx10^(23) molecules of a nonelectrolyte substance in 250 g of water freeze?(K_(t)=1.86).

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269.28 K
271.14 K
271 K
276.72 K

Answer :A
30.

At what temperature can carbon be used to reduce ferric oxide?

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SOLUTION : > 1073 K
31.

At what temperature carbon monoxide reduces ferric oxide to Fe ?

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900 K
1200 K
2000 K
500 K

Answer :A
32.

At what relative humidity will copper sulfate pentahydrate lose its waters of hydration when the air temperature is 30°C? What is K_(p) for this process at this temperature?

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SOLUTION :From the table, we see that the vapor PRESSURE of the hydrate is 12.5 torr, which corresponds to a relative humidity of 12.5/31.6 = 0.40 or 40%. This is the humidity that will be maintained if the hydrate is placed in a CLOSED container of dry air.
For this hydrate,` Kp = P_(H_(2)O)^(5)` , so the PARTIAL pressure of water vapor that will be in equilibrium with the hydrate and the dehydrated solid (remember that both solids MUST be present to have equilibriumhArr), expressed in atmospheres, will be `(12.5//760)^(5) = 1.20 xx10^(–9)`.
33.

At what pH will 1 xx 10^(-3)M solution of an indicator with K_b = 1 xx 10^(-10) change colour?

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SOLUTION :The indicator changes COLOUR when the conjugates are of EQUAL concentration.
34.

At what pH the potential of hydrogen electrode will be 0.059 V ?

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ANSWER :1
35.

At what pH the oxidation potential of hydrogen electrode will be -0.413 V ?

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Solution :`E = (-0.0591)/(1) LOG""(1)/([H^(OPLUS)]), (-(E xx 1)/(0.0591)) = - log(H^(oplus)) = P^(H) , P^(H) = 6.988 ~= 7`
36.

At what pH of HCl slutionw ill hydrogen gas electrode show electrode potential of -0.118V? H_(2) gas is bubbled at 298 K and 1 atm pressure.

Answer»

Solution :Writing ELECTRODE REACTION as REDUCTION reaction, `H^(+)+e^(-)to(1)/(2)H_(2)`
APPLYING nernst equation, `E_(H^(+)//H_(2))=E_(H^(+)//H_(2))^(@)-(0.0591)/(1)"log"(1)/([H^(+)])=0+0.0591log[H^(+)]`
or `-0.118=-0.0591pH` or `pH=2`(`becausepH=-log[H^(+)])`
37.

At what pH an indicator with pk_(b) = 4changes colour :

Answer»

4
8
10
12

Answer :C
38.

At what partial pressure, oxgyenwill have a solubility of "0.05 g L"^(-1) in water at 293 K ? Henry's constant (K_(H)) for O_(2) in water at 293 K is 34.86 kbar. Assume the density of the solution to be same as that of the solvent .

Answer»

<P>

SOLUTION :Calculation of mole fraction `(x_(O_(2)))`
MASS of 1 L of solutio = 1000 g`""(because d ="1 g mL"^(-1))`
`therefore"Mass of solvent (water) "=1000g-0.05 g~=1000g`
`therefore""n_(H_(2)O)=("1000 g")/("18 g MOL"^(-1))="55.5 moles,"n_(O_(2))=("0.05 g")/("32 g mol"^(-1))=1.56 xx10^(-3)" mole"`
`therefore""x_(O_(2))=(n_(O_(2)))/(n_(O_(2))+n_(H_(2)O))~=(n_(O_(2)))/(n_(H_(2)O))=(1.56xx10^(-3))/(55.5)=2.81xx10^(-5)`
Calculation of partial pressure. Applying Henry's law,
`p_(O_(2))=K_(H)xx x_(O_(2))=(34.86xx10^(3)" bar")xx(2.81xx10^(-5))="0.98 bar."`
39.

At what partial pressure, nitrogen will have a solubility of 0.05 g L^(-1) in water at 293 K ? Given that k_(H) for N_(2) at 293 K is 76.48 k bar. Assume that the density of the solution is the same as that of the pure solvent.

Answer»


Solution :Step I. Calculation of `x_(N_(2))`
`"Mass of 1 L of solution"=1000mLxx1" g mL"^(-1)=1000 g`
`"Mass of solution (water)=(1000-0.05)~~1000 g`
`"No. of moles of water "=((1000g))/((18g MOL^(-1)))=55.5 mol`
`"No. of moles of NITROGEN"=((0.05g))/((28" g mol"^(-1)))~~1.79xx10^(-3)mol`
`X_(N_(2))=(n_(N_(2)))/(n_(N_(2))+n_(H_(2)O))=((1.79xx10^(-3)mol))/((1.79xx10^(-3)+55.5 mol))=3.22xx10^(-5)`
Step II. Calculation os partial pressure of `N_(2)`
`rho_(N_(2))=K_(H)xxX_(N_(2))=(76.48xx10^(3)"bar")xx(3.22xx10^(-5))=246.26"bar"`
40.

At what concentration of copper sulphate solution , the potential of Cu^(2+), Cu becomes zero? The standrd reduction potential of Cu^(2+), Cu is 0.34 V.

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Solution :`E = E^@ + (0.059)/n log C "(or)" 0 = 0.34 + (0.059)/2 log [Cu^(2+)]`
`0.0295 log [Cu^(2+)] = -0.34 " (or) " log [Cu^(2+)] = -(0.34)/(0.0295) = -11.57`
CONCENTRATION of `CuSo_4 = 10^(-11.57) = 10^(-12 + 0.43) = 2.9 xx 10^(-12)M`.
41.

At what concentration of Cu^(2+) in a solution of CuSO_4 will be electrode potential be zero at 25^(@)C Given: E^0(Cu|Cu^(2+))=-0.34V

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ANSWER :A::B::C
42.

At what angle for a first order diffraction, the distance between two adjacent planes of crystal is equal to the wavelength of x-rays used

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`30^(@)`
`60^(@)`
`90^(@)`
`45^(@)`

ANSWER :A
43.

At what concentration of copper sulphate solution, the potential of Cu^(2+), Cubecomes zero? The standard reduction potential of Cu^(2+), Cu is 0.34 V.

Answer»

SOLUTION :`E = E^(@) +(0.059)/(n) LOG C ` (or)
`0=0.34+(0.059)/(2) log [Cu^(2+)]`
`0.0295 log [Cu^(2+)]=-0.34`
(or) `log[Cu^(2+)]=- (0.34)/(0.0295) = -11.57`
`[CuSO_(4)]=10^(-11.57) = 10^(-12+0.43) = 2.9xx10^(-12)M`.
44.

At very low pressure, what is the equation of Langmuir Adsorption Isotherm ?

Answer»

`(x)/(m)=Kp^((1)/(n))`
`(x)/(m)=(AP)/(1+bp)`
`(x)/(m)=ap`
`(x)/(m)=(a)/(B)`

Solution :`(x)/(m)=ap`
45.

At w hich tem perature, Fe_(3)O_(4) is a fe rri­magnetic solid converted to a paramagnetic solid ?

Answer»

850 K
300 k
400 k
600 k

ANSWER :A
46.

At two dimensional solid pattern formed by two different atoms X and Y is shown below. The black and white squares represent atoms X and Y respectively. The simplest formula for the compound based on the unit cell from the pattern is ……………….. .

Answer»

`XY_(8)`
`X_(4)Y_(9)`
`XY_(2)`
`XY_(4)`

Solution :`XY_(8)`
47.

At very high pressure the langmuir adsorption isotherm takes the form of

Answer»

<P>`x/m = KP`
`x/m = a/b`
`x/m = 1/(1 + ap)`
`x/m = p`

Answer :B
48.

At T(K), 100 L of dry oxygen is present in a sealed container. It is subjected to silent electrical discharge till the volumes of oxygen and ozone become equal What is the volume (in L) of ozone formed at T(K)?

Answer»

50
60
30
40

Solution :`2SO_(2)+O_(3) to 2SO_(3), 3SnCl_(2)+6HCl+O_(3) to 3SnCl_(4)+3H_(2)O`
49.

At times the solution of lime water appears milky. Comment.

Answer»

SOLUTION :The lime water lying in the reatgent bottle absorbs `CO_(2)` gas from the atmosphere and produces WHITE insoluble `CaCO_(3)Ca(OH)_(2)+CO_(2)toCaCO_(3)(s)+H_(2)O`
50.

At time a white ppt. is obtained in group VI even in the absence of Mg. explain.

Answer»

SOLUTION :The CATION of group V which escape PRECIPITATION now precipitate in group VI.