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 equilibrium the rate of dissolution of a solution solute in a volatile liquid solvent is ……

Answer»

Less than the RATE of crystallisation
Greater than the rate of crystallisation
Equal to the rate of crystallisation
Zero

Solution :EXPLANATION : This happens as per conditions attained at EQUILIBRIUM state , i.e, rate of forward reaction (DISSOLUTION) = rate of BACKWARD reaction (crystallisation).
2.

Which is correct?

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`/_\T=0`
`/_\S=0`
`/_\H=0`
`/_\G^0=0`

SOLUTION :GENERAL CONCEPT
3.

At equilibrium the rate of dissolution of a solid solute in a volatile liquid solvent is

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less than the rate of CRYSTALLISATION
greater than the rate of crystallisation
equal to the rate of crystallisation
zero

Solution :Is the CORRECT answer.
The solution is saturated. In the solution, the rate of DISSOLUTION is equl to RATIO of crystallisation.
4.

At equilibrium the rate of dissolution of a solid solute in a volatile liquid solvent is …………..

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LESS than the RATE of CRYSTALLISATION
greater than the rate of crsystallisation
equal to the rate of crystallisation
zero

Solution :At EQUILIBRIUM, rate of DISSOLUTION = rate of crystallisation.
5.

At equilibrium the free energy change (Delta G) is :

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`Delta G = 0`
`Delta G GT 0`
`Delta G lt 0`
MAY have any value but is not equal to zero

Answer :A
6.

At equiliberium , the amount of HI in a 3 litre vessel was 12.8 g. Its equiliberium concentraction is :

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4.267 M
0.033 M
0.1 M
0.2 M

Answer :B
7.

At elevated temperatures, S_2 is the dominant species and is _________like O_2

Answer»

SOLUTION :PARAMAGNETIC
8.

At elevated temperature, HI decomposes according to the chemical equations: 2HI(g) to H_(2)(g) + I_(2)(g) a)Determine (i) the order of reaction and (iii) Write the rate expression. b) Calculate the rate constant and give its units.

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Solution :From this data, it is QUITE evident that when (HI) is increased from 0.005 to 0.01 (made twice),
rate = `(30 xx 10^(-4))/(7.5 xx 10^(-4)) =4`
When (HI) is increased from 0.01 to 0.02 (made twice),
rate `= (12 xx 10^(-3))/(3 xx 10^(-3)) =4`
a) `therefore` ORDER of reaction=2
Rate expression, (r)=`k[HI]^(2)`
b) Calculation of rate constant:
For EXPT. (1), `k(0.005 mol L^(-1))^(2)=(7.5 xx 10^(-4)mol^(-1)s^(-1)`
`k=(7.5 xx 10^(-4) mol L^(-1)s^(-1))/(2.5 xx 10^(-5)mol^(2)L^(-2))=30 mol^(-1)Ls^(-1)`.
9.

At different condition nitration of phenol gives

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o-nitrophenol
p-nitrophenol
Picric acid
All of these

Answer :D
10.

At CMC, the surfactant molecules :

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Dissociate
Associate
Become completelysoluble
Decompose

Answer :B
11.

At critical micelle concentration (CMC)

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the ions of surfactant molecules UNDERGO ASSOCIATION to FORM clusters
the turbidity of solution INCREASES abruptly
substances like grease, FATS, etc. dissolve colloidally
colligative properties increase suddenly.

Answer :A::B::C
12.

At constant volume, for a fixed number of moles of a gas, the pressure of the gas increases with rise of temperature due to:

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INCREASE in AVERAGE molecular speed
increased rate of COLLISIONS amongst molecules
increase in molecular attraction
decrease in MEAN free path

Answer :A::B
13.

At constant volume, for a fixed number of moles of a gas the pressure of the gas increases with rise of temperature due to

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1. INCREASE in average molecular speed
2. INCREASED rate of collision amongst molecules
3. increase in molecular attraction
4. decrease in mean free PATH

ANSWER :A
14.

At constant volume at 27^(@) C, 2C_(6)H_(6)(g)+150_(2)(g) to 12CO_(2)(g)+6H_(2)O(l), Delta u=-1600 kcal 2C_(2)H_(2)(g)+5O_(2)(g) to 4CO_(2)(g)+2H_(2)O (l), Deltau=-620 kcal Calculate the heat of polymerisation of acetylene to benzene at constant pressure.

Answer»


ANSWER :(-131.2 KCAL)
15.

At constant teperature, the equlimbeijm constnt (K_(p)) for the decomposition reaction N_(2)O_(4)hArr2NO_(2) is expressed by K_(p)=(4x^(2)P)(1-x^(2)), where P = pressure, x= ectent of decomposition. Which one of the following statements is true

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`K_(p)` increases with increase of P
`K_(p)` increase with increase of X
`K_(p)` increase with decrease of x
`K_(p)` REMAINS CONSTANT with change in P and x

Solution :With change of pressurr,will change in such a WAY that `K_(p)` remains a constant.
16.

At constant volume and temperature conditions the rates of diffusion D_(A) and D_(B) of gases A and B having densities rho_(A) and rho_(B) are related by the expression

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`D_(A)=[D_(B).(rho_(A))/(rho_(B))]^(1//2)`
`D_(A) . [(rho_(B))/(rho_(A))]^(1//2)`
`D_(A)=D_(B)((rho_(A))/(rho_(B)))^(1//2)`
`D_(A)=D_(B)((rho_(B))/(rho_(A)))^(1//2)`

SOLUTION :`(D_(A))/(D_(B))=SQRT((rho_(B))/(rho_(A)))` i.e. `D_(A)=((rho_(B))/(rho_(A)))^(1//2)`
17.

At constant volume , for a fixed number of mole of a gas , the pressure of the gas increase with rise of temperature due to :

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INCREASE in AVERAGE molecular speed
Increase in number of mole
Increase in molecular attraction
Decrease in mean free path

Answer :A
18.

At constant temperture and pressure which one of the following statements is correct for the reaction ? CO(g)+ 1/2O_(2)(g) to CO_(2)(g)

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`DeltaH = DELTAE`
`DeltaH lt DeltaE`
`DeltaH gt DeltaE`
`DeltaH` is independent of PHYSICAL state of reactant

Solution :As we know , `DeltaH=(- 1348.9 xx10^(3))+ 11xx2xx298=-1342.34 kcal`
for the reaction , `CO(G)+1/2O_(2)(g) to CO_(2)(g)`
` Deltan=1-(1+1/2) =-1/2`
`DeltaH=DeltaE-1/2 RT`
`DeltaH lt DeltaE`
19.

At constant temperature, the pressure of V mL of a dry gas was increased from 1 atm to 3 atm. The new volume will be :

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3V
`V//3`
`V^(3)`
`2V^(3)`

Solution :`p_(1) = 1 atm,p_(2)` = 3 atm, `V_(1) `= V ml, `V_(2) = ?`
`p_(1) V_(1) = p_(2) V_(2)` ( at constant temperature )
`1 XX V = 3 xx V_(2)`
or `V_(2) = V//3`
20.

At constant temperature, the equilibrium constant (K_(p)) for the decomposition reaction, N_(2)O_(4)iff2NO is expressed by K_(p)=((4x^(2)P))/((1-x^(2))), where P = pressure, x = extent of decomposition. Which one of the following statements is true?

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`K_(p)` increases with increase of P
`K_(p)` increases with increase of x
`K_(p)` increases with DECREASE of x
`K_(p)` remains constant with change in P and x

Solution :`K_(p)` is temperature dependent only. By CHANGING P or x equilibrium shifts as per LE Chatelier.s principle, but after some time it establishes a new state of equilibrium at which `K_(p)` is MAINTAINED constant.
21.

At constant temperature, the equilibrium constant (K_(p)) for the decomopsition reaction N_(2)O_(4)hArr2NO_(2) is expressed byK_(p) = ((4x^(2)P))/((1-x^(2)), where P = pressure, x = extent of decomposition. Which one of the following statement is true ?

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<P>`K_p` remains constant with change in P
`K_p` INCREASES with decrease of X
`K_p` increases with increase of x
`K_p` increases with increase of P

Answer :A
22.

At constant temperature, the equilibrium constant (K_p)for the decomposition reaction, N_2O_4 hArr 2NO_2 is expressed by K_p=4x^2P//(1-x^2) , where P = pressure and x = extent of decomposition. Which of the following statements is true?

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`K_p`increases with increase in P.
`K_p`increases with increase in X.
`K_p`increases with decrease in x.
`K_p` REMAINS constant with CHANGE in P and x.

Solution :The equilibrium constant does not change with change in concentration, volume, pressure and presence of a catalyst. It CHANGES only with change in temperature of the system.
23.

At constant temperature ,osmotic pressure of an aqueous solution of 1.5 M NH_(4)NO_(3) and x N Al_(2)(SO_(4))_(3) are equal, thenmention the value of X. ( Assume that ionic solid substances completely dissociates in the solution . )

Answer»

0.1
3.6
1.2
0.5

Solution :`n_(1)M_(1) = n_(2)M_(2)`
For`NH_(4)NO_(3)`VALUES of `n_(1)` and `M_(1)` are 2 and 1.5respectively
For `Al_(2)(SO_(4))_(2)n_(2) = 5 `
`:. M_(2) =( 2)/(5) xx 1.5 = 0.6 `
Normality of `Al_(2)(SO_(4))_(3)= 6 xx ` molarity `= 6 xx 0.6 = 3.6 N `
24.

AT constant temperature, the equilibrium constant (K_p) for the decomposition reaction N_2O_4 hArr 2NO_2 is expressed by K_p=(4x^2P)//(1-x^2), where P= pressure ,x = extent of decomposition . Which one of the following statements is true ?

Answer»

`K_p` INCREASES with increases of P
`K_p` increases with increases of x
`K_p` increases with DECREASE of x
`K_p` remains CONSTANT with change in P and x

Answer :D
25.

At constant temperature in one litre vessel when the reaction 2SO_2 (g) ⇌ 2SO_2 (g) + O_2 (g) is at equiliberium the SO_2 concentration is 6.0 M, initial concentration of SO_3 is 1 M. calculate the equiliberium constant.

Answer»

2.7
1.36
0.34
0.675

Answer :D
26.

At constant temperaturein a litre vessel , when the reaction 2SO_(3)(g) iff 2SO_(2)(g) + O_(2)(g)the SO_(2) concentration is 0.6M , initial concentration ofSO_(3) is 1M . The equilibrium constant is

Answer»

1.36
0.34
2.7
0.675

Answer :D
27.

At constant temperature and volume, X decomposes as 2" X "(g) to 3" Y "(g)+2" Z "(g). P_(x) is the partial pressure of X. {:("Observation No.",,,"Time (in minutes)",,,P_(x)("in mm of Hg")),(1,,,0,,,800),(2,,,100,,,400),(3,,,200,,,200):} (i)What is the order of reaction with respect with respect of X ? (ii) Find the time for 75% completion of the reaction. (iii) Find the total pressure when pressure of X is 700 mm of Hg.

Answer»

Solution :(i) As pressure of X is changing with time, it cannot be a zero order REACTION. LET us now check it for 1st order.
`{:("At",,,t=100" min,",,,k=(2.303)/(100)log""(P_(0))/(P_(t))=(2.303)/(100)log""(800)/(400)=6.932xx10^(-3)"min"^(-1)),("At",,,t=200" min,",,,k=(2.303)/(200)log""(800)/(200)=(2.303)/(800)log4=6.932xx10^(-3)"min"^(-1)):}`
As k comes out to be constant, hence it is a reaction of 1st order.
(II) `t_(75%)=(2.303)/(k)log""(100)/(100-75)=(2.303)/(6.932xx10^(-3)"min"^(-1))log4=200" min".`
(iii) `2X(g) to 3" Y"(g)+2" Z"(g)`
`{:("Initial Pressure",,,800" mm",,,0,,,0),("Pressure after time t",,,800-2" p",,,3" p",,,2" p"):}`
When pressure of X is 700 mm, `800-2" p"=700" or "p=50" mm"`
Total pressure `=(800-2" p")+3" p"+2" p"=800+3" p"=800+3xx50=950" mm"`
28.

At constant temperature , if pressure increases by 1% , the percentage decrease of volume is :

Answer»

`1%`
`100 //101%`
`1//101%`
`1//100%`

Solution :Let `P_(1) = 100` mm Hg
INCREASE in PRESSURE
`= 100 XX ( 1)/( 100) = 1 `
`p_(2) = 100 +1 = 101 mm Hg `
`V_(1) = V` and `V_(2) = ?`
`p_(1) V_(1) = p _(2) V_(2)`
or `V_(2) = ( p_(1) V_(1))/(p_(2)) = ( 100 xx V )/( 101) = ( 100)/( 101) V `
Decrease in volume`= ( V - ( 100)/( 101) V )/( V )`
` = V ( 1 - (100)/( 101))/( V ) = ( 1)/( 101)%`
29.

At constant temperature , in a given mass of an ideal gas

Answer»

The RATIO of pressure and volume ALWAYS remains constant
Volume always remains constant
Pressure always remains constant
The product of pressure and volume always remains constant

Solution :According to BOYLE's LAW`V prop (1)/(P)`
`V = ("Constant")/(P) , VP =` Constant
30.

At constant temperature and pressure, correct order of entropy of gases for their equal moles is

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`He gt NE gt Ar gt Kr gt XE`
`H lt O_2 lt CO_2 lt PCI_3`
`He = Ne = 0,_2lt Ar lt Kr = Xe`
`O_2 = 0_3 lt N_2 lt Cl_2`

Answer :2
31.

At constant temperature and 1 atm pressure, PCl5 dissociate 2% then at what pressure PCl5 dissociate4% :- (Use PCl_(5)(g) hArrPCl_(3)(g) + Cl_(2)(g))

Answer»

`1/8 ` atm
`1/(16) atm`
`(1)/(2) atm`
`1/4 atm`

Solution :`{:(PCL5 , hArr,PCl 3, + ,CL2),( 1,,0,,0),(1-alpha , ,alpha ,, alpha ):}`
`KP=( alpha^(2))/((1-alpha ))xx[(p)/((1+alpha))]^(1)`
`KP= (alpha^(2) P)/(1-alpha ^(2)) ~~- alpha ^(2)P `
` alpha_(1) ^(2)P_(1)= alpha _(2) ^(2)P_(2) `
`( 0.02 ) xx 1 = ( 0.04 ) ^(2).Pa`
`impliesP2= 0.25atm `
`=(1)/(4) atm`
32.

At constant temperature and pressure, which of the following statement is true for the reaction.CO[g]+1/2O_2[g]-rarrCO_2[g]

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`/_\H=/_\E`
`/_\Hlt/_\E`
`/_\Hgt/_\E`
`/_\H is INDEPENDENT to PHYSICAL state.

Answer :B
33.

At constant temp. the osmotic pressure (pi) and the molarity (M) of the solution are related as

Answer»

`PI prop M`
`pi prop 1/M`
`pi prop SQRT(M)`
`piprop 1/sqrt(M)`

Solution :A/c to van't Hoff's equation, `:. pi prop M`
`pi=CRT, C=Molarity`
34.

At constant T and P, which one of the following statements is correct for the reaction, CO(g) +(1)/(2)O_(2)(g) rarrCO_(2)(g)

Answer»

`DeltaH` is independent of the PHYSICAL state of the reactants of that compound
`DeltaH GT DELTAE`
`DeltaH lt DeltaE`
`DeltaH = DeltaE`

Solution :`Deltan_(g)=1-(3)/(2)=(-1)/(2)`,As `Deltan_(g)` is negative, thus `DeltaH lt DeltaE`.
35.

At constant pressure, the heat of formation of a compound is not dependent on temperature, when

Answer»

`DELTA C_P = 0`
`Delta C_V = 0`
`Delta C_P gt 0`
`Delta C_P lt 0`

Solution :ACCORDING to kirchhoff.s equation
`Delta H_2 = Delta H_1+ Delta C_P (Delta T)`
when `Delta C_P = 0` then `Delta H` does not DEPEND on temperature .
36.

At constant T and P which one of the following statements is correct for the reaction : CO(g) + (1)/(2) O_(2)(g) rarr CO_(2)(g)

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`Delta H = Delta U`
`Delta H lt Delta U`
`Delta H GT Delta U`
`Delta H` is INDEPENDENT of the PHYSICAL state of the REACTANT of that compound

Answer :B
37.

At constant P and T which statement is correct for the reaction, CO(g)+1//2O_2(g)rarrCO_2(g):

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`triangleH=triangleU`
`triangleHlttriangleU`
`triangleHgttriangleU`
`TRIANGLE H` is independent for PHYSICAL state of reaction.

Answer :B
38.

At constant P and T which of the following statement is correct for, C (s) + O_(2)(g) rarr CO_(2) (g)

Answer»

`DELTAH = DeltaE`
`Delta H LT DeltaE`
`Delta H gt DeltaE`
`DeltaH` is INDEPENDENT of the physical state of reactants

Answer :A
39.

At CMC the surfactant molecules

Answer»

DECOMPOSES
BECOMES COMPLETELY SOLUBLE
associates
dissociates

Answer :C
40.

At CMC, the surfactant moleculesdecompose,become completely soluble,associate,dissociate.

Answer»

decompose
become COMPLETELY soluble
associate
dissociate

Answer :C
41.

At CMC the surfact molecules:

Answer»

decompose
associate
become COMPLETELY soluble
dissociáte

Answer :B
42.

At CMC, surfactant molecules

Answer»

HYDROLYSE
dissociate
ASSOCIATE
DISSOLVE competely

Answer :C
43.

At CMC (Critical Micellisation Conc.) the surface molecules

Answer»

Associate
Dissociate
Decompose
Become COMPLETELY soluble.

Answer :A
44.

At certain temperature, vapour pressure of pure element A and B has 108 and 36 torr respectively. If solution has equal mole of A and B elements, then vapour pressure of solution is …….

Answer»

144 torr
72 torr
90 torr
125 torr

Answer :B
45.

At certain temperature, the r.m.s. velocity for CH_4gas molecules is 100 m/sec. This velocity for SO_2 gas molecules at same temperature will be

Answer»

400 m/sec
200 m/sec
50 m/sec
25 m/sec

Answer :C
46.

At certain temperature, the solution of benzene in toulene exihibits the vapour pressure (in m bar) representes as P= 150x + 65, where 'x' is mole fraction of benzene, the vepour pressure of pure benzene is

Answer»

150 m BAR
65 m bar
90 m bar
215 m bar

ANSWER :D
47.

At certain temperature a solution of 3mol of a liquid A(P^(0)=500 torr) and 1 mol of liquide B(P^(0)=340 torr) has a vapour pressure of 435 torr , we can conclude that

Answer»

a solution of 5 mL of A and 50 mL of B must have a volume of 100 mL .
the solution shows negative DEVATION from Raoult's law
on mixing the two LIQUIDS heat be given out so as to OBTAIN solution at the same temperature
entropy of mixing is zero .

Answer :B::C
48.

At certain temperature and infinite dilution , the equivalent conductances of sodium benzoate , hydrochloric acid , and sodium chloride are 240, 349 and 229 Omega^(-1) cm^(2) eq^(-1) respectively . The equivalent conductance of benzoic acid in Omega^(-1) cm^(2) eq^(-1) at the same condition is

Answer»

80
328
360
408

Solution :`Lambda_(C_(6)H_(5) COOH)^(@) = Lambda_(C_(6)H_(5) COONa)^(@) + Lambda_(HCl)^(@) - Lambda_(NACL)^(@)`
`Lambda_(C_(6)H_(5) COOH)^(@) = 240 + 349 - 229`
`Lambda_(C_(6)H_(5) COOH)^(@)= 240 + 349 - 229`
`Lambda_(C_(6)H_(5)COOH)^(@) = 360 Omega^(-1) CM^(2) eq^(-1)`
49.

At certain temperature 1.6% solution of an unknown substance is isotonic with 2.4 % solution of Urea. If both the solutions have the same solvent and both the solutions have same density 1 gm/cm^(3), what will be the molecular mass of unknown substance in gm/mol. [Mole mass of urea = 60 gm/mol]

Answer»

40
90
80
30

Solution :For, isotonic solution
`((% w//V)/(M))_("UREA")=((%w//V)/(M))_("unknown")`
`(2.4)/(60)=(1.6)/(M)`
`THEREFORE M=(1.6xx60)/(2.4)=40` gram/mole
50.

At cathode, the electrolysis of aqueous Na_(3)SO_(4) gives

Answer»

`NA`
`H_(2)`
`SO_(3)`
`SO_(2)`

Solution :Calomel electrode USED asa reference electrdrodeuses `Hg_(2)Cl_(2)`.