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.

In AB_(4)L_(x), Molecule is non polar then value of x is : L = Lone pair, A, B = Type of atom

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0
1
2
(A) & (C ) both

Solution :Con. `H_(2)SO_(4)` behave as DEHYDRATING agent
2.

In a zero order reactionn 47.5% of the reactant remains at the end of 2.5 hours. The amount of reactant consumed in one hour is

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0.105
0.32
0.526
0.21

Solution :`C_(0)-C_(t)=K.t`……………(1), `x=Kxx1`….2 `52.5/x=2.5/1impliesx=52.5/2.5=21`
3.

In A^(=)B^(-) ionic compound radii of A^(=) and B^(-) ions are 180 pmand 187 pm respectively .The crystal structure of this compound will be

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`NaCI` type
`CACI` type
`ZnS` type
SIMILAR to diamond

SOLUTION :`r_(+) // r_(-) = (180)/(187) = 0.962 `which lies in the range of `0.732 - 1.000` hence coordination number `= 8 `i.e. the STRUCTURE is `CaCI` type.
4.

In a zero order reaction, for every 10^(@) rise of temperaure, reaction rate is doubled. If the temperature 10^(@)C to 100^(@)C, the reaction rate will become:

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256 times
512 times
64 times
128 times

Solution :b) For every `10^(@)` rise in TEMPERATURE, REACTION rate increase by 2.
`(r_(100^(@))C)/(r_(10^(@))C) = 2^((100-10))/(10) = 2^(9)=512` times.
5.

In a zero order reaction, the time taken to reduce the concentration of reactant from 50% to 25% is 30 minutes. What is the time required to reduce the concentration from 25% to 12.5%?

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SOLUTION :15 MINUTES
6.

In a zero order reaction, A toProducts, starting with 0.5 mol L^(-1), if 0.4 mol L^(-1) are present after 10 min., the rate constant of the reaction will be

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`0.05 MOL L^(-1) mi n^(-1)`
`0.04 mol L^(-1) mi n^(-1)`
`10^(-2) mol L^(-1) mi n^(-1)`
cannot be DETERMINED with the given data

Solution :For zero order reaction,
`k = 1/t[[A]_0-[A]]= 1/(10min)[0.5 - 0.4] mol L^(-1)`
`= 10^(-2) mol L^(-1) mi n^(-1)`
7.

In a zero-order reaction for every 10^(@) rise of temperature, the rate is doubled. If the temperature is increased from 10^(@)C to 100^(@)C . The rate of the reaction will become

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`64` TIMES
`512` times
`256` times
`128` times

Answer :B
8.

In which of these processes platinum is used as a catalyst ?

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OXIDATION of ammonia to form `HNO_(3)`
Hardening of OILS
Production of synthetic rubber
Synthesis of methanol

Answer :A
9.

In a voltaic cell, ar 298K if activity of both electrolyte is unify then according to Nernst'sequation

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`E=E^(0)`
`E gt E^(0)`
`E lt E^(0)`
`E NE E^(0)`

SOLUTION :Defination of STANDARD ELECTRODE.
10.

In a vessel, two equilibrium are simultaneously established at the same temperature as follows: N_(2)(g) + 3H_(2) (g)hArr2 NH_(3) (g) ...(1) N_(2)(g) + 2H_(2) (g)hArrN_(2)H_(4) (g) ...(2) Initially the vessel contains N_(2) and H_(2 )in the molar ratio of 9 : 13. The equilibrium pressure is 7P_(0) , in which pressure due to ammonia is P_(0) and due to hydrogen is 2P_(0). Find the values of equilibrium constants (K_(P)’s) for both the reactions

Answer»

SOLUTION :`K_(P_(1))=1/(20O_(0)^(2)),K_(P_(2))=3/(20P_(0)^(2))`
11.

In a vessel containing SO_3, SO_2 and O_2 at equilibrium,some helium gas is introduced os that the total pressure increases while temperature and volume remain constant. According to Le chatelier's principle the dissociation of SO_3 :

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Increases
decreases
Reamains unaltered
Changes unpredictably

Answer :C
12.

In a vessel containing SO_(3),SO_(2)and O_(2) at equlibrium, some helium gas is introduced so that the total pressure increases while temperature and volume remain constnat. According to Le-Chatelier principle the dossociation of SO_(3)

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Increases
DECREASES
REMAINS unaltered
Changes unpredicatable

Answer :C
13.

In a vessel containing SO_(3) , SO_(2) and O_(2) at equilibrium , some He gas is introduced so that total pressure increases while temperature and volume remain constant . According to Le Chatelier's principle , the dissociation of SO_3 :

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increases
decreases
does not CHANGE
CHANGES unpredictably

Answer :C
14.

In a vessel 2 g of O_2 2g Of H_2 and 2 g of N_2 are present. Which has the largest no of atoms?

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`O_2`
`H_2`
`N_2`
All have EQUAL no of atoms

Answer :B
15.

In a tyre of "Ferrari" car of Mr. Obama, a tube having a volume of 12.3 litres is filled with air at a pressure of 4 atm at 300 K.Due to travelling, the temperature of the tube and air inside it raised to 360 K and pressure reduced to 3.6 atm in 20 minutes. If the porosity (number of pores per unit area) of the tube material is 5xx10^3 pores/cm^2 and each pore can transfer air from inside to outside of the tube with the rate of 6.023xx10^8 molecules per minute.Calculate the total surface area (m^2) of the tube.(R=0.082 Lt-atm/mole-K) Give your answer divide by 100.

Answer»


Solution :Initial MOLES of air inside the tube `=(4xx12.3)/(0.082xx300)=2`
moles of air after 20 minutes`=(3.6xx12.3)/(0.082xx360)=1.5`
`:.` moles of air leaked in 20 minutes =2-1.5=0.5
`:.` MOLECULES of air leaked in 20 minutes `=0.5 N_A`
Let the total surface AREA of tube =`A cm^2`
Total molecules ESCAPED in 20 minutes`=(5xx10^6)A xx 6.023xx10^(8)xx20=0.5 N_A`
`:. A=(0.5xx6.023xx10^23)/(5xx10^5xx6.023xx10^8xx20)=5xx10^7 cm^2 =5000 m^2`
16.

In a transition series, with the increase in atomic number, the paramagnetism

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increases gradually
DECREASES gradually
FIRST increases to a MAXIMUM and then decreases
first decreases to a maximum and then increases

Answer :C
17.

In a transition series, as the atomic number increase paramagnetism:

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INCREASE gradually
DECREASE gradually
first increase to a MAXIMUM and then decrease
first decrease to a MINIMUM and then increase

Answer :C
18.

In a tetrahedral crystal field, the d-orbitals split into …… orbitals collectively called …… and …… orbitals collectively called ….. .

Answer»

Solution :`d_(X^(2)-y^(2))` and `d_(z^(2)), E and d_(XY), d_(YZ) and d_(xz), t_(2)`
19.

In a test-tube, there is 18 g of glucose (C_(6)H_(12)O_(6)) 0.08 mole of glucose is taken out. Glucose left in the test tube is

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0.10 G
0.02 g
0.10 mol
3.60 g

Solution :Molar mass of glucose, `C_(6)H_(12)O_(6) = 180 g mol^(-1)`
18 g glucose = 0.10 mol
Moles of glucose LEFT = 0.10-0.08 = 0.02 mol
Mass of glucose = `(0.02 mol)XX(180 g mol^(-1)) = 360G`
20.

In a tetragonal unit cell:

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a = b = c, `ALPHA = BETA = gamma NE 90^(@)`
a = b ne c, `alpha = beta = gamma = 90^(@)`
a = b = c, `alpha = beta = gamma = 90^(@)`
a = b ne c, `alpha = beta = 90^(@), gamma = 120^(@)`

ANSWER :D
21.

In a Tetragonal unit cell :

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a = b = c, `ALPHA = beta = GAMMA ne 90^(@)`
a = b ne c, `alpha = beta = gamma = 90^(@)`
a = b = c, `alpha = beta = gamma = 90^(@)`
a = b ne c, `alpha = beta = 90^(@), gamma = 120^(@)`

Answer :A
22.

In a system A(s) hArr 2B(g) +3C (g) if the conc. Of C at equilibrium is increased by a factor of 2, it will cause the equilibrium concentration of B to change to

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TWO times the original value
one half of its original value
`2sqrt2` time the original value
`1//2sqrt2` time the original value

Solution :`A(s) hArr 2B(g) +3C(g)`
Let, X and y be the concentrations of B and C at equilibrium respectively.
`therefore K_c =X^2Y^3` …………..(1)
Now, the concentration of C is changed from y to y. such that `y.=2y`. If x. is thenew concentration of B
`therefore K_c=(X.)^2(y.)^3=(x.)^2(2y)^3`........... (2)
From Eqs. (1) and (2)
`(x.)^2(8y^3)=x^2y^3`
`therefore x.=sqrt((x^2)/8)=x/(2sqrt2)`
`therefore` Equilibrium concentration of B changes to `1/(2sqrt2)` times the original value .
23.

In a system A(s) hArr 2B (g) +3C(g) if the conc. Of C at equilibrium is increased by a factor of , it will cause the equilbrium concentration of B to change to

Answer»

two times the original VALUE
one half of its original value
`2SQRT2` time the original value
`1//2sqrt2` times the original value

ANSWER :D
24.

In a system : A(s) hArr 2B(g) + 3C(g) If the concentration of C at equillibrium is increased by factor 2 then predict the equillibrium concentration of B in terms of original val,ue .

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Two TIMES of its original value
One half of its original value
`2sqrt2` times of its original value
`(1)/2 sqrt2` times of its original value

Answer :D
25.

In a successive radioactive disintegration underset(N_(1))(A)overset(lamda_(1))tounderset(N_(2))(B)overset(lamda_(2))toC …. No. of nuclides after time t. in which the parent has a longer but not much longer half-life than the daughter. Which of the following expressions is correct?

Answer»

`(N_(1))/(N_(2))=lamda_(2)/lamda_(1)`
`(N_(1))/(N_(2))=(lamda_(2)-lamda_(1))/lamda_(1)`
`N_(1)/N_(2)=lamda_(1)/lamda_(2)`
`N_(1)/N_(2)=(lamda_(2)-lamda_(1))/lamda_(2)`

Answer :B
26.

In a study of the reaction 2A + 2BhArr3C + D, A and B are mixed in a vessel at t^(@)C. The initial conc. of A is twice the initial conc. of B. After equilibrium is reached, the conc of C is three times the conc. of B. Calculate the equilibrium constant K_(C) in terms of P/Q and report P + Q.

Answer»


ANSWER :43
27.

In a system A(g) rarr 4B (g) + 3C(g) If the conc. of C at equilibrium is increased by a factor of 2, it will cause the equilibrium concentration of B to change to

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TWO times the original VALUE
ONE half of its original value
`2 sqrt(2)` times the original value
`(1)/(2) sqrt(2)` times the original value

Answer :D
28.

In a study of equilibrium: H_(2)(g) + I_(2)(g)

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SOLUTION :`{:(H_(2),+,I_(2), `RARR K =x^(2)/((1-x/2)(3-x/2))`………(1)
`(1-x/2) + 2-x/2 (3-x/2) -x/2, x+x`
`rArr 4-x, 3-x, 2x rArr K=(4x^(2))/(3-x)^(2)`
`therefore` From (1) and (2) x=1.5
`therefore` From (2) `K=(4(1.5)(1.5))/(3-1.5)^(2) rArr K=4`
29.

In a standard hydrogen electrode, the pressure of hydrogen gas and H^(+)concentration respectively are

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1 ATM, 1 M `H^(+)`
1 atm, 0.01 M `H^(+)`
10 atm, 1 M `H^(+)`
10 atm, 0.1 M `H^(+)`

ANSWER :A
30.

In a standard hydrogen electrode , the concentration of H^(+) is :

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0.1 M
1 M
10 M
Not fixed.

Answer :B
31.

In a square planar complex of the type [Mabex], the number of geometrical isomers can be :

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No GEOMETRICAL ISOMERS
THREE
Two
Four

Solution :It has three geometrical isomers. For details CONSULT SECTION 9.7.
32.

In a spontaneouse process, the system suffers

Answer»

INCREASE in INTERNAL ENERGY
lowering in FREE energy
no energy change
decrease in entropy

Answer :B
33.

In a spontaneouse irreversible process, the total entropy of the system and surroundings:

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REMAINS constant
Increases
Decreases
Zero

Answer :B
34.

In a spontaneous process, the entropy of the system and its surroundings

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EQUALS zero
Decreases
Increases
Remains constant

Solution :`DeltaS=+ve` than PROCESS is SPONTANEOUS.
35.

In a solution of the miscible volatile liquids A and B, the plots of their partical V.P. V_(s) their mole fractions is given by (Assume, V.P. of pure Agt V.P. of pure B)

Answer»




Solution :According to Raoult's lawof partial vapour PRESSURE.
Partial V.P.(P) of a component in a solution =V.P. of pure component` (P) xx `its mole fraction(X) in the solution i.e.`P_(S)=P_(O) xx x.`
This is an EQUATION of a straight line. Hence if `P_(O)` is plotted against x a straight line passing through the origin and with positive slope is obtained.
Hence OPTIONS (c) and (d) are incorrect. Now in option (a) and (B), points O and P represent V.P. of pure A and B respectively, and as V.P. of pure A gt V.P. of pure B. The correct option is (a) .
36.

In a solution of acid H^(+) concentration is 10^(-10)M. The pH of this solution will be

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8
6
Between 6 and 7
Between 3 and 6

Solution :AS the solution is ACIDIC `pH lt 7`. This is because `[H^(+)]` from `H_(2)O[10^(-7)]` cannot be neglected in COMPARISON to `10^(-10) M HCL`.
37.

In a solution of pH = 5, more acid is added in order to reduce the pH = 2. The increase in hydrogen ion concentration is

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100 times
1000 times
3 times
5 times

Solution :`[H^(+)]_(I) = 10^(-5) = [H^(+)]_(II) = 10^(-2)`.
THUS INCREASE in `[H^(+)] = (10^(-2))/(10^(-5)) = 1000` times.
38.

In a solution of acetic acid, sodium acetate is added, then its pH value

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Decreases
Increases
Remains unchanged
(a) and (B) both are correct

SOLUTION :Solution of `CH_(3)COONa` on ADDITION to ACID shows a decrease in DISSOCIATION of acid due to common ion effect. To decrease in `[H^(+)]` or increase pH.
39.

In a solution of 7.8 g benzene (C_6H_6) and 46.0 g toluene (C_6H_5CH_3) the mole fraction of benzene is:

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1/6
1/5
1/2
1/3

Answer :A
40.

In a solution of 7.8 gm benzene C_(6)H_(6) and 46.0 gm toluene (C_(6)H_(5)CH_(3)), the mole fraction of benzene in this solution is

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`1//6`
`1//5`
`1//2`
`1//3`

SOLUTION :Mole fraction at `C_(6)H_(6)=((7.8)/(78))/((7.8)/(78)+(46)/(92))=(1)/(6)`.
41.

In a solution of 7.8 g benzene (C_(6)H_(6)) and 46.0 g toluene (C_(6)H_(5)CH_(3)) the mole fraction of benzene is :

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`1//6`
`1//5`
`1//2`
`1//3`

SOLUTION :MOLE fraction of`C_(6)H_(6)=(7.8//78)/(7.8/78+46/92)=1/6`
42.

In a solution if the amount of solvent is doubled, keeping the amount of solutes same, the share of solute in the solution

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become half
WOULD decrease but not a half
remain unchanged
CHANGE unpredictably

SOLUTION :No. of moles of solvent are increacsing but total no. of moles of solute solvent are not BECOMING double, However the no. of moles of solute are same. HENCE the share of solute in the solution would decrease but not to half.
43.

In a solution containing 1 mol of ethyl alcohol and 4 mol of water, the mole fraction of water is :

Answer»

0.25
`0.20`
0.75
`0.80`

SOLUTION :MOLE FRACTION of WATER `=(4)/(4+1)=0.8`
44.

In a solution , 0.02 M acetic acid is 4% dissociation . The [OH^(-)] in the solution is

Answer»

`8 xx 10^(-4)`
`2xx10^(-14)`
`8xx10^(10)`
`1.25 xx 10^(-11)`

Solution :`[H^(+)]=sqrt(alpha^(2)C^(2))=sqrt((0.02)^(2)*(0.04)^(2))=8xx10^(-4)`
`[H^(+)][OH^(-)]=10^(-14)`
`:. [OH^(-)]=(10^(-14))/(8xx10^(-4))=1.25xx10^(-11)M`
45.

In a solid, oxide ions are arranged in CCP. Cations A occupy one-sixth of the tetrahedral voids and cations B occupy one-third of the octahedral voids. The formula of the compound is -

Answer»

`ABO_(3)`
`AB_(2)O_(3)`
`A_(2)BO_(3)`
`A_(2)B_(2)O_(3)`

Answer :A
46.

In a solid lattice the cation has left a lattice site and is located at an interstitial position. The lattice defect is known as -

Answer»

INTERSTITIAL DEFECT
Valency defect
Frenkel defect
Schottky defect

Answer :C
47.

In a solid lattice the cation has left a lattice site and is located at an interstitial position, the lattice defect is

Answer»

metal EXCESS DEFECT
vacancy defect
FRENKEL defect
schottky defect

ANSWER :C
48.

In a solid, atom M occupies ccp lattice and (1/3)^(rd) of tetrahedral voids are occupied by atom N. The formula of compound is .........

Answer»

`M_3N_2`
`M_2N_3`
`M_4N_3`
`M_3N_4`

Solution :Total number of M-atom = 4
Total number of TETRAHEDRAL voids = 8
Total number of N-atoms = `(8 xx 1/3) = 3`
RATIO of M-atoms to N-atoms = `M:N = 4:8/3`
` =3 :2`.
49.

In a solid lattice the cation and anion both have left a lattive site. The lattice defect is known as

Answer»

intestitial DEFECT
vancancy defect
Frenkel defect
schottky defect

ANSWER :D
50.

In a solid atom M occupies ccp lattice and ((1)/(3)) of tetrahedral voids = 2n Given that ((1)/(3))rd of tetrahedral voids are occupied i.e., ((1)/(3))xx2n are occupied by6 N atoms therefore M:NrArr n:((2)/(3)) 1:((2)/(3))3:2 rArr M_(3)N_(2)

Answer»

MN
`M_(3)N`
`MN_(3)`
`M_(3)N_(2)`

Solution :Hint : If the TOTAL number of M atoms is n, then the number of tetrahedral voids = 2n
Given that `((1)/(3))`rd of tetrahedral voids are occupied i.e., `((1)/(3))xx2n` are occupied by N atoms. `therefore M:N RARR n:((2)/(3))`
`1:((2)/(3))3:2 rArr M_(3)N_(2)`