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 a compound, nithrogen atoms (N)make cubic close pakced lattice and metal atoms (M)occupyone- third tetrahedral voids present. Determine the formula of the compound formed by M andN ?

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2.

In a compound, C, H and N are present in the ratio 9 : 1: 3.5 by weight. If the molecular weight. If the molecular weight of the compound is 108, what is the number of hydrogen atoms present in the molecular formula of the compound ?

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SOLUTION :Step I. Empirical FORMULA of the COMPOUND
`{:("Element","Percentage/weight (ratio)","Gram atoms (Moles)","Atomic ratio (Molar ratio)","Simplest whole no. ratio"),("C",9,(9)/(12)=0.75,(0.75)/(0.25)=3,3),("H",1,(1)/(1)=1.0,(1.00)/(0.25)=4,4),("N",3.5,(3.5)/(14)=0.25,(0.25)/(0.25)=1,1):}`
Empirical formula of the compound`= C_(3)H_(4)N`
Step II. Molecular formula of the compound
Empirical formula mass `= 3 xx 12 + 4 xx 1 +14 = 54 u`
Molecular mass = 108 u (given)
`n=("Molecular mass")/("Empirical formula mass")=((108u))/((54u))=2`
Molecular formula of the compound `= n xx` Empirical formula
`:.` Number of HYDROGEN atoms PRESENT = 8.
3.

In a compound C , H and N atoms are present in 9 : 1 : 3.5 by weight . Molecular weight of compound is 108. Molecular formula of compound is

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`C_(2) H_(6) N_(2)`
`C_(3) H_(4) N`
`C_(6) H_(8)N_(2)`
`C_(9) H_(12) N_(3)`

Answer :C
4.

In a compound, C, H and N are present in the ratio 9 : 1: 3.5 by weight. If the molecular weight. If the molecular weight of the compound is 108. Molecular formula of the compound is :

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`C_(2)H_(6)N_(2)`
`C_(3)H_(4)N`
`C_(6)H_(8)N_(2)`
`C_(9)H_(12)N_(3)`

Solution :Percentage
`C=(9)/(13.5)xx100=66.6`
`H=(1)/(13.5)xx100=7.4`
`N=(3.5)/(13.5)xx100=26.0`
Atomic ration
`C=underset(=5.5)((66.6)/(12)),H=underset(=7.4)((7.4)/(1)),N=underset(=1.86)((26.0)/(14))`
Simplest ratio
`C = 3,H=4,N=1`
Empirical formula `= C_(3)H_(4)N`
Empirical formula MASS `= 36 + 4 + 14 = 54`
`n=("Molecular mass")/("Empirical formula mass")`
`=(108)/(54)=2`
Molecular formula `= C_(6)H_(8)N_(2)`.
5.

In a compound, atoms of element Y form ccp lattice and those of element X occupy the corners, Y atoms in the body centred positions and Z atoms at the centres of faces of the unit cell. What is the empiricalformula of the compound ?

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`XY_(2)Z_(3)`
`XYZ_(3)`
`X_(2)Y_(2)Z_(3)`
`X_(8)YZ_(6)`

SOLUTION :No. of ATOMS of X per unit CELL `=8times 1/8 = 1`(at corners)
No. of atoms of Y per unit cell = 1 (at body centre)
No. of atoms of Z per unit cell =`6times 1/2 = 3` (at face centres)
`therefore "formula is "XYZ_(3)`
6.

In a compound , atoms of element Y form ccp lattice and those of element X occupy 2/3rd of tetrahedral voids.The formula of the compound will be

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` X_(2)Y`
` X_(3)Y_(4)`
`X_(4) Y_(3)`
`X_(2)Y_(3)`

ANSWER :D
7.

In a compound, atoms of element Y form ccp lattice and those of element X occupy 2//3^(rd) of tetrahedral voids. The formula of the compound will be

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`X_(3)Y_(4)`
`X_(4)Y_(3)`
`X_(2)Y_(3)`
`X_(2)Y`

Solution :NUMBER of Y atoms per unit CELL in ccp lattice (N) = 4
Number of tetrahedral voids `= 2N=2xx4=8`
Number of tetrahedral voids occupied by `X=(2)/(3)rd` of the tetrahedral void `=(2)/(3)xx8=(16)/(3)`
HENCE, the formula of the COMPOUND will be `X_(16//3)Y_(4)=X_(4)Y_(3)`
8.

In a complex compound ligand acts as

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Lewis acid
Lewis BASE
Lowry-Bronsted acid
ARRHENIUS base

SOLUTION :` SiF_4`has vacant d- orbitals but cannot add or GIVES`H^(+) `
` CO_2`cannotadd or give ` H^(+) `
` SO_2`has lone pair
` NH_3`can give and TAKE `H^(+) `and has lone pair
9.

In a colloidal solution of AgCl in water the AgCl particles

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do not CARRY any CHARGE
carry a +ve charge
carry a -ve charge
may carry a +ve or -ve charge depending upon conditions.

Solution :AgCl particles in colloidal solution canadsorb both `Ag^(+)` or `CL^(-)` IONS depending upon which EVER is present in excess.
10.

In a collection of H-atoms, all the elctrons jump from n = 5 to ground level finally (directly or indirectly), without emitting any line in Balmer series. The number of possible different photons are

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10
8
7
6

Solution :NUMBER of SPECTRUM LINES `=(N(n-1))/(2)`
11.

In a closed vessel the rate of solubility of sugar and rate of crystallization is same. What it indicate ?

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SOLUTION :At this temperature SATURATED solution of SUGAR in WATER and solubility of sugar remains constant.
12.

In a closed vessel at 25^(@)C temperature 4 mole O_(2), 3 mole Cl_(2) and 3 mole N_(2) are mixed and the total pressure found is 50 bar. Find the partial pressure of each gas.

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<P>

Answer :`p_(O_(2))=20` BAR, `p_(Cl_(2))=15` bar, `p_(N_(2))=15` bar
13.

In a closed vessel at 448^@C 0.5 mol H_2 and 0.5 mol I_2 react and from hydrogen iodide. Reaction H_(2(g)) + I_(2(g)) hArr2HI_((g)) of K_c = 50. (i) At equilibrium the moles of I_2 which are unreacted. (ii) Calculate K_p.

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SOLUTION :`K_c=K_p`=50, 0.11 MOL `I_2`
14.

In a closed system A(S) hArr 3 B(g)+3C(g) If partial pressure of C is doubled, then partial pressuer of B will be ........ Time the original value.

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SOLUTION :`(1)/(SQRT(2))`
15.

In a closed system : A_((s))hArr2B_((g))+3C_((g)) if the partial pressure of C is doubled then partial pressure of B will be

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TWICW the ORIGNAL PRESSURE
Half of its orignal pressure
`(1)/(2sqrt(2))` times, the original pressure
`2sqrt(2)` times its original pressure

Answer :C
16.

In a closed room of 1000m^(3) a perfume bottle is opened up. The room develops a smell. This is due to which property of gases?

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Viscosity
Density
DIFFUSION
None

Solution :PARTICLE of gas are FREELY to MOVE. Particles spread throughout the move due to the diffusion.
17.

In a closed system: A(s) hArr 2B(g)+3C(g), if the partial pressure of C is doubled at equillibrium, then partial pressure of B will be :

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TWO TIMES the original value
one-half of its original value
`(1)/(2sqrt2)` times the original value
`2sqrt2` times the original value

Answer :C
18.

In a closed container two reactions take place simultaneously (i) PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)and (ii)CO(g)+Cl_(2)(g)hArr COCl_(2)(g) On adding more CO into the container, select the correct option : (P) Degree of dissociation of PCl_(5)(g) decreases. (Q) Conc. of CO(g) at new equilibrium position is less than that of at initial equilibrium conc. (R) Degree of dissociation of PCl_(5)(g) increases.

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<P>Only (P)
Only (R)
Both (P) and (Q)
Both (Q) and (R)

ANSWER :B
19.

In a closed container at 1 atm pressure, 2 mole of SO_(2) (g) and 1 mole of O_(2)(g) were allowed to react to form SO_(3)(g) under the influence of a catalyst. The following reaction occurred: 2SO_(2)(g)+O_(2)(g) hArr 2SO_(3)(g) At equilibrium it was found that 50% of SO_(2)(g) was converted to SO_(3)(g). The partial pressure of O_(2)(g) at equilibrium will be :

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0.66 atm
0.493 atm
0.33 atm
0.2 atm

Answer :D
20.

In a closed container following equilibrium will be attained- A(s)+B(g)hArrAB(g) B(g)+C(g)hArrBC(g) On adding He gas (inert) to the above system at constant pressure & temperature

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Amount of `AB(g)` will be increased surely
Amount of `B(g)` will be decreased surely
Amount of `C(g)` will be decreased surely.
Amount of `BC(g)` will be decreased surely.

Solution :On addign INERT gas at content temperature & pressure `2^(nd)` reaction will be shifted in backward direction.
21.

In a closed container and at constant temperature 0.3 mole of SO_(2) & 0.2 mole of O_(2) gas at 750 torr are kept with a catalyst . If at equilibrium 0.2 mole of SO_(3) is formed the partial pressure of SO_(2) is …torr

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375
187
360
150

Answer :D
22.

In a close vessel PCl_(5(g))is obtained by the chemical reaction between PCl_3 and Cl_2. If the equilibrium concentration in this vessel of PCl_3, Cl_2 and PCl_5 at 500 K tempe. is 1.59 M, 1.59 M and 1.41M respectively. Than find equilibrium constant. PCl_(3(g)) + Cl_(2(g)) hArr PCl_(5(g))

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SOLUTION :`K_c="0.558 mol L"^(-1) , K_P = 1.3xx10^(-2)`BAR
23.

In a chemical reaction, chlorine atom undergoes reduction and aluminium atom undergoes oxidation. Will this redox reaction affect their initial number of protons, neutrons and electrons?

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Solution :Chlorine atom on accepting an electron BECOMES `Cl^(-)` and aluminium atom after donating an electron becomes `Al^(3+)`
This changes will affect their number of protons, neutrons and ELECTRONS.
`Cl^(-)` :
No. of electrons = 17 + 1 = 18
No. of protons = 17
No. of neutrons = 35 - 17 = 18
`Al^(3+)` :
No. of electrons = 13 - 3 = 10
No. of protons = 13
No. of neutrons = 27 - 13 = 14
24.

In a chemical equilibrium, the rate constant for the forward reaction is 2.5xx10^(2) and the equilibrium constant is 50. The rate constant for the reverse reaction is.....

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11.5
5
`2XX10^(2)`
`2xx10^(-3)`

Solution :`K_(f)=2.5xx10^(2)`,`K_(C)=50` K=?
`K_(C)=(K_(f))/(K_(R))rArr50=(2.5xx10^(2))/(K_(r))rArrK_(r)=5`
25.

In a chemical equilibrium, the rate constant for the forward reaction is 2.5xx10^(2) and the equilibrium constant of 50. The rate constant for the revese reaction is.

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11.5
5
`2XX10^(2)`
`2xx10^(-3)`

ANSWER :B
26.

In a certain sample of gas at 25^(@)C the number of molecules having speeds between 4 km sec^(-1) and 4.1 km "sec"^(-1) is N. if the total number of gas molecules at the same temperature are doubled what will happen?

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Value of most probable velocity will change
Area under the Maxwell's curve for distribution of speeds will INCREASES by four times.
No of MOLECULES between 4 km `"SEC"^(-1)` and 4.1 km `"sec"^(-1)` will become 2N.
No of molecules between 4km `"sec"^(-1)` amd 4.1 km `"sec"^(-1)` will remain same.

Answer :c
27.

In a certain polar solvent, PCl_(5) undergoes ioonisation as follows: 2PC l_(5) hArr PCl_(4)^(+)+PCl_(6)^(-), predict geometrical shapes of all the species involved.

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Solution :The geometrical shapes of the molecules or ions can be PREDICTED from the hybridisation state of the central atom.
for `PC l_(5):H=(1)/(2)[5+5-0+0]=(10)/(2)=5`
THEREFORE, central P-atom is `SP^(3)d`-hybridised. Consequently, the molecule is trigonal bipyramidal in shape.
For `PCl_(4)^(+):H=(1)/(2)[5+4-1+0]=(8)/(2)=4`.
Therefore, central P-atom is `sp^(3)`-hybridised. consequently, the geometrical shape of this ion is tetrahedral.
For `PCl_(6)^(-):H=(1)/(2)[5+6-0+1]=(12)/(2)=6`.
Thereofore, the central P-atom is `sp^(3)d^(2)` hybridised. consequently, the geometrical shape of the ion is octahedral.
28.

In a certain electronic transition from the quantum level, 'n' to the ground state in atomic hydrogen in one or more steps, no line belonging to the Bracket series is observed. What wave number may be observed in the Balmer series ? (R = Rydberg Constant)

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Answer :BAR(vartheta) = (3R)/(16); bar(vartheta) = (5R)/(36)`
29.

In a ccp structure of X atoms, Y atoms occupy all the octahedral holes. If 2X atom are removed from corners and replaced by Z, then the formula of the compound will be

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`X_(15)Y_(16)Z`
`X_(7)Y_(8)Z`
`X_(7.5)Y_(8)Z`
`X_(8)Y_(8)Z_(3)`

Answer :A
30.

In a carius determination, 0.234g of an organic substance gave 0.334g of barium sulphate. Calculate the weight percentage of sulphur in the given compound

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ANSWER :0.1966
31.

In a buffer solution consisting NaH_2PO_4 and Na_2HPO_4select the correct statement among the following

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`NaH_2PO_4 ` is ACID and `Na_2HPO_4` is SALT
The PH of solution can be calculated using the relation ` :pH =pK_2+ log _(10) "" ([HPO_4^(2-)])/([H_2PO_4^(-) ]) `
`Na_2HPO_4` is acid and `NaH_2PO_4` is salt
The pH cannot be calculated

Solution :`H_2 PO_4^(-)to HPO_4 ^(2- )`
` A "" " salt of Wa" `
` pH =pKa_2 +log ""(["salt"])/(["acid"]) `
32.

In a bonded molecule, the order of repulsion between the bonded and non-bonded electrons is

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LONE PAIR - lone pair `GT` BOND pair - bond pair `gt` lone pair - bond pair
bond pair - bond pair `gt` lone pair - lone pair `gt` lone pair - bond pair
lone pair - lone pair `gt` lone pair - bond pair `gt` bond pair - bond pair
bond pair - bond pair `gt` lone pair - bond pair `gt` lone pair - lone pair.

Answer :C
33.

In a bomb calorimeter , volume is constant. Do you think thatDeltaH = DeltaU? Why or why not ?

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Solution :No, `DeltaH cancel(=) DeltaU`. This is BECAUSETHE relation `DeltaH = DeltaU + P DeltaV` holds good only at constant PRESSURE. This may be seen as FOLLOWS `:`
`H =U +P Delta V :. Delta H= Delta U + Pdelta V + V DeltaP`
At constant pressure, `Delta p = 0 ,:.DeltaH = Delta U +P DeltaV`
At constant VOLUME, `Delta V = 0 , ,' DeltaH = Delta U + VDeltaP`
34.

In a Bohr's model of atom when an electron jumps from n=1 to n=3 , how much energy will be emitted or absorbed

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`2.15xx10^(-11)` ERG
`0.1911xx10^(-10)` erg
`2.389xx10^(-12)` erg
`0.239xx10^(-10)` erg

Solution :According to Bohr's model `DeltaE=E_1-E_3`
`=2.179xx10^(-11)-(2.179xx10^11)/9`
`=8/9xx2.179xx10^(-11)=1.91xx10^(-11)=0.191xx10^(-10) ` erg
Since ELECTRON is GOING from n=1 to n=3 HENCE energy is absorbed
35.

In a body-centred cubic crystal of an element , the ratio of the readius of the radius of the atom toi the edge of the unit cell is ……

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ANSWER :No. of atoms PER unitcell, AVOGADRO, NUMBER
36.

In a body-centred cubic crystal of an element, the ratio of the radius of the atom to the edge of the unit cell is __________

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ANSWER :0.433
37.

In a body centred cubic cell, an atom at the body of centre is shared by:

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1 unit CELL
2 unit CELLS
3 unit cells
4 unit cells

ANSWER :A
38.

In a binary solution.

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SOLVENT my be LIQUID
solvent may be SOLID
SOLUTE may be gas
any of THSES

Answer :D
39.

In a basic buffer, 0.0025 mole of NH_(4)Cl and 0.15 mole of NH_(4) OH are present. The pH of the solution will be (pK_(a)=4.74)

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11.04
10.24
6.62
5.48

Solution :From Henderson's equation,
`POH = pK_(b) + log. (["Salt"])/(["Base"])`
`=4.74+log.(0.0025//V)/(0.15//V)=4.74+log. (1)/(60)`
`=4.76-1.78=2.96`.
HENCE, pH = 14 - pOH = 14 - 2.96 = 11.04.
40.

In a bcc-arrangement which of the marked planes have maximum spatial density of atoms ?

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1
2
3
4

Answer :A
41.

In a balanced redox reaction net gain of electron (s) is equal to net loss of electrons (s).n_("factor") isa reaction specific parameter and for intermolecular redox reaction n-factor of oxidising reducing agent is the no. of moles of electron gained /lost by one mole of compound. 50mL 0.1 M CuSO_(4) are mixed with 50mL of 0.1 M KI then, number of moles of electrons involved in the reaction will be:

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4
2.5
`2.5xx10^(-3)`
NONE of these

Answer :C
42.

In a balanced equation H_(2)SO_(4)+xHIrarrH_(2)S+yI_(2)+zH_(2)O. The values of x,y and z are

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3, 5 & 2
4, 8 & 5
8, 4 & 4
5, 3 & 4

Answer :C
43.

In a attempt to prepare propane by Wurtz reation 1 mole of methyl bromide and 1 mole of ethyl bromide are treated with sodium . Assuming equal probability for all possible reaction . How many 'g' of propane will be obtained ?

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`44G`
`22G`
`33G`
`14.67g`

ANSWER :A::D
44.

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

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NaCl type
CsCl type
ZnS type
similar to diamond

Solution :(b) `(R^(+))/(r^(-))=(180)/(187)=0.96`, this VALUE lies in the RANGE of 0.732-1.000
45.

In a 7.0 L evacuated chamber, 0.50 mole H_(2) and 0.50 mole I_(2) react at 427^(@) C. According to reaction H_(2)(g)+I_(2)(g)hArr2HI(g) At the given temperature, K_(c)=49 for the reaction. What is the partial pressure (atm) of HI in the equilibrium mixture?

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6.371
12.77
40.768
646.58

Answer :A
46.

In a 7.0 L evacuated chamber, 0.50 mole H_(2) and 0.50 mole I_(2) react at 427^(@) C. According to reaction H_(2)(g)+I_(2)(g)hArr2HI(g) At the given temperature, K_(c)=49 for the reaction. What is the value of K_(p)?

Answer»

7
49
24.35
None of these

Answer :B
47.

In a 7.0 L evacuated chamber, 0.50 mole H_(2) and 0.50 mole I_(2) react at 427^(@) C. According to reaction H_(2)(g)+I_(2)(g)hArr2HI(g) At the given temperature, K_(c)=49 for the reaction. How many moles of the iodine remain unreached at equilibrium?

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0.388
0.112
0.25
0.125

Answer :B
48.

In a 7.0 L evacuated chamber, 0.50 mole H_(2) and 0.50 mole I_(2) react at 427^(@) C. According to reaction H_(2)(g)+I_(2)(g)hArr2HI(g) At the given temperature, K_(c)=49 for the reaction. What is the total pressure (atm) in the chamber?

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83.14
831.4
8.21
None of these

Answer :C
49.

In a 7.0 Levacuated chamber, 0.50 mol H_(2) and 0.50 mol I_(2) react at 427^(@) C. H_(2) (g) + I_(2) (g) hArr HI (g)+ Heat , At the given temperature, K_(c) = 49for the reaction.What is the total pressure (atm) in the chamber ?

Answer»

<P>83.14
831.4
8.21
None of these

Solution :PV=nRT
`IMPLIES p=(nRT)/(V)=(1 xx 0.0821 xx 700)/(7)=8.21` atm
50.

In a 7.0 Levacuated chamber, 0.50 mol H_(2) and 0.50 mol I_(2) react at 427^(@) C. H_(2) (g) + I_(2) (g) hArr HI (g)+ Heat , At the given temperature, K_(c) = 49for the reaction.what is the value of K_(p) ?

Answer»

<P>7
49
24.5
None of these

Solution :`K_(P)=K_(c), DELTA ng =0`