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.

A solution containing 1MXSO_4(aq) and 1MYSO_4(aq) is electrolysed. If conc. Of X^(2+) is 10^(-z)M when deposition Y^(2+) and X^(2+) starts simultaneously, calculate the value of Z. Given: (2.303RT)/F=0.06 E_(x^(2+)"|"X)^(@)=-0.12V,E_(Y^(2+)"|"Y)^(@)=-0.24V

Answer»


ANSWER :D
Discussion
2.

A solution consists of 0.2 MNH_4 OH and 0.2 MNH_4Cl .If K_b of NH_4 OH is 1. 8 xx 10 ^(-5), the [OH^(-) ] of the resulting solution is

Answer»

`0.9 XX 10 ^(-5) M`
` 1.8 xx 10 ^(-5) `
` 3.2 xx 10 ^(-5) M`
` 3. 6 xx 10 ^(-5) ` M

SOLUTION :`POH =P^(K_b)+ log ""( [S])/([B]) `
If ` [S] =[B] rArr pOH = P^(K_b) rArr [OH^(-) ] =K_b`
Discussion
3.

A soltuion contains mixture of H_(2)SO_(4) and H_(2)C_(2)O_(4), 25mL of N//10 NaOH for neutralization and 23.45 mL fo N//10 KMnO_(4) for oxisation. Calcualte: (i) Normality of H_(2)C_(2) O_(4) and H_(2)SO_(4) (ii) Strength of H_(2)C_(2)O_(4) and H_(2)SO_(4). Assume molecular weight of H_(2)C_(2)O_(4) = 126

Answer»


ANSWER :(i) `H_(2)C_(2)O = 0.0938N, H_(2)SO_(4) = 0.0482N`,
(II) `H_(2)SO_(4) = 2.362g//"litre"`
Discussion
4.

Solubility of NaCl in heavy water is

Answer»

Both A and R are TRUE and R is the correct EXPLANATION of A
Both A and R are true but R is not the correct explanation of A
A is true but is false
A is false but R is true

ANSWER :D
Discussion
5.

A solid substance AB has a rock salt geometry. What is the coordination number of A and B ? Ho amy atoms oif A and B are presnet in the unit cell ?

Answer»

SOLUTION :6, 6, : 4,4,
Discussion
6.

A solid substance AB has a rock salt geometry . What is the coordination number of A and B ? How many atoms of A and B are present in the unit cell ?

Answer»

SOLUTION :6, 6, , 4 ,4
Discussion
7.

A solid mixture(5.000 g) consisting of lead nitrate and sodium nitrate was heated below 600°C until the weight of the residue was constant. If the loss in weight is 28%, find the amount of lead nitrate and sodium nitrate in the mixture.

Answer»


Solution :Suppose, the given mixture CONTAINS x g of lead nitrate.
`therefore` The amount of sodium nitrate in the mixture = `5.0 - x` g The corresponding equations are:
`2Pb(NO_(3))_(2) to UNDERSET(2 xx 331.22 g)(2PbO) + 4NO_(2) uarr + underset(2 xx 223.2 g)(O_(2)) uarr`
`2NaNO_(3) to underset(2.85 g)(2NaNO_(2)0 + underset(2 xx 69 g)(O_(2)) uarr`
According to these equations, weight of PBO formed by heating x g of `Pb(NO_(3))_(2)`
`=(2 xx 223.2)/(2 xx 331.22) xx x = 0.6739 g`
and weight of `NaNO_2` obtained by heating `(5.0 -x) g` of `NaNO_(3)`
`=(2 xx 69)/(2 xx 85) xx (5.0-x) = 0.8118(5.0 -x)`g of `NaNO_(3)`
`therefore` Total weight of the residue = = 0.6739 x + 0.8118 (5.0 - x) g
Since, the mixture loses 28% on heating, the residue obtained on heating 5.0 g of mixture
`=(100-28)/100 xx 5.000 = 3.600 g`
The amount of `Pb (NO_3)_2` in the mixture = 3.328 g and the amount of `NaNO_3` in the mixture = 5.0 - 3.328 = 1.672 g
Discussion
8.

A solid is made up of two elements P and Q. Atoms Q are in ccp arrangement while atoms P occupy all the tetrahedral sites. What is the formula of the compound ?

Answer»


Solution :SUPPOSE number of ATOMS Q=N
Then number of tetrahedral sites =2 n
`THEREFORE` Number of atoms P=2n
`therefore` Ratio P:Q=2n:n =2 :1 , i.e., FORMULA is `P_2Q`
Discussion
9.

A solid element has specific heat 1 I gm^(-1)k^(-1).If the equivalent weight of the element is 9. Identify the valency and atomic weight of element.

Answer»

2, 6
3, 27
9, 28
6, 27

Solution :At WT = `(6.4)/("SP. Heat")=26.75~=27`
Valency = `("atomic WEIGHT")/("EQUIVALENT weight")=(26.75)/(9)~=3`
Discussion
10.

A solid compound XY has NaCl structure. If the radius of the cation is 100 pm, the radius of the anion (Y^-) will be

Answer»

275.1
322.5 PM
241.5 pm
165.7 pm

Solution :NACL has face-centred cubic arrangement of `Cl^-` IONS and `NA^+` ions are present in the octahedral voids. Hence, for such a SOLID radius of cation =0.414 xradius of the anion ( or `r_+/r_(-)`=0.414)
i.e., `r_+= 0.414xxr_(-)`
or `r_(-)=r_+/0.414=100/0.414`=241.5 pm
Discussion
11.

A solid compound 'X' on heating gives CO_(2)gas and a residue. The residue when mixed with water forms 'Y'. On passing an excess of CO_(2)through 'Y' in water a clear solution 'Z' is obtained. On boiling 'Z' compound 'X' is reformed. The compound 'X' is.....

Answer»

`Ca(HCO_(3))_(2)`
`CaCO_(3)`
`Na_(2)CO_(3)`
`K_(2)CO_(3)`

ANSWER :B
Discussion
12.

A solid compound XY has NaCl structure. If the radius of the cation is 100 pm, the radius of the anion (y^(-))will be

Answer»

275.1 PM
322.5 pm
241.5 pm
165.7 pm

Solution :In NaCl, Cl' ion from ccp and `Na^(+)` IONS OCCUPY octahedral voids.
`R^(+)//r=0.414 or 100/r=0.414 ,`
`r^(-)=(100"pm")/0.414=241.5 " pm"`
Discussion
13.

A solid compound 'X' on heating gives CO_(2)gas and a residue. The residue mixed with water forms 'Y'. On passing an excess of CO_(2)through Y' in water, a clear solution 'Z' is obtained. On boiling 'Z', compound 'X' is reformed. The compound 'X' is....

Answer»

`Na_(2)CO_(3)`
`K_(2)CO_(3)`
`Ca(HCO_(3))_(2)`
`CaCO_(3)`

Solution :The given compound X must be `CaCO_(3)`. It can be explained by FOLLOWINGREACTIONS,
`underset((X))(CaCO_(3)) overset(Delta)to underset(("Residue"))(CaO+CO_(2)UARR),CaO+H_(2)underset((Y))to Ca(OH)_(2)`
`Ca(OH)_(2)+CO_(2)+H_(2)O to underset((Z))(Ca(HCO_(3)))`
`Ca(HCO_(3))_(2) overset(Delta)to underset((X))(CaCO_(3))+CO_(2)uarr+H_(2)O`
Discussion
14.

A solid compound contains X, Y and Z atoms in a cubinc lattice with X atoms occupying ht corners. Y atoms in the body centred positions and Z atoms at tjee centres of faces of the unit cell. What is the empricial formula of the comound?

Answer»

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

SOLUTION :(b) `Z_(x)=(1)/(8)xx8=1` `Z_(y)=1` "["At BODY centre']", `Z_(3)=3` "["At face centre"]"
Discussion
15.

A solid compound contains X,Y and Z atoms in a cubic lattice with X atoms occupying the corners, Y atoms in the body centred positions and Z atoms at the centres of faces of the unit cell. What is the empirical formula of the compound ?

Answer»

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

SOLUTION :Atoms of X per UNIT CELL `=(8times1)/8=1`
Atoms of Y per unit cell =1
Atoms of Z per unit cell `=6times1/2=3`
`therefore` the formula of the compound is `XYZ_(3)`
Discussion
16.

A solid acts as an adsorbent because it has

Answer»

A definite shape
Small PORES in it
Unsaturated valencies
A HIGH LATTICE energy

Answer :C
Discussion
17.

A solid AB has NaCl structure. If the radius of the cation A is 100 pm, what is the radius of the anion B

Answer»


SOLUTION :As NACL has OCTAHEDRAL STRUCTURE, `r_A^+=0.414xxr_B^-`
Discussion
18.

A solid AB has NaCl structure. If the radius of the cation A is 100 pm, what is the radius of the anion B ?

Answer»


SOLUTION :As NaCl has octahedral STRUCTURE` r_(A +) = 0.414 xx r_(B^(-))`
Discussion
19.

A solid AB has NaCl structure. If the radius of cation A^+ is 170 pm, calculate the maximum possible radius of the anion B^-

Answer»


SOLUTION :The formula `r_+=0.414xxr_(-)` GIVES MAXIMUM POSSIBLE radius of ANION for a given cation for close packing.
Discussion
20.

A solid AB has NaCl structure. If the radius of cation A^(+) is 170 pm, calculate the maximum possibleradius of the anionB^(-)

Answer»


Solution :The FORMULA ` r_(+) = 0.414 xx r_(-)` GIVES maximum possible radius of ANION for a given CATION, for close packing
Discussion
21.

A solid AB has CsCl type structure. The edge length of the unit cell is 404 pm. Calculate the distance of closest approach between A^+ and B^- ions?

Answer»


SOLUTION :Distance of closest APPROACH is EQUAL to the distance between the nearest neighbours (d). As CSCL has BCC lattice,
`d=sqrt3/2a=1.732/2xx404` pm =349.9 pm
Discussion
22.

A solid AB has CsCl typestructure. The edge length of the unit cell is404 pm.Calculate the distance of closest approach between A^(+)and B^(-)ions ?

Answer»


Solution :Distance of closest approach is EQUAL to the distance between the nearest NEIGHBOURS (d)As CsCl has BCC latttic .,
`d = sqrt3/ 2 a = 1.732/2 XX 404"pm"= 349.9 ` pm
Discussion
23.

A solid A^+ B^- has NaCl type close packed structure. If the anion has a radius of 250 pm, what should be the ideal radius of the cation ? Can a cation C^+ having a radius of 180 pm be slipped into the tetrahedral site of the crystal A^+ B^- ? Give reason for your answer.

Answer»


Solution :`(r_(+)(C^+))/(r_(-)(B^-))="180pm"/"250pm"=0.72`
It does not lie in the RANGE 0.225-0.414 . HENCE, `C^+` cannot be slipped into the tetrahedralsite.
Discussion
24.

A solid A^(+)B^(-) has NaCl type close packed structure. If the anion has a radius of 250 pm,what should be the ideal radius of cation ? Can a cationC^(+)having a radius of 180 pm be slipped into the tetrahedral site of the crtstalA^(+) B^(-)?Given reason for your answer .

Answer»


SOLUTION :`(r_(+)(C^(+)))/(r_(-)(B^(-))) = (180"pm")/(250"pm") = 0.72`
It does not ,lie in the RANGE ( 0.225 - 0.414 . Hence`C^(+)`cannot be slipped into the tetrahedral site.
Discussion
25.

A solid A^+B^- has NaCl type close packed structure .If the anion has a radius of 241.5 pm , what should be the ideal radius of the cation ? Can a cation C^+ having radius of 50 pm be fitted into the tetrahedral hole of the crystal A^+ B^- ?

Answer»

Solution :As `A^+ B^-` has NACL structure, `A^+` ions will be present in the octahedral voids. Ideal radius of the cation will be equal to the radius of the octahedral void because in that case. It will touch the anions and the arrangement will be CLOSE packed. Hence, Radius of the octahedral void =`r_(A^+)=0.414 xxr_(B^-)=0.414xx241.5` pm =100.0 pm
Radius of the tetrahedral void =`0.225 xxr_(B^-)` =0.225 x 241.5 pm =54.3 pm
As the radius of the cation `C^+` (50 pm) is smaller than the size of the tetrahedral void, it can be placed into the tetrahedral void (but not exactly FITTED into it)
Discussion
26.

A solid A^(+)B^(-) has NaCl close packed structure. The radius of thecation when the radius of the anion is 250 pm is

Answer»

103.5 pm
207 pm
69 pm
276 pm

Solution :Since the solid `A^(+) B^(-)`has NaCl type close packed structure, it belongs to a system with coordination number 6. In thesecases the ratio of the cation to the anion radii is GIVEN by
`R^(+)/r^(-)ge0.414`
`"Now" r^(-)=250"pm"`
`thereforer^(+)=250times0.414=103.5"pm"`
Discussion
27.

A soldium salt of unknown anion when treated with MgCl_(2) gives white precipitate only on boiling. The anion is:

Answer»

`SO_(4)^(2-)`
`HCO_(3)^(-)`
`CO_(3)^(2-)`
`NO_(3)^(-)`

SOLUTION :The salt is `NaHCO_(3)`
`2NaHCO_(3) + MgCl_(2) rarr underset(("White ppt"))underset(MgCO_(3))underset(darr "heat")(Mg(HCO_(3))_(2))+ 2NaCl`
Discussion
28.

(A) :Soil becomes infertile due to acid rain (R) :Due to acid rain pH of the soil increases

Answer»

Both A and R are true , and R is correct EXPLANATION of A
Both A and R are true , and R is not the correct explanation of A
A is true but R is FALSE
A is false but R is true

Answer :C
Discussion
29.

A sodium street light gives off yellow light that has a wavelength of 600 nm. Then (For energy of a photon take E=(12400eV-"Å")/(lambda("Å")):

Answer»

frequency of this LIGHT is `7xx10^(14)s^(-1)`
frequency of this light is `5xx10^(14)s^(-1)`
wave number of the light is `3XX10^(6) m^(-1)`
energy of the photon is APPROXIMATELY 2.07 eV

Answer :B::D
Discussion
30.

A sodium salt on treatment with MgCl_(2) gives white precipitate on heating. The anion of the sodium salt is:

Answer»

`CO_(3)^(2-)`
`HCO_(3)^(Θ)`
`SO_(4)^(2+)`
`NO_(3)^(Θ)`

Solution :Given, `NaX+MgCl_(2)overset(Delta)(rarr)`White ppt.
`NaHCO_(3)+MgCl_(2)rarrunderset("Soluble")(MG(HCO_(3))_(2))+NaCl`
`Mg(HCO_(3))_(2)overset(Delta)(rarr)underset("White ppt".)(MgCO_(3))+H_(2)O+Co_(2)`
Hence, anion of the SODIUM salt is `HCO_(3)^(Θ)`.
Discussion
31.

A sodium salt of unknown anion when treated with MgCl_(2) gives white precipitateonly on boiling . The anion is

Answer»

`SO_(4)^(2-)`
`HCO_(3)^(-)`
`CO_(3)^(2-)`
`NO_(3)^(-)`

Solution :BICARBONATES on heating FORM insoluble CARBONATES
`MgCl_(2) + 2 NaHCO_(3) to NaCl + Mg(HCO_(3))_(2)`
`Mg(HCO_(3))_(2) OVERSET(Delta)(to) MgCO_(3) darr + CO_(2) + H_(2)O`
Discussion
32.

A sodium salt on treatment with MgCI_(2) gives white precipitate only on heating. The anion of sodium salt is

Answer»

`HCO_(3)^(-)`
`CO_(3)^(2-)`
`NO_(3)^(-)`
`SO_(4)^(2-)`

SOLUTION :`2NaHCO_(3) + MgCI_(2)overset(Delta) RARR Mg(HCO_(3))_(2) + 2NaCI`
Discussion
33.

A sodium salt of unknown anion when treated with MgCl_(2)gives white precipitate only on boiling. The anion is ..........

Answer»

`SO_(4)^(2-)`
`HCO_(3)^(-)`
`CO_(3)^(2-)`
`NO_(3)^(-)`

ANSWER :B
Discussion
34.

(A) : Sodium metals is stored in kerosene (R): The density of sodium is less than water

Answer»

Both A and R are correct and 'R' is the correct EXPLANATION of A
Both A and R are correct and 'R' is not the correct explanation of A
A is correct but R is FALSE
A is false but R is correct 

Answer :B
Discussion
35.

(A) Sodium fusion extract of a compound gives black precipitate with lead acetate. (R) Sulphur containing compounds form Na_(2)S in sodium fusion extract.

Answer»

If both ASSERTION and reason are CORRECT and reason is the correct EXPLANATION of the assertion
If both assertion and reason are correct but reason is not the correct explanation of the assertion
If assertion is correct but reason is INCORRECT
If assertion is incorrect but reason is correct

Answer :B
Discussion
36.

A sodium fire in the laboratory is extinguished by

Answer»

Water
Petrol
Alcohol
CC14

Solution :SINCE NA does not REACT with `CCI_(4)`
Discussion
37.

A: SO_(2) molecule has unsymmetrical shape R: The dipole moment of SO_(2) molcule is equal to zero.

Answer»

Both (A) and (R) are true and (R) is the correct EXPLANATION of (A)
Both (A) and (R) are true and (R) is not the correct explanation of (A)
(A) is true but (R) is FALSE
(A) is false but (R) is true

ANSWER :A
Discussion
38.

(A) Soap and detergent are macro-molecular colloids . (R) soap and detergent are molecular of large size.

Answer»

IF both (A) and (R) are CORRECT and (r) is the correct EXPLANATION for (a).
If both (a) and (r) are correct but (r) is not the correct explanation for (a).
IF (a) is correct but (r) is INCORRECT.
If (a) is incorrect but (r) is correct.

ANSWER :D
Discussion
39.

A smuggler could not carry gold by depositing iron chemically on the gold surface since

Answer»

GOLD is denser
iron rusts
gold has HIGHER REDUCTION potential than iron
gold has lower reduction potential than iron

Solution :Gold having higher `E_("red")^(@)` oxidises Fe to `Fe^(2+)`
Discussion
40.

A smaple of pyrolusite weighing 0.5 g is distilled with conc. HCl. The evolved Cl_(2) when passed through a solution of KI liberates sufficient I_(2) to react with 125 mL of N//12.5 hypo (Na_(2)S_(2)O_(3). 5H_(2)O). Calculate the percentage of MnO_(2) in pyrolusite.

Answer»


SOLUTION :N//A
Discussion
41.

A small quantity of gaseous NH_(3) and HBr are introduced simultaneously into the opposite ends of an open tube which is one metre long. Calculatethe distance of the white solid NH_(4)Br formed from the end which was used to introduce NH_(3).

Answer»


SOLUTION :SIMILAR to SOLVED PROBLEM 5.
Discussion
42.

A small particle of massm moves in such a way that the potential energy U = ar^(2) where a is a constant and r is the distance of the particle from the origin. Assuming Bohr's model of quantization of angular momentum and circular orbits, find the radius of n^(th) allowed orbit.

Answer»

`n^2`
`n`
`sqrtn`
`1/(sqrtn)`

SOLUTION :`-1/2 xx P.E.= K.E`
`=-1/2 (- 1/2 mkr^2) = 1/2 mv^2 , mvr = (NH)/(2PI),`
`v^2 = (n^2 h^2)/(4pi^2 m^2 r^2) ,r^4 = (n^2 h^2)/(2pi^2 m^2 k) ` or `r^00 sqrtn`
Discussion
43.

A small amount of NH_(4)HS is placed in a flask already containing ammonia gas at a certain temperature and 0*50 atm pressure. Ammonium hydrogensulphide decomposes to yield NH_(3) andH_(2)S gass in the flask . When the decomposition reaction reaches equilibrium , the total pressure in the flask rises to 0.84 atm. The equilibrium constant for NH_(4)HS decomposition at this temperature is

Answer»

`0.30`
`0*18`
`0*17`
`0*11`

Solution :`{:(,NH_(4)HS (s),HARR,NH_(3) (g),+,H_(2)S (g)), ("Intial",a " moles",,0.5" atm",,), (" At .eqm.",(a-X),,(0.5+p),,p " atm".):}`
i.e., if x moles of `NH_(4)HS` decompose , increase in pressure due to `NH_(3)` = increase in pressure due to `H_(2)S` = p atm .
Total pressure at equilibrium
` 0.5 + p+ p = 0.5 + 2 p ` atm
` 0.5 + 2 p = 0.84 " atm" or p=0.17` atm
` :. p_(NH_(3)) = 0.5 + 0.17 = 0.67` atm
` p_(H_(2)S) = 0.17 ` atm
` K_(p) =p_(NH_(3)) xx p_(H_(2)S)`
` =0.67 xx 0.17 = 0.1139`
Discussion
44.

A small amount of CaCO_(3) completely neutralises 525 ml of (N)/(10) HCl and no acid is left at the end after converting all calcium chloride to CaSO_(4). How much plaster paris (CaSO_(4)(1)/(2)H_(2)O) can be obtained

Answer»

`1.916g`
`5.827g`
`7.53g`
`3.81g`

Solution :No. of Meq of HCl = `525xx(1)/(10)=52.5`
No. of Meq of HCl = No. of mecq of `CaCl_(2)` =
No. of Meq of PLASTER of PARIS
So No. of Meq of plaster of paris = 52.5
`52.5=("wt of plaster of paris")/("GEW of plaste of paris")xx1000`
`52.5=(X)/(72.5)xx1000impliesx=3.81g`
Discussion
45.

A single electron system has ionization energy 11180 kJ mol^(-1). Find the number of protons in the nucleus of the system.

Answer»


Solution :`IE=(Z^(2))/(N^(2))xx21.69xx10^(-19)J`
`(11180xx10^(3))/(6.023xx10^(23))=(Z^(2))/(1^(2))xx21.69xx10^(-19)`
`Z~~3`.
Discussion
46.

A single compound of the structure is obtained from ozonolysis of twhich of the followiing cyclic compounds ?

Answer»




SOLUTION :
Discussion
47.

A simplified application of MO theory to the hypothetical "molecule" OF would give its bond order as

Answer»

2
`1.5`
`1.0`
`0.5`

ANSWER :B
Discussion
48.

A Simple aromatic hydrocarbon A reacts with Cl_2 to give B of molecular formula C_6H_5Cl. B on reaction with ethyl chloride along with sodium metal gives C of formula C_8H_10.C alone reacts with Na metal in the presence of ether to give D C_12 H_10. Identify A,B,C & D.

Answer»

Solution : (i) The simple aromatic HYDROCARBON is `C_6H_6` Benzene (A)
(ii) Benzene reacts with `Cl_2` to give chlorobenzene.
`C_6H_6 + Cl_2 overset(FeCl_2) to underset("benzene")underset("Chloro")(C_6H_5Cl) + HCl_B`
(III) Chlorobenzene reacts with ethyl CHLORIDE to FORM ethyl benzene, C as the product.
`C_6H_5Cl + 2 Na+ C_2H_5Cl overset("ether")to underset("Ethyl benzene C")(C_6H_5- C_2H_5) + 2NaCl`
(iv) Chlorobenzene on reaction Na metal gives Biphenyl compound D as the product.
`C_6H_5Cl + 2Na + C_6H_5Clto underset("Biphenyl D")(C_6H_5-C_2H_5) + 2NaCl`
Discussion
49.

A similarity between optical and geometrical isomerism is that

Answer»

each forms equal number of isomers for a GIVEN compound
if in a compound, ONE is PRESENT then so is the other
both are INCLUDED in stereoisomerism
they have no similarity

Answer :C
Discussion
50.

(A): Silicones are synthetic organosilicon compounds (R) : Silicones contain Si-O-Si linkages

Answer»

A and R are true, R explains A 
A and R are true, R does not explain A 
A is true, but R is false 
A is false, but R is true 

ANSWER :B
Discussion