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.

Solubility product of AgBr is 5.0xx10^(-13). The quantity of KBr (M = 120) to be added to 1 litre of 0.05 M solution of AgNO_(3) to start the precipitation of AgBr is

Answer»

`1.2xx10^(-10)G`
`1.2xx10^(-9)g`
`6.2xx10^(-5)g`
`5.0xx10^(-8)g`

Solution :Wt. in g = no. of MOLES `xx` MOL. Wt.
Discussion
2.

Solubility product of a sulphide MS is 3 xx 10^(-25) and that of another sulphide NS is 4 xx 10^(-40). In ammoniacal solution

Answer»

Only NS gets precipitated
Only MS gets precipitated
No sulphide precipitates
Both sulphides precipitate

Solution :The concentration of `S^(2-)` IONS in group IV is lowered by MAINTAINING acidic medium in the presence of `NH_(4)Cl`. The ionization of `H_(2)S` is supressed due t common ION effect. So the ionic product is less than SOLUBILITY product.
Discussion
3.

Solubility product of Ag_2CrO_4 is 1xx10^(-12) What is the solubility of Ag_2 CrO_4 in 0.01 M AgNO_3 solution?

Answer»

SOLUTION :Given that `K_(sp)=1XX10^(-12)`
`Ag_2 CrO_4(s) hArr 2ag_((aq))^(+)+CrO_((aq))^(-2)`
`underset(0.01M)(AgNO_(3(s))) hArr underset(0.01M)(Ag_((aq))^(+))+underset(0.01M)(NO_(3(aq))^(-))`
`K_(ap)=[AG^+]^2[CrO_(4)^(2-)]""{:[([Ag^+]=2s+0.01),(""0.01gtgt2s),(therefore[Ag^+]=0.01M),([CrO_(4)^(2-)]=s)]:}`
`1xx10^-12=(0.01)^2(s)`
`(s)=(1xx10^(-12))/((10^(-2)^2)=1xx10^-8 M`
Discussion
4.

Solubility product of a salt AB is 1xx10^(-8)M^(2) in a solution in which the concentration of A^(+) ions is 10^(-3)M. The salt will precipitate when the concentration of B^(-) ions is kept

Answer»

`10^(-4)M`
`10^(-7)M`
`10^(-6)M`
`10^(-5)M`

SOLUTION :`[A^(+)][B^(-)]=K[AB]=K_(sp)`
`[10^(-3)][B^(-)]=10^(-8)`
Hence, the salt will precipitate when the conc. Of B becomes more than `10^(-5)` M.
Discussion
5.

Solubility product of a saltAB is 1 xx 10^(-8) in a solution in which concentration of A is 10^(-3) M. The salt will precipitate when the concentration of B becomes more than

Answer»

`10^(-4) M`
`10^(-7) M`
`10^(-6) M`
`10^(-5) M`

Solution :`[B] = (K_(sp)AB)/([A]) = (1 XX 10^(-8))/(10^(-3)) = 1 xx 10^(-5) M`
Where ionic PRODUCT `gt K_(sp)`, ppt formed
`:.` B should be more then `10^(-5)M`.
Discussion
6.

Solubility product constant (K_(sp)) of salts of types MX_(1), MX_(2) and M_(3)X at temperature 'T' are 4.0 xx 10^(-8), 3.2 xx 10^(-14) and 2.7 xx 10^(-15), respectively. Solubilities (mol dm^(-3)) of the salts at temperature 'T' are in order

Answer»

`MX_(1) gt MX_(2) gt M_(3)X`
`M_(3)X gt MX_(2) gt MX_(1)`
`MX_(2) gt M_(3)X gt MX_(1)`
`MX_(1) gt M_(3)X gt MX_(2)`

Solution :`{:(MX1,MX2,M3X,),(S_(1)^(2) = 4 xx 10^(-8),4S_(2)^(3) = 3.2 xx 10^(-14),27 S_(3)^(4) = 2.7 xx 10^(-15),),(S_(1) = 2 xx 10^(-4),S_(2)^(3) = 8 xx 10^(-15),S_(3)^(4) = 10^(-16),),(,S_(2) = 2 xx 10^(-5),S_(3) = 10^(-4),),( rArr S_(1) gt S_(3) gt S_(2),rArr MX_(1) gt M_(3)X gt MX_(2),,):}`
Discussion
7.

Solubility product for salt AB_(2) is 4 xx 10^(-12). Calculate solubility

Answer»

`1 XX 10^(-3)` gm mol/litre
`1 xx 10^(-5)` gm mol/litre
`1 xx 10^(-4)` gm mol/litre
`1 xx 10^(-2)` gm mol/litre

Solution :`{:(AB_(2)hArr,A^(2+),+,2B^(-)),(,(S),,(2S)^(2)):}`
`K_(sp) = 4S^(3)`
`S = 3sqrt((K_(sp))/(4)) = 3sqrt((4 xx 10^(-12))/(4)) = 1 xx 10^(-4)` gm. mol/litre
Discussion
8.

Solubility product is

Answer»

The IONIC PRODUCT of an electrolyte in its SATURATED solution
The product of the SOLUBILITIES of the IONS of the electrolyte
The product of solubilities of the salts
The product of the concentration of the ions

Answer :A
Discussion
9.

Solubility of which of the following salts can be increased by adding some HNO_(3) solution?

Answer»

`BaF_(2)`
`CH_(3)COOAg`
`BaSO_(4)`
`PbS`

Answer :A::B::D
Discussion
10.

Solubility produce of Ag_2CrO_4 is 1 times 10^-12. What is the solubility of Ag_2CrO_4 in 0.01 M AgNO_3 solution?

Answer»

Solution :Solubility product of `Ag_2CrO_4`
`K_(SP)=1 times 10^-12`
`Ag_2CrO_4 leftrightarrow 2Ag^(+)+CrO_4^(-2)`
`AgNO_3 leftrightarrow Ag^(+) + NO_3^(-)`
`K_(sp)=[Ag^+]^2 [CrO_4^(2-)]`
`[Ag^+]=2s+0.01`
`because 0.01 gt gt 2s`
`therefore [Ag^+]=0.01 M`
`[CrO_4^(2-)]=s`
`K_(sp)=(0.01)^2. (s)`
`1 times 10^-12=(0.01)^2.(s)`
`s=(1 times10^-12)/(10^-2)^2=(1 times10^-12)/10^-4=1 times10^-8 M`
Discussion
11.

Solubility of which among the followingsolids in water changes slightly with temperature?

Answer»

`KNO_(3)`
`NaNO_(3)`
KBr
NaBr

Solution :SOLUBILITY of `KNO_(3), NaNO_(3)` and KBr increases with increases in TEMPERATURE. While solunility of NaBr CHANGE slighltly with temperature.
Discussion
12.

Solubilityof whichamong the followingsolidsin waterchangesslightly with temperature /

Answer»

`KNO_(3)`
`NaNO_(3)`
`KBR`
`NABR`

Solution :The SOLUBILITY of NaBr CHANGES slightly with temperature.
Discussion
13.

Solubility of the alkaline earth's metal sulphates in water decreases in the sequence

Answer»

CagtSrgtBagtMg
SrgtCagtMggtBa
BagtMggtSrgtCa
MggtCagtSrgtBa

Solution :Solubility of ALKALINE earth metal sulphates decreases down the GROUP `MgSO_(4) GT CaSO_(4) gt SrSO_(4) gt BaSO_(4)`
Discussion
14.

Solubility of solutes which dissolve with the liberation of heat decreases with:

Answer»

DECREASE in temperature
Increase in temperature
No CHANGE in temperature
None

Answer :B
Discussion
15.

Solubility of PbI_(2) is 0.005 M. Then, the solubility product of PbI_(2) is

Answer»

`6.8 xx 10^(-6)`
`6.8 xx 10^(6)`
`2.2 xx 10^(-9)`
None of these

Solution :`PbI_(2) hArr Pb + I_(2)`.
`K_(sp) = 4S^(3) = 4 xx (0.005) xx (0.005) xx (0.005)`
`K_(sp) = 4 xx 5 xx 5 xx 5 xx 10^(-9)`
`K_(sp) = 500 xx 10^(-9)`
`K_(sp) = 5.0 xx 10^(-7)`.
Discussion
16.

Solubility of sodium sulphate in water

Answer»

increases with INCREASE in temperature
decreases with increase in temperature
decreases with decrease in temperature
INDEPENDENT of temperature

Solution :Dissolution of sodium sulphate is an exothermic process. Increasing temperautre solubility decreases.
Discussion
17.

The solubility of noble gases in water shows the order :

Answer»

`He gt NE gt AR gt KR gt Xe`
`He gt Ne gt Kr gt Ar gt Xe`
`Xe gt Kr gt Ar gt Ne gt He`
none

Answer :C
Discussion
18.

Solubility of MX_(2) type electrolyte is 0.5 xx 10^(-4) mole/litre. The value of K_(sp) of the electrolyte is

Answer»

`5 XX 10^(-13)`
`25 xx 10^(-10)`
`1.25 xx 10^(-13)`
`5 xx 10^(12)`

SOLUTION :`MX_(2) hArr underset((S))(M^(2+)) + underset((2S)^(2))(2X^(-))`
`4S^(3) = 4 xx (0.5 xx 10^(-4))^(3) = 5 xx 10^(-13)`.
Discussion
19.

Solubility of NaCl in heavy water is:

Answer»

More than ORDINARY water
Less than ordinary water
Same as in ordinary water
None of these

Answer :B
Discussion
20.

Solubility of iodine in CS_2 is due to :

Answer»

Van der Waal's forces
Ionic BOND formation
Covalent bond ATTRACTION
Coordinate bond interaction

ANSWER :A
Discussion
21.

Solubility of iodine in water is greatly increased by the addition of iodide ions because of the formation of

Answer»

`I_(2)`
`I_(3)`
`I_(3)^(-)`
`I^(-)`

SOLUTION :`I_(2) + I^(-) to I_(3)^(-)`
Discussion
22.

Solubility of gas at 1 bar pressure and 300K is 2xx10^(-3) M. What will be its solubility at same pressure and 400K if DeltaH_("solution") of the gas is -9.96 kJ. [ Take : R=8.3J/K mole,e=2.72]

Answer»

`2xx10^(-3)` M
`7.35xx10^(-4)`M
`6.67xx10^(-4)` M
`5xx10^(-4)M`

ANSWER :B
Discussion
23.

Solubility of deliquescent substances in water is generally:

Answer»

high
Low
Moderate
Can not be said

Answer :A
Discussion
24.

solubility of calcium acetate_____ with increase in temperature.

Answer»


ANSWER :DECREASES
Discussion
25.

Solubility of CaF_(2) in terms of its solubility product is given by

Answer»

`s=(K_(SP))^(1//3)`
`s=(K_(sp)//2)^(1//3)`
`s=(K_(sp)//4)^(1//3)`
`s=(K_(sp)//2)^(1//2)`

ANSWER :C
Discussion
26.

Solubility of BaF_(2) in a solution of Ba(NO_(3))_(2) will be represented by the concentration term :

Answer»

`[BA^(2+)]`
`[F^(-)]`
`(1)/(2)[F^(-)]`
`2[NO_(3)^(-)]`

ANSWER :C
Discussion
27.

Solubility of CaF_(2) in terms of its solubility product is given by :

Answer»

`s=(K_(sp))^((1)/(3))`
`s=((K_(sp))/(2))^((1)/(3))`
`s=((K_(sp))/(4))^((1)/(3))`
`s=((K_(sp))/(2))^((1)/(2))`

Solution :`CaF_(2)hArr underset(s)(Ca^(2+))+underset(2S)(2F^(-))`
`K_(sp)=(s)(2s)^(2)`
`K_(sp)=4S^(3)`
`s=((K_(sp))/(4))^((1)/(3))`
Discussion
28.

Solubility of BaF_2 in a solution of Ba(NO_3)_2 will be represented by the concentration term:

Answer»

`[BA^(2+)]`
`[F^-]`
1/2`[F^-]`
`2[NO_3^-]`

ANSWER :C
Discussion
29.

Solubility of BaF_(2) in a solution Ba(NO_(3))_(2) will be represents by the concentration term

Answer»

`[BA^(++)]`
`[F^(-)]`
`(1)/(2)[F^(-)]`
`2[NO_(3)^(-)]`

ANSWER :C
Discussion
30.

Solubility of Al(OH)_(3) = S, K_(sp) will be

Answer»

`108 S^(3)`
`27 S^(3)`
`4 S^(2)`
`27 S^(4)`

Solution :`{:(AL(OH)_(3)(s),HARR,Al^(3+)(aq),+,3OH^(-)(aq)),(,,s,,3s):}`
`k_(sp) = [Al^(3+)][OH^(-)]^(3) = (S)(3S)^(3)`
`K_(sp) = 27 S^(4)`.
Discussion
31.

Solubility of an AB_(2) type electrolyte is 5.0xx10^(-5) mol L^(-1). K_(sp) for the electrolyte AB_(2) is :

Answer»

`5xx10^(-12)`
`25xx10^(-10)`
`1XX10^(-13)`
`5xx10^(-13)`

ANSWER :D
Discussion
32.

Solubility of alcohols decreases with increase in alkyl group because of

Answer»

decrease in hydroge bonding.
decrease in VAN der Waals forces.
increase in HYDROCARBON part.
decrease in POLAR part.

Answer :C
Discussion
33.

Solubility of alcohol in water is due to

Answer»

HYDROPHOBIC R-group
hydrophillic OH-group
hydrophobic OH-group
hydrophilic R-group

Answer :B
Discussion
34.

Solubility of AgCl will be minimum in

Answer»

0.001 M `AgNO_(3)`
Pure water
0.01 M `CaCl_(2)`
0.01 M NaCl

Solution :0.01 M `CaCl_(2)` gives maximum `Cl^(-)` ions to keep `K_(sp)` of AGCL constant, decrease in `[Ag^(+)]` will be maximum.
Discussion
35.

Solubility of AgCl will be minimum in :

Answer»

0.01 M NACL
`0.01 M CaCl_(2)`
PURE WATER
`0.001 M AgNO_(3)`

Answer :B
Discussion
36.

Solubility of AgCl in the solution containing 0.01 M NaCl is [given, K_sp (AgCL)= 1.6 x 10^-10]

Answer»

`8` X `10^-9`
`1.6 `x `10^-9`
`2` x `10^-8`
`1.6` x `10^-8`

ANSWER :44
Discussion
37.

Solubility of AgCl is least in

Answer»

0.1 M NaCl
PURE water
0.1 M `BaCl_(2)`
0.1 M `AlCl_(3)`

Solution :(d): Concentration of common ion `prop (1)/("SOLUBILITY")` In 0.1 M `AlCl_(3)`, the concentration of common ion i.e., `Cl^(-)` is high, therefore, AgCl will be least SOLUBLE in this solution.
Discussion
38.

Solubility of AgCl at 20^(@)C is 1.435 xx 10^(-3) gm per litre. The solubility product of AgCl is

Answer»

`1 xx 10^(-5)`
`1 xx 10^(-10)`
`1.435 xx 10^(-5)`
`108 xx 10^(-3)`

Solution :`S = 1.435 xx 10^(-3) g//l, (1.435 xx 10^(-3))/(143.5) = 10^(-5) m`
`K_(SP) = S xx S = 10^(-10)`.
Discussion
39.

Solubility of a solute in a solvent ( say H_(2)O ) is dependent on temperature as given by S = Ae^(- Delta H // RT ) where Delta H is heat of reaction Solute + H_(2)O hArr Solution , Delta H = +- X For a given solution variation of logs with temperature is shown graphically. Hence solute is

Answer»

`CuSO_(4). 5H_(2)O`
`NACL`
Sucrose
CaO

Answer :D
Discussion
40.

Solubility of a subsatnce which dissolves with a decrease in volume and absorption of heat will be favoured by:

Answer»

<P>HIGH P and High T
LOW P and low T
High P and low T
Low P and high T

Answer :A
Discussion
41.

Solubility of a solid in to water increases as the temperature is raised . Match the process listed in column-I with the changes in appropriate properties listed in column-II. (m=molality) {:("COLUMN-I","COLUMN-II"),((A) "Gas+Water" rarr "solution",(p)DeltaHgt0","DeltaSgt0),((B)"solid+water"rarr"solution",(q)DeltaH=0","Deltam=0),((C)"Saturated solution+solid solute",(r)DeltaHlt0","Deltamlt0),((D)"Super Saturated solution+solid solute",(s)DeltaHlt0","Delta slt0):}

Answer»


ANSWER :A::B::C::D
Discussion
42.

Solubility of a gas in liquid, _____ with rise of temperature .

Answer»

SOLUTION :DECREASES
Discussion
43.

Solubility of a gas in water:

Answer»

INCREASES with temperature
Decreases with pressure
Decreases with temperature
None

Answer :C
Discussion
44.

Solubility of a gas in liquid increaes on :

Answer»

ADDITION of a catalyst
Increasiing the PRESSURE
DECREASING the pressure
Increasing the temperature

Answer :B
Discussion
45.

Solubility of a gas in liquid, increases with rise of temperature .

Answer»

SOLUTION :DECREASES .
Discussion
46.

Solubility of a gas in a liquid increases with

Answer»

increase of PRESSURE and increase of TEMPERATURE
decrease of pressure and increase of temperature
increase of pressure and decrease of temperature
decrease of pressure and decrease of temperature

Solution :Increasing pressure of gas solubilityincreases. Decreasing temperature of gas SOLUBILITY INCREASES.
Discussion
47.

Solubility of a gas in liquid decreases with rise in temperature

Answer»


ANSWER :TRUE
Discussion
48.

Solubility curve of Na_(2)SO_(4).10H_(2)O in water with temperature is given as

Answer»

`DeltaH` is -ve after `34^(@)C`
solution process is EXOTHERMIC till `34^(@)C` and ENDOTHERMIC after `34^(@)C`
solution process is endothermic till `34^(@)C` and exothermic THEREAFTER
`DeltaH` is -ve after `34^(@)C`

ANSWER :A::C::D
Discussion
49.

Solubility curves of four ionic salts X,Y,Z,W are given below. In which case the value of DeltaH_(sol) lt 0?

Answer»

X
Y
Z
W

Answer :A
Discussion
50.

Solubility curve of a hydrated salt in water with temperature is given. The curve indicates that the solution process is :

Answer»

exothermic
proceding to MAINTAIN same `DeltaH`
endothermic
endothermic till `60^(@)C` and exothermic agterwards

Solution :It is endothermic till `60^(@)C` as colubility INCREASES with rise in temperature and then becomes exothermic as the solubility decreases with further rise in temperature for example, `Na_(2)SO_(4)10H_(2)O` behaves with same n=manner, For details, consult section 3.
Discussion