Explore topic-wise InterviewSolutions in Current Affairs.

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1.

Lassaigne's test is used for detection of elements N, S, X and P. The sodium fused extract is tested for these elements with appropriate reagents. The observation made in to identify the element present in the organic compound.The sodium fusion of extract of an organic compound containing S, N on treatment FeCl_(3) solution gives a coloured solution. The colour of the solution is

Answer»

blue
GREEN
yellow
blood RED

Solution :In Lassaigne.s TEST, organic compound containing both N & S gives blood red COLOUR.
Discussion
2.

Lassaigne's test is used for detection of elements N, S, X and P. The sodium fused extract is tested for these elements with appropriate reagents. The observation made in to identify the element present in the organic compound.Sulphur is organic compound on fusion with metallic sodium will be converted into

Answer»

`Na_(2)SO_(4)`
`Na_(2)SO_(3)`
`Na_(2)S`
`H_(2)SO_(4)`

Solution :In Lassaigne.s extract, S of organic COMPOUND CONVERTS into `Na_(2)S`.
Discussion
3.

Lassaigne's test is used for detection of elements N, S, X and P. The sodium fused extract is tested for these elements with appropriate reagents. The observation made in to identify the element present in the organic compound.The sodium fusion extrat is treated with FeSO_(4), followed by H_(2)SO_(4), gave blue colour. The element present in the compound is

Answer»

sulphur
nitrogen
CHLORINE
Phosphorus

Solution :In Lassaigne.s test, ORGANIC COMPOUND containing only N gives purussian Blue COLOUR.
Discussion
4.

Lassaigne's test is not shown by diazonium salts. Why ?

Answer»

Solution :DIAZONIUM salts usually LOSE `N_(2)` on heating much before they have a chance to react with FUSED sodium metal. Therefore, diazonium salts do not show positive LASSAIGNE's test for nitrogen.
Discussion
5.

Lassaigne's test for the detectionof nitrogen will fail in the case of

Answer»



`NH_(2)-NH_(2).HCL`

ANSWER :C
Discussion
6.

Lassaigne's test for the detection of nitrogen fails in

Answer»

`NH_(2)CONHNH_(2)*HCl`
`NH_(2)NH_(2)*HCl`
`NH_(2)CONH_(2)`
`C_(6)H_(5)NHNH_(2)*HCl`

ANSWER :B
Discussion
7.

Lassaigne's test fails in

Answer»

`NH_2NH_2`
`H_2NCONH_2`
`C_6H_2NHNH_2`
`NH_2OH`

ANSWER :A
Discussion
8.

Lassaigne's solution is boiled with HNO_3before testing for halogens because

Answer»

SILVER halides are INSOLUBLE in `HNO_(3)`
`Na_(2) S and NaCN ` are decomposed by `HNO_(3)`
` Ag_(2)S`is SOLUBLE in `HNO_3`
AGCN is soluble in `HNO_3`

Answer :A
Discussion
9.

Lassaigne's test for the detection of nitrogen fails in ............. .

Answer»

`H_(2)N-CO-NH.NH_(2).HCl`
`NH_(2)-NH_(2).HCl`
`C_(6)H_(5)-NH-NH_(2).HCl`
`C_(6)H_(5)CONH_(2)`

ANSWER :C
Discussion
10.

Lassaigne's solution gives violet colouration with sodium nitroprusside. It indicates the presence of

Answer»

nitrogen
sulphur
halogen
both N and S.

Answer :B
Discussion
11.

Lassaigne's extract is prepared in distilled water and not in tap water. Explain.

Answer»

Solution :Tap water generally contains in it chloride IONS `(Cl^(-))`. Therefore, it cannot be used to TEST the presence of chlorine in an organic COMPOUND. Keeping this in view, the Lassaigne's EXTRACT is normally PREPARED in distilled water.
Discussion
12.

Largest ion among the following is

Answer»

`NA^(+)`
`O^(-2)`
`S^(-2)`
`CL^(-)`

ANSWER :C
Discussion
13.

Larger the size of the atom ,lesser is the ionisation energy . Explain.

Answer»

Solution :Reason :The ionization energy decreases with the INCREASING size of ATOM. This is due to the fact , in LARGER atoms electrons are loosely bound to the NUCLEUS , i.e LESSER energy is required for removal of electrons from larger atoms than the smaller one.Hence ionization energy is lower for larger atoms.
Discussion
14.

Largest bond angle is present in

Answer»

`CH_4`
`C_2 H_6`
`C_2 H_4`
`C_2 H_2`

ANSWER :D
Discussion
15.

Larger number of oxidation states are exhibited by the actinoids than those by the lanthanoids. This can be best explained as

Answer»

lesser ENERGY difference between 5F and 6d than between 4f and 5d ORBITALS
more energy difference between 5f and 6d than between 4f and 5d orbitals
more reactive nature of the actinoids than the lanthanoids
4f orbitals more DIFFUSED than the 5f orbitals

Answer :A
Discussion
16.

Large -scale source of florine is

Answer»

`CaF_(2)`
`MgF_(2)`
`BaF_(2)`
`BeF_(2)`

Answer :1
Discussion
17.

Lanthanoids used in glass blower's goggles are

Answer»

PR and Nd
Eu and GD
TB and DY
EM and Sm

Answer :A
Discussion
18.

Lanthanum belongs to

Answer»

s-block
p-block
d-block
f-block

Answer :C
Discussion
19.

Lanthanoidsare

Answer»

14 elementsin thesixthperiod(atomicno= 90 to103 ) thatare filling4f- sublevel
14elementsin theseventhperiod ( atomicno= 90to 103 ) that arefilling 5f-sublevel
14 element in the SIXTH period( atomicno = 58to 71 )that arefilling4f- sublevel
14elementsin theseventhperiod( atomicno =58to 71 )that arefilling4f- sublevel

Solution :NA
Discussion
20.

Lanthanoids and actinoids are placed in separate rows at the bottom of the periodic table .Explain.

Answer»

Solution :LANTHANOIDS and actinoids are placed in separate rows at the bottom of the periodic table to MAINTAIN its structure and to preserve the principle of classification by keeping elements with similar PROPERTIES in a single column.
Discussion
21.

Lanthanoids and actinoidsare placed inseparate rowsat thebottomof theperiodictable,Explain the reason for thisarrangement.

Answer»

SOLUTION :These have beenplacedseparatelyat the bottomof theperiodictable FORCONVENIENCE . If theyare placedwithin the bodyof the periodictable inserialorder to increasingatomic NUMBERTHE periodictablewill becomeextremely longand cumbersome.
Discussion
22.

Lanthanoid contraction occurs because

Answer»

the 4f electrons, which are gradually added, CREATE a strong SHIELDING effect
the 4f orbitals are greater in size than the 3d and 3f orbitals
the 5f orbitals STRONGLY penetrate into the 4f orbitals
the poor shielding effect of 4f electrons is coupled with INCREASED attraction between the nucleus and the added electrons.

Answer :D
Discussion
23.

Lanthanides are separated best by

Answer»

FRACTIONAL crystallisation
Solvent EXTRACTION
Complex FORMATION USING EDTA
Ion exchange resins

Answer :D
Discussion
24.

Lanthanides are characterized by the filing of the

Answer»

penultimate 4F energy LEVEL
ANTEPENULTIMATE 4f energy level
penultimate 5f energy level
antepenultimate 5f energy level

Answer :B
Discussion
25.

Laminar flow of a liquid means

Answer»

Regular gradation of VELOCITY for LAYERS in passing from one layer to the NEXT layer of a liquid
Showing constancy in the velocity of layers of a liquid
Increase in the velocity of layers from surface to BOTTOM of a liquid
All

Solution :It means all layers move with same velocity.
Discussion
26.

Lakes containing excess of nutritious substance are called

Answer»

Polluted lakes
Eutrophic lakes
DEAD lake
Green Lake

Answer :B
Discussion
27.

Lakes containing excess of nuritious substance are called

Answer»

POLLUTED LAKES
EUTROPHIC lakes
Dead lake
Green Lake

Solution :Eutrophic lakes
Discussion
28.

Laevulinic acid is a

Answer»

`ALPHA-`KETO acid
`beta-`Keto acid
`gamma-`Keto acid
`delta-`Keto acid

Solution :`gamma-`keto acid
Discussion
29.

Lactic acid overset(X)rarr Lactile chloride. The compound X is

Answer»

`PCl_(5)`
`Cl_(2)`
`NACL`
`ICL`

Solution :`underset("Lactic ACID")(CH_(3) overset(O)overset(|)(C)H COOH) overset(PCl_(5))rarr underset("Lactyl chloride")(CH_(3) overset(Cl)overset(|)(C)HCOCl) + POCl_(3) + HCl`
Discussion
30.

Lactic acid on heating with dill. H_(2)SO_(4) gives

Answer»

Acetic acid
Propionic acid
Acrylic acid
Formic acid

Solution :`CH_(3) - overset(OH)overset(|)(CH) - COOH overset(H^(+))RARR HCOOH + CH_(3) CHO`
Discussion
31.

Lactic acid on oxidation by alkaline postassium permanganate gives

Answer»

Tartaric ACID
Pyruvic acid
Cinnamic acid
Propionic acid

Solution :`CH_(3) - UNDERSET("Lactic acid")(CHOH - COOH) + [O] OVERSET(KMnO_(4))rarr underset("Pyruvic acid")(CH_(3) - CO - COOH) + H_(2)O`
Discussion
32.

Lactic acid on heating with conc. H_(2)SO_(4) gives

Answer»

ACETIC ACID
Propionic acid
Acrylic acid
Formic acid

Solution :SEE LACTIC acid
Discussion
33.

Lactic acid is:

Answer»

propionic ACID
`alpha-`HYDROXY propionic acid
`BETA-`hydroxy propionic acid
None of these

Solution :SEE Lactic acid
Discussion
34.

Laboratory grade concentrated sulphuric acid has a density 1.82g cc^(-1) .Weight percentage of acid is 98. Calculate the normality of solution.

Answer»

SOLUTION :NORMALITY of solution is given as N `=(10xx%xxd)/(GEW)=(10xx1.82xx98)/(49)`
Normality of solution =36.4 eq `L^(-1)`
Discussion
35.

L. fetimes of the molecules in the excited state are often measured by using pulsed radiation soruce of duration nearly in the nano second range. If the radiation source has the duration of 2 ns and the number of photons emitted during the pulse source is 2.5 xx 10^(15), calculate the energy of the source.

Answer»

SOLUTION :FREQUENCY `= (1)/(2 xx 10^(-9) s) = 0.5 xx 10^(9) s^(-1)`
Energy `= N HV = (2.5 xx 10^(5)) (6.626 xx 10^(-34) Js) (0.5 xx 10^(9) s^(-1)) = 8.28 xx 10^(-10) J`
Discussion
36.

L_(1)is the length between two adjacent carbon atoms in a layer and L_(2)is the length in-between two layers of graphite. The approximate ratio between L_(1) and L_(2)

Answer»

`1:1`
`2:5`
`5:2`
`1:5`

ANSWER :B
Discussion
37.

l = 3, then the values of magnetic quantum numbers are

Answer»

`pm 1 , pm 2 , pm 3`
`0 , pm 1 , pm 2 , pm 3`
`-1 , -2 , -3`
`0, +1 , +2 , + 3`

SOLUTION :`l=3 , m= pm 3, pm 2, pm 1, 0`
Discussion
38.

K_w=1xx10^(-14) calculate the concentration of [H_3O^+] in 10^(-8)M HCl.

Answer»

`1.0xx10^(-8)` M
`1.0525xx10^(-7)` M
`1.0xx10^(-6)` M
`9.525xx10^(-8)`M

Solution :`[HCl]=[H_3^+ O]= 1xx10^(-8)` M
`10^(-8) M LT 1 xx 10^(-7)M` of `H_3O^+`
`THEREFORE` Total `[H_3O^+]=(1xx10^(-8)+1xx10^(-7))`
`=1.1xx10^(-7) approx 1.0525xx10^(-7)` M
`[H^+][OH^-]=10^(-14)`
`therefore (10^(-8)+X)x=10^(-14)=0`
`therefore x=9.5xx10^(-8)`
Total `[H^+]=1xx10^(-8)+9.5xx10^(-8)`
`=1.05xx10^7` M
Discussion
39.

What is the minimum volume of water required to dissove 1g of calcium sulphate at 298K ? (For calcium sulphate, K_(sp) is 9.1xx10^(-6)).

Answer»

Solution :Solubility product of calcium sulphate, `K_(sp)` for `CaSO_4 = 9.1 XX 10^(-6)`
Solubility of `CaSO_4= SQRT(K_(sp))= 3.016 xx 10^(-3)mol L^(-1)`
Solubility of `CaSO_4 = 3.016 xx 10^(-3)xx136 = 0.41 g L^(-1)`
0.41 gram of `CaSO_4` is dissolved in a solution of VOLUME one litre.
Volume of solution in which 1 gram of `CaSO_4` is dissolved
`=" weight " xx (1)/("solubility ")=1 xx (1)/(0.41 ) = 2.441` .
Discussion
40.

K_(sp) of PbI_2 is 1.4xx10^(-8) . The molecular mass of PbI_2 is 461 g mol^(-1) . Then molecular mass of Pb(NO_3)_2 is 331.9 mol^(-1)So, (a) In 500 mL water (b) 500 mL 0.10 M KI (c) What is the weight of PbI_2 when soluble in 1.33 g Pb(NO_3)_2 containing 500 mL solution ?

Answer»

SOLUTION :(a)0.346 G , (B)`0.322xx10^(-3)` g , (b) 0.15 g
Discussion
41.

K_(sp) of PbI_2 is 1xx10^(-8) .Then what is the solubility of PbI_2 in 0.1 M KI ?

Answer»

SOLUTION :`y=K_(sp)/[I^-]^2=(1xx10^(-8))/(0.1)^2=1xx10^(-6)` M (Solubility in pressure of COMMON ION)
Discussion
42.

K_(Sp)of MX =K_(sp)" of "MX_2. Which is more soluble ?

Answer»

Mx
`Mx_2`
Equal
`K_(SP)` value is REQUIRED

Solution :Ksp of MX = Ksp of `MX_ 2 = K `

` x^(2)=K and 4y^(3)=K`
` x= sqrt(K)and y =3sqrt( (K)/(4)) because K LT 1, y GT x `
Discussion
43.

K_(sp) of M(OH)_(x), is 27 xx 10^(-12) and its solubility in water is 10^(-3) mol litre^(-1) . Find the value of X

Answer»


SOLUTION :` M(OH) _x HARR underset(S) M^(X+) +X underset( xs) OH^(-) `
` K_(sp)=S(XS)^(x) `
` 27 XX 10 ^(-12)= (10 ^(-3)) ( X 10 ^(-3)) ^(x) `
` 27 xx 10 ^(_12)= 10 ^(-3-3x), X^(x) rArr 3^(3) =27 ,x=3`
Discussion
44.

K_(sp) of BaSO_4=1.05xx 10^(-10) at same temperature is the concentration of Ba^(2+) and SO_4^(2-) in saturated solution.

Answer»

Solution :The ionic equilibrium in concentratedof `BaSO_4` is under.
`BaSO_(4(s)) hArr Ba_((AQ))^(2+) + SO_(4(aq))^(2-)`
So, `K_(sp)=[BA^(2+)][SO_4^(2-)]`
but `[Ba^(2+)]=[SO_4^(2-)]`= solubility S MOL `L^(-1)`
`K_(sp)=(S) (S) =S^2 =1.05xx10^(-10)`
`therefore S=[Ba^(2+)]=[SO_4^(2-)]`
`=(1.05xx10^(-10))^(1/2)`
`=1.0247xx10^(-5) "mol L"^(-1)`
Discussion
45.

Ksp of AgBr, AgCl,Ag_(2)CO_(3) and Ag_(3)AsO_(4) are 5 xx 10^(-13), 1.8 xx 10^(-10), 8.1 xx 10^(-12) and 10^(-22) respectively. Silver nitrate is added to the solution each having 0.1 M concentration of Br^(-),Cl^(-),CO_(3)^(+) and AsO_(4)^(3-). Select the correct statements

Answer»

`AgBr` will be precipitated before `AGCL`
`Ag_(2)CO_(3)` will be precepitated AFTE `Ag_(3)AsO_(4)`
`Ag_(3)AsO_(4)` will be precipitated after `AgCl`
`Ag_(3)AsO_(4)` will be precipitated after `Ag_(2)CO_(3)`

Solution :N//A
Discussion
46.

K_(sp) of A_2B_(3)in water at 25^(@) C.is 1.1 xx 10^(-23)concentration of A^(+)ions is

Answer»

`1xx10^(-5)`
` 2xx10^(-5)`
` 3xx10^(-5) M`
` 5xx10 ^(-11)`

Solution :` A_2 B_3 HARR 2A + 3B `
` KsP=108S^(5) RARR 108S^(5)=1.1 XX 10 ^(-23) `
` S ~~1 xx 10 ^(-5), [A] =2S = 2xx 10 ^(-5)`
Discussion
47.

Determine the mass by mass percentage concentration of a 100 g salt solution which contains 20 g salt.

Answer»

` 2XX10^(-6) M`
` 1xx10^(-4) M`
` 1.6 xx10^(-4) M`
` 4xx10^(-10) M`

Solution :` MX_2 hArr M + 2X `
` KsP=4S^(3)=4 xx 10 ^(-12)rArr S =10 ^(-4)= [M^(+2)]`
Discussion
48.

K_(sp)of a salt , having the generalformula MX_2in water 4xx10^(-12)The concentration of M^(2+)ions inthe aqueous solution of the saltis

Answer»

` 2xx10^(-6) M`
` 1xx10^(-4) M`
` 1.6 xx10^(-4) M`
` 4xx10^(-10) M`

Solution :` MX_2 hArr M + 2X `
` KsP=4S^(3)=4 XX 10 ^(-12)rArr S =10 ^(-4)= [M^(+2)]`
Discussion
49.

K_(sp)for Ca(OH)_(2) is 5.5xx10^(-6). What is the maximum pH that can be attained in a sewage tank treated with slaked lime ?

Answer»

`9.35`
`10.35`
`11.35`
`12.35`

SOLUTION :`CA(OH)_(2)hArrCa^(2+) + 2 OH^(-)`
`K_(SP)=[Ca^(2+)][OH^(-)]^(2)=5.5xx10^(-6)` (Given)
`:.s(2s)^(2)=5.5xx10^(-6)`
or `4s^(3)=5.5xx10^(-6) ` or `s^(3)=1.375xx10^(-6)`
`:. S=1.11xx10^(-2)M`
`:. [OH^(-)]=2s=2.22xx10^(-2)M`
`:.pOH=-log(2.22xx10^(-2))=2-0.3464=1.65`
`:.pH=14-1.65=12.35`
Discussion
50.

K_(sp) (25^(@)C)of Ag_(2)CrO_(4) is 4 xx 10^(-12). Which of the following mixer will bring about precipitatin of Ag_(2)CrO_(4).

Answer»

`10 mL 3 xx 10^(-4) M AgNO_(3) + 10 mL10^(-4) M Na_(2)CrO_(4)`
Adding `10` milli moles of `Na_(2)CrO_(4)` to `100` mL ofsaturated `AgCl(K_(SP) = 2 xx 10^(-10))` solution
Mixing equal VOLUMES of `10^(-4) M AgNO_(3)` with `10^(-4) M Na_(2)CrO_(4)` solution
`10 mL 10^(-3) M AgNO_(3) + 15 mL10^(-3) M Na_(2)CrO_(4)`

Solution :`K_(sp) Ag_(2)CrO_(2) = 4 xx 10^(-12)`
In B & D option
`[AG^(+)]^(2) [CrO_(4)^(2-)] gt K_(sp)`
`:.` Precipitation occurs in case of `B` & D
Discussion