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.

How would you explain the following observations? (i) BeO is almost insoluble but BeSo_(4) is soluble in water (ii) BaO is soluble but BaSO4, is insoluble in water (iii) Lil is more soluble than KI in ethanol.

Answer»

SOLUTION :Lattice ENERGY of BeO is comparatively higher than the hydration energy is almost insoluble in water. Whereas `BeSO_(4)` is ionic in nature and its hydran dominates the lattice energythe hydration energy
Both BaO and `BaSO_(4)` are ionic compounds but the hydration energy of BaO is higher than utice energy, therefore it is SOLUBLE in water
Since the size of `Li^(+)` ion is very small in comparison to `K^(+)` ion, it polarises the electron ud of ion to a GREAT extent. Thus Lil dissolves in ethanol more easily than the kI
Discussion
2.

How would you explain the following observations? a. BeO is almost insoluble but BeSO_(4) is soluble in water. b. BaO is soluble but BaSO_(4) is insoluble in water. c. Lil is more soluble than Kl in ethanol.

Answer»

Solution :a. Because of SMALLER size, higher IONISATION enthalpy and higher electronegativity, `BeO` is essentially covalent and hence is insoluble in water. In constrast,`BaSO_(4)` is ionic. Further because of small size of `Be^(2+)` ion, the hydration enthalpy of `BaSO_(4)` is much higher than is lattice enthalpy and hence `BaSO_(4)` is highly SOLUBLE in water.
b. Both `BaO` and `BaSO_(4)` are ionic compounds. However, the size of `O^(2-)` ion is much smaller than that of the `SO_(4)^(2-)` ion. SINCE a bigger anion stabilises a bigger cation more than a smaller anion stabilises a bigger cation, THEREFORE, the lattice enthalpy of `BaO` is much smaller than that of `BaSO_(4)` and hence `BaO` is soluble while `BaSO_(4)` is insoluble in water.
c. `Li^(o+)` is much smaller than `K^(o+)` ion. Therefore, according to Fajan's rule, `Li^(o+)` ion can polarise bigger `I^(ө)` ion to a greater extent than `K^(o+)` ion. As a result, `Lil` is more covalent that `Kl` and hence is more soluble in organic solvents like ethanol.
Discussion
3.

How would you explain the following observation? BeO is almost insoluble but BeSO_(4) is soluble in water.

Answer»

Solution :(i) BeO is almost insoluble in water because `Be^(2+)` is a small cation with high POLARISING power and `O^(2-)` is small anion.
`therefore` the LATTICE energy very high.
When BeO is dissolved in water, the hydration energy of its ions is not sufficient to overcome high lattice energy.
(II) On the other hand, `Be^(2+)` is a small cation and `SO_(4)^(2-)` is a large anion. Hence `Be^(2+)` can EASILY polarize `SO_(4)^(2-)` ions, making `BeSO_(4)` unstable the lattice energy of `BeSO_(4)` is not very high and so it is soluble in water.
Discussion
4.

How would you explain the following observation ? BeO is almost insoluble but BeSO_4 is soluble in water.

Answer»

Solution :(i) BeO is almost insoluble in water because `Be^(2+)` is a small cation with high polarising power and `O^(2)` is small anion.
`:.` The lattice ENERGY very high when BeO is dissolved in water, the hydration energy of its ions is not SUFFICIENT to overcome high lattice energy.
(II) On the other hand `Be^(2+)` is a small cation and `SO_4^(2-)` is a large anion. HENCE `Be^(2+)` can easily polarize `SO_4^(2-)` ions, making `BeSO_4` unstable the lattice energyof `BeSO_4` is not very high and so it is soluble in water.
Discussion
5.

How would you explain the following observations ? (i) BeO is almost insoluble but BeSO_(4) is soluble in water (ii) BaO is soluble but BaSO_(4) is insoluble in water (iii) LiI is more soluble than KI in ethanol

Answer»

Solution :Both BaO and `BaSO_4` are ionic in nature but the size of `SO_4^(2-)` ions is much larger than that of `O^(2-)` ions. Since a larger ANION STABILISES a larger cation more EFFICIENTLY than a smaller anion stabilises a larger cation, the lattice ENERGY of `BaSO_4` is much higher than that of `BaO`. This is why BaO is SOLUBLE, whereas `BaSO_4` is insoluble in water.
Discussion
6.

Howwould youexplainthe fact thethat thefirstionizationenthalpyof sodiumis lowerthan thatofmagnesium but itssecondionizationenthalpyis higherthan that of magnesium ?

Answer»

Solution :The E.C.Of Na is `1s^(2)2s^(2)2p^(6)3s^(1)` and E.C.Of Mg is `1s^(2)2s^(2)2p^(6)3s^(2)` . In boththe casesfirstelectronis to be removedfrom the 3sorbitalbut nuclearcharge ofMg (+2)is hgiherthan thatof Na (+11). Furthermore 3s-orbital in Mg iscompletelyfilled(morestable )while INNA itis onlyhalf- filled (less STABLE ) . THEREFORE`Delta_(i) H` of Na islowerthan thatof Mg.
`Na (1 s^(2)2s^(2)2p^(6) 3s^(1))OVERSET(Delta_(i)H_(1))(to)Na^(+) (1s^(2)2s^(2)2p^(6)) overset(Delta_(i)H_(2))(to)Na^(2+) (1s^(2) 2s^(2) 2p^(5))`
Neongasconfiguration
(morestable )
`Mg (1s^(2) 2s^(2)2p^(6)3s^(2)) overset(Delta_(i) H_(1))(to)Mg^(+) (1s^(2)2s^(2)2p^(6)3s^(1))overset(Delta_(i)H_(2))(to)Mg^(2+)(1s^(2) 2s^(2) 2p^(6))`
Neongas CONFIGURATION
(more stable)
Afterthe removal of thethe firstelectron`Na^(+)` acquiredthe morestable neon gas configuration. Thus for the secondelectronfrom Nais to beremovedfrom thestablenoble(i.e.,neon ) gasconfigurationbutthe lossof secondelectronfrom Mggivesthe morestableneongas configuration.Thus `Delta_(i) H_(2)` of Na is more than`Delta_(i)H_(2)` of Mg.
Discussion
7.

How would you explain the fact that the first ionization enthalpy of sodium is lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium ?

Answer»

Solution :The electronic configuration of Na is `[Ne]3s^(1) " and that of Mg is " [Ne]3s^(2)`.
The configuration of Mg is more stable (being completely filled) than of Na.
Therefore, first ionization enthalpy of Mg is more than that of Na.
After the LOSS of an electron from Na, it acquires the electronic configuration of noble gas, Ne i.e., `1s^(2) 2s^(2) 2P^(6)`.
On the other hand in case of Mg atom, the electronic configuration becomes `[Ne]3s^(1)` .
Thus the electronic configuration of `Na^(+)`is more stable than `Mg^(+)`and HENCE the second ionization enthalpy of Na is much larger than that of Mg.
`Na([Ne]3s^(1)) TONA^(+)([Ne]) to Mg^(2+) ([Ne])`
`Mg([Ne]3s^(2)) to Mg^(2+)([Ne]3s^(1))toMg^(2+)([Ne])`
`IE_(1)(Na) lt IE_(1)(Mg) "" IE_(2)(Na) gt IE_(2)(Mg)`
Discussion
8.

How would you explain the fact that the second ionisation potential is always higher than first ionisation potential ?

Answer»

Solution :(i) Second ionization potential is always higher than first ionization potential.
(ii) Removal of one electron from the valence orbit of a NEUTRAL gaseous atom is easy so first ionization energy is less. But from a UNIPOSITIVE ion, removal of one more electron becomes difficult due to the more forces of attraction between the excess of PROTONS and less numberof electrons.
(iii) Due to GREATER nuclear attraction, second ionization energy is higher than first ionization energy
Discussion
9.

Why the first ionisation enthalphy of sodium is lower than that of magnesium while its second ionisation enthlpy is higher than that of magnesium ?

Answer»

Solution :Electronic configuration of Na and Mg are
`Na=1s^(2)2s^(2)2p^(6)3s^(1)`
`Mg=1s^(2)2s^(2)2p^(6)3s^(2)`
First electron in both cases has to be removed from 3s-orbital but the nuclear charge of Na (+11) is lower than that of Mg (+12) therefore first ionization ENERGY of sodium is lower than that of magnesium.
After the LOSS of first electron, the electronic configuration of
`Na^(+)=1s^(2)2s^(2)2p^(6)`
`Mg^(+)=1s^(2)2s^(2)2p^(6)3s^(1)`
Here electron is to be removed from inert (neon) gas configuration which is very stable and HENCE REMOVAL of second electron requires more energy when compared to Mg.
Therefore, second ionization enthalpy of sodium is higher than that of magnesium.
Discussion
10.

How would you explain the fact that first ionisation enthalpy of sodium is lower than that of magnesium but its second ionisation enthalpy is higher than that of magnesium ?

Answer»

Solution :emoving 1 electron from the outer orbital of sodium atom, the ion formes the most stable configuration of inert atom gas neon. The SECOND electron is REMOVED from ONE of the 2p-orbitals which are completely filled i.e., have a total of 6 electron and are closer to the nucleus.
Discussion
11.

How would you explain ? (i) BeO is insoluble but BeSO_(4) is soluble in water . (ii) BaO is soluble but BaSO_(4) is insoluble in water (iii) LiI is more soluble than KI in ethanol.

Answer»

Solution :(i) Because of smaller SIZE , higher ionization enthalpy and higher electronegativity , BeO is essentially covalent and hence is insoluble in WATER . In contrast , `BeSO_(4)` is ionic . Further because of small size of `Be^(2+)` ion , the hydration energy of `BeSO_(4)` is much higher than its lattice energy and hence `BeSO_(4)` is highly soluble in water .
(ii) Both `BaO` and `BaSO_(4)` are ionic compounds . however , the size of `O^(2-)` ion is much smaller than that of the `SO_(4)^(2-)` ion . Since a bigger anion stabilizes a bigger cation more than a smaller anion stabililzes a bigger cation , therefore , the lattice energy of BaO is much smaller than that of `BaSO_(4)` and hence BaO is soluble while `BaSO_(4)` is insoluble in water .
(iii) `Li^(+)` is much smaller than `K^(+)` ion . Therefore , according to Fajan RULE , `Li^(+)` ion can polarize bigger `I^(-)` ion to a greater extent than `K^(+)` ion . As a result , LiI is more covalent than KI and hence is more soluble in organic solvents like ethanol.
Discussion
12.

How would you explain that the first ionisation energy of Na is lawer than that of Mg but is second inoisation enthalpy is higher than that of Mg ?

Answer»

Solution :First ionisation enthalpy of NA is LAWER than that of Mgbecause the nuclear charge is LESSER for Na than Mg. in the second ionisation energy for Na an electron is to be removed from N `a^+` which requires large energy than the removel of s electron from M `g^+` because `Na^+(1s^2 2s^2 2p^6)` has mare table electronic CONFIGURATION than `Mg^+ (1s^2 2s^2 2p^6 3s^1)`.
Discussion
13.

How would you expect the metallic hydrides to be useful storage? Explain

Answer»

Solution :Some metals likepalladium (Pd), platinum (Pt), etc.have a capacity to adsorb large volume of hydrogen on their surface forming hydrides. In FACT. Hydrogen dissociates on the surface of METAL as H atoms which are adsorbed. In order to accommodate these atoms, the metal lattice expands and becomes rather unstable UPON heating. The hydride releases hydrogen and changes back to the metallic state. The hydrogen evolved in this MANNER can be used as a fuel. The metals listed above (belonging to transition metals) can be used to store as well as transport hydrogen which is to be used as a fuel. Thus, metal hyderides have a ver useful role in hydrogen ECONOMY.
Discussion
14.

How would you distinguish gem-dihalides and vicinal dihalides?

Answer»

Solution :(i) Gem-dihalides on hydrolysis with aqueous KOH gives an aldehyde or a ketone WHEREAS vicinal dihalides on hydrolysis with aqueous KOH give glycols.
(ii)`underset("Ethylidene dichloride ")(CH_3-CHCl_2) + KOH(AQ.) underset(-2KCl)to [CH_3-CH (OH)_2] overset(H_2O)to underset("Acctaldehyde")(CH_3-CHO)`
(iii) `underset("Ethylene dichloride ")(underset(Cl)underset(|)(CH_2) - underset(Cl)underset(|)CH_2) + 2KOH (aq.) underset(2KCl)to underset("Ethylene glycol ")(underset(OH)underset(|)(CH_2)-underset(OH)underset(|)(CH_2))`
The above reaction can be used to distinguish between gem-dihalides and vicinal dihalides.
Discussion
15.

How would you determine molar mass from relative lowering of vapour pressure.

Answer»

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Solution :(i) The measureement of relative lowering of vapour pressure can be used to determine the molar mass of a non-volatile solute.
(ii) A known mass of the solute is dissolved in a known quantity of solvent. The relative lowering of vapour pressure is measured experimentally.
(iii) According to Raoult.s law, the relative lowering of vapour pressure is
`(P_("solvent")^(0) -P_("solution"))=x _(B)`
`W _(A) = weight of solvent, `W_(B)=` weight of solute
`M _(A)=` Molar mass of solvent,` M_(B)=` molar mass of solute
`therefore x _(B) = (n_(B))/(n _(A)+ n _(B))`
where `n _(A)=` NUMBER of moles of solvent, `n _(B)=` number of moles of solute.
For dilute solution, `n _(A) gt gt n _(B), n _(A) + n _(B) ~~ n _(A).`
Then,` x _(B) = (n_(B))/(n _(A))`
Number of moles of solvent and solute are
`n _(A) = (W_(A))/(M_(A)) , n _(B) = (W_(B))/( M _(B))`
`therefore x _(B) = ((W _(B))/(M _(B)))/((W _(A))/(M _(A)))`
Thus, relative lowering of vapour PRESURE ((W _(B))/(M _(B)))/((W _(A))/(M _(A)))`
Relative lowering of vapour pressure ` = (P^(@)-P)/( P^(@))`
`(P^(@) -P)/(P^(@)) = (W_(B) xxM _(A))/(W _(A) xx M _(B))`
From the above equation, molar mass of the folute `M_(B)` can be calculated using the knonw values of `W_(A), W_(B), M_(A)` and the measured relative lowering of vapour pressure.
Discussion
16.

How would you detect the halogens in an organic compound?

Answer»

Solution :(i) To a portion of the Lassaigne.s filtrate. DIL. `HNO_(3)` is added, warmed gently and `AgNO_(3)` solution is added.
(ii) Appearance of curdy white precipitate SOLUBLE in AMMONIA solution indicates the presence of chlorine.
(iii) Appearance of pale YELLOW precipitate sparingly soluble in ammonia solution indicates the presence of bromine.
(iv) Appearance of yellow precipitate insoluble in ammonia solution indicates the presence of iodine.
`Na+Xoverset(Delta)rarrNaX` (where X=Cl, Br, I)
`AgNO_(3)+NaXrarrAgXdarr+NaNO_(3)`
Discussion
17.

How would you detect sulphur?

Answer»

Solution :(i) To a portion of the Lassaigne.s extract, freshly prepared sodium nitroprusside solution is added. If deep violet or purple colour is formed, the presence of sulphur is confirmed.
`UNDERSET("Lassaigne.s extract")(Na_(2)S) + underset("Sodium nitro-prusside")(Na_(2)[Fe(CN)_(5) NO]) rarr underset("Purple colour")(Na_(4)[Fe(CN)_(5)NOS])`
(ii) To another portion of Lassaigne.s extract, acetic acid and lead acetate solution are added. If black PRECIPITATE is formed, sulphur presence is confirmed.
`underset("Leadacetate")((CH_(3)COO)_(2)Pb) + Na_(2)S rarr underset("Black precipitate")(PbSdarr) + 2CH_(3)COONa`
Discussion
18.

How would you convert the following compounds into benzene ? (i) Ethyne (ii) Ethene (iii) Hexane

Answer»

SOLUTION :
Discussion
19.

How would you convert the Ethynecompounds into benzene?

Answer»

SOLUTION :
Discussion
20.

How would you convert the following compounds into benzene? (i) Ethyne (ii) Ethene (iii) Hexane

Answer»

SOLUTION :
Discussion
21.

How would you convert n-butane to iso-butane?

Answer»

SOLUTION :n-butane `UNDERSET(AlCl_3 - HCl)OVERSET(448K , 35 ATM)(to)` iso-butane
Discussion
22.

How would you campare Raoult's law and Henry's law.

Answer»

Solution :(i) According to Raoult.s law, for a solution containing a non volatile SOLUTE,
`P _("solute")=P _("solute"^(0) x _("solute")`
(ii) According to Henry.s law, `P_("solute")=K_(H),x_("solute")` in solution
(iii) The difference betwwn the abouve two LAWS is the proportionality constant `P^(@)` (Raoult.s law) and `K_(H)` (Henry.s law).
(iv) Henry.s law is applicable to solution containing gaseous solute in liquid solvent, while Raoult.s law is applicable to non volatile solid solute in the liquid solvent.
(v) If the solute is non volatilethen the Henry.s law constant will BECOME equal to the vapour pressure of puire solvent `P^(@),` thus Raoult.s law becomes a special case of Henry.s law.
(v) For very dilute solutions, the solvent OBEYS Raoult.s law and the solute obeys Henry.s law.
Discussion
23.

How would you calculate the equivalent mass of anhydrous oxalic acid and hydrated oxalic acid.

Answer»

Solution :In ACID medium,

16 g of oxygen is used for oxidation of 90 g of oxalic acid.
`:.` 8 g of oxygen will oxidize = `90/16 xx 8 = 45 g eq^(-1)`
EQUIVALENT MASS of Anhydrous oxalic acid = 45 g `eq^(-1)`
Equivalent mass of `underset(COOH)(COOH)xx2H_(2)O = ("Molar mass")/("Basicity")`
= `126/2`
= 63 g `eq^(-1)`
Discussion
24.

How would you attribute the structure of PH_(3) molecule using VSEPR model?

Answer»

SOLUTION :PYRAMIDAL with ONE LONE PAIR
Discussion
25.

Find the sum of the following (""^(15)C_(1))/(""^(15)C_(0))+2(""^(15)C_(2))/(""^(15)C_(1))+3(""^(15)C_(3))/(""^(15)C_(2))+......+15(""^(15)C_(15))/(""^(15)C_(14))

Answer»

SOLUTION :`B.O =(N_(b)-N_(a))/(2)`
Discussion
26.

How will your account for 104.5^@ bond angle in water ?

Answer»

Solution :In water OXYGEN undergoes `sp^3` HYBRIDISATION so H-O-H bond angle should have been `109^@28.` but In water, the oxygen atomis SURROUNDED by two shared pairs two lone pairs ELECTRONS. H And ACCORDING to VSEPR theory, lone pair - lone pair repulsions are stronger than bond pair-bond pair repulsions. As a result, the HOH bond angle in water slightly decreases from the regular bond angle of `109^@ 28.` to `104.5^@`.
Discussion
27.

How will you test supphur present is organic compounds?

Answer»

SOLUTION :To aportion of the lassaigne.s extract freshly PREPARED sodium nitropusside solution is added if deep violet or purple cplour is FORMED the presence of sulphur is confirmed
`Na_(2)S+Na_(2)[Fe(CN)_(5)NO]toNa_(4)[Fe(CN)_(5)NOS]`
Discussion
28.

How will you store H_(2)O_(2) ?

Answer»


Answer :`H_(2)O_(2)` is STORED in wax - coated GLASS or plastic container in DARKNESS.
Discussion
29.

How will you separate iodine from sodium chloride ?

Answer»

Solution :EITHER by sublimation or by EXTRACTION with `C Cl_(4)` FOLLOWED by evaporation.
Discussion
30.

How will you separate the constituents from a mixture of H_(2)S and SO_(2) ?

Answer»

SOLUTION :In order to separate the CONSITUENTS, pass the gaseous mixture through alkaline solution of ferrous sulphate. As a result, a black PRECIPITATE of ferrous sulphide is formed due to `H_(2)S` taking part in the REACTION.
`FeSO_(4)+H_(2)S rarr underset("(Black ppt.)")(FeS)+H_(2)SO_(4)`
`SO_(2)` will not react and will escape from the solution in pure form. In order to regenerate `H_(2)S`, treat the black precipitate with dilute `HCl`.
`FeS(s)+2HCl (dil) rarr FeCl_(2)(aq)+H_(2)S(g)`
Discussion
31.

How will you separate propene from propyne ?

Answer»

Solution :By passing the MIXTURE through ammoniacal `AgNO_3` solution or ammoniacal CUCL solution when propyne reacts while PROPENE PASSES over.
Discussion
32.

How will you separate mixture of two gases ?

Answer»

Fractional distillation technique
Grahm's law
Osmosis
CHROMATOGRAPHY

Solution :Fractional distillation is used to separate a mixture of miscrible liquids which differ in their boiling points , e.g. , `(C_(6) H_(6) + CHCl_(3))`
Mixture of two gases can be separated by DIFFUSION technique . According to Graham.s law of diffusion or effusion : At constant temperature and pressure , the rate of diffusion of a gas is inversely proportional to the square root of its density (4) or molecular MASS (M) .
i.e., `r prop (1)/(sqrtd) or (1)/(sqrtM)` (at constant T and p)
`or (r_(1))/(r_(2)) = sqrt((d_(1))/(d_(2))) = sqrt((M_(2))/(M_(1))) ( M = 2 xx d)`
Thus , the lighter gas will diffuse faster and the one with higher molecular mass is left behind .
Osmosis is the simple process of movement of only solvent molecules from the region of their higher concentration to the region of the lower concentration through a semipermeable membrane .
Chromatography is used to separate a mixture of miscible liquids with differ in their rate of adsorption on the stationary PHASE .
Discussion
33.

How will you separate ether from water ?

Answer»

SOLUTION :By SEPARATING PROCESS.
Discussion
34.

How will you separate a mixture of urea and sodium chloride ?

Answer»

SOLUTION :Both urea and NACL are soluble in water but urea is soluble in alcohol but NaCl is not. THEREFORE, this MIXTURE can be separated by SHAKING with alcohol when urea goes into solution while NaCl remains undissolved. The mixture.suspension thus obtained is filtered. Evaporation/distillation of the filtrate gives urea while NaCl remains as a residue on the filter paper.
Discussion
35.

How will you separate a mixture of two solids which differ in their solubilities in the same solvent ?

Answer»


ANSWER :By FRACTIONAL CRYSTALLISATION
Discussion
36.

How will you separate a mixture of two organic compounds which have different solubilities in the same solvent ?

Answer»

SOLUTION :By FRACTIONAL CRYSTALLIZATION.
Discussion
37.

How will you separate a mixture of two organic compound which have different solubilities in the same solvent?

Answer»

SOLUTION :By FRACTIONAL CRYSTALISATION
Discussion
38.

How will you separate a mixture of benzoic acid and naphthalene ?

Answer»

Solution :Both these solids sublime on heating. Therefore, these cannot be separated by sublimation. The mixture is heated with water when only benzoic acid will dissolve. Upon filtration, naphthalene is separated and the solution upon cooling gives crystals of benzoic acid.
Alternatively the mixture is reacted with strong aqueous sodium hydroxide when benzoic acid is CONVERTED to SOLUBLE sodium benzoate whie NAPTHALENE is unaffected. It can be separated by filtration. the filtrate is then treated with dilute `HCL` to regenerate benzoic acid.
`underset("Benzoic acid")(C_(6)H_(5)COOH)(s)+NaOH(aq) rarr underset("Sod. benzoate")(C_(6)H_(5)COONa(aq))+H_(2)O`
`C_(6)H_(5)COONa(aq) + HCl (aq) rarr underset("(White ppt.)")(C_(6)H_(5)COOH (s))+NACL (aq)`
Discussion
39.

How will you separate a mixture of ethane, ethylene and acetylene ?

Answer»

Solution :The mixture can be separated into its comstituents by the following steps :
Step 1 . Pass the mixture of gases through Tollen's REAGENT when acetylene will form WHITE precipitate of disilver acetylide while ethane and ethylene will pass through .
`underset"Acetylene"(HC-=CH)+underset"Tollen's reagent "(2[Ag(NH_3)_2]^(+) OH^-) to undersetunderset"(White ppt.)""Disilver acetylide"(Ag-C-=C-Ag)+4NH_3+ 2H_2O`
Separate the white ppt. by FILTRATION and treat it with dil. `HNO_3` to regenerate . Collect it in a separate container.
`Ag-C-=C-Ag + 2HNO_3 to HC-=CH+2 AgNO_3`
Step 2.Pass the remaining mixture of ethane and ethylene through cold conc. `H_2SO_4` when ethylene will be absorbed as ethyl HYDROGEN sulphate while ethane escapes . The ethane thus obtained is collected in a separate container.
`underset"Ethylene"(CH_2=CH_2)+ H_2SO_4(conc.) to underset"Ethyl hydrogen sulphate"(CH_3CH_2OSO_2OH)`
The ethyl hydrogen sulphate thus obtained is heated with conc. `H_2SO_4` to 433-443 K when ethylene is obtained which is collected in a separate container.
`CH_3CH_2OSO_3H overset"433-443 K"to CH_2=CH_2+H_2SO_4`
Discussion
40.

How will you separate a mixture of o-nitrophenol and p-nitropheonol ?

Answer»

SOLUTION :Steam distillation. O-Nitrophenol being steam VOLATILE being steam volatile DISTILS over along with water while p-nitrophenol being non-volatile remains in the FLASK.
Discussion
41.

How will you relate the Van't Hofffactors to association and dissociation?

Answer»

Solution :The degree of dissociation or association can be RELATED to Van't HOFF factor (i) using the following relationships
` alpha _( "dissociation ") =(i-1)/( n-1) `
( where n is number ions/ species FORMED by the dissociation of a SINGLE MOLECULES)
` alpha _( "association ") =( ( 1-i ) n)/( n=1) `
(here n is the number of solute involved in association.)
Discussion
42.

How will you regenerate zeolite in ion-exchange method?

Answer»

SOLUTION :In ion exchange method, when all the `NA^+` IONS of zeolite are exchange by Ca/Mg, the zeolite can be regenerated by TREATING with a strong solution of NACL.
`M-Z+2NaCltoNa_2-Z+MCl_2`
Discussion
43.

How will you purify impure sample of benzene ?

Answer»


ANSWER :By DISTILLATION
Discussion
44.

How will you purify impure naphthalene ?

Answer»

SOLUTION :By SUBLIMATION PROCESS.
Discussion
45.

How will you purify a simple of benzoic acid containing sodium chloride as impurity ?

Answer»


ANSWER :By CRYSTALLISATION
Discussion
46.

How will you purify impure benzene ?

Answer»

SOLUTION :By DISTILLATION PROCESS.
Discussion
47.

How will you purify essential oils ?

Answer»

SOLUTION :Essential oils are volatile and are INSOLUBLE in water. THEREFORE, they are PURIFIED by steam DISTILLATION.
Discussion
48.

How will you purify a liquid having non-volatile impurities ?

Answer»

Solution :Simple distillation will give US the pure liquid while the non-volatile IMPURITIES will remain in the flask as residue. For example, SEA water containing non-volatile impurities of chlorides and sulphates of Na, K, Ca and Mg can be PURIFIED by simple distillation.
Discussion
49.

How will you prepare propane from a sodium salt of fatty acid

Answer»

SOLUTION :`CH_(3)- CH_(2) -CH_(2) COONaoverset(NaOHCaO)(to)CH_(3)- CH_(2)-CH_(3) + Na_(2) CO_(3)`
SODIUM saltof butyricacid on heatingwithsodalinegivespropane
Discussion
50.

How will you prepare propyne using alkylene dihalide?

Answer»

SOLUTION :
Discussion