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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 88201. |
A overset(Ph-SO_(2)Cl)rarrB overset("KOH")rarrCoverset(C_(2)H_(5)I)rarrD 'C' is water soluble Correct structure of A and D are |
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Answer» `R-NH_(2)","Ph-SO_(2)-NR-(C_(2)H_(5))_(2)^(+)I^(-)` |
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| 88202. |
(A) overset(NH_(3))to (B) underset(P_(2)O_(5))overset(Delta)to (C) overset(LiAlH_(4)) to (D)overset(HNO_(2)) to (E) WhereEis the alcohol used to denature ethylalcohol . Identify (A) . |
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Answer» HCO OH |
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| 88203. |
A overset(NH_(2)OH)(rarr)B overset(H_(2)SO_(4))(rarr) C underset(H_(2)O)overset(260-270^(@)C)(rarr)D If 'D' is a polymer nylon–6 then 'A' can be : |
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Answer»
 .
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| 88204. |
A overset(KCN"added slowly")(to) underset(("Reddish brown precipitate"))(B) overset("In excess KCN precipitate dissolves")(to)underset(("Yellow solution"))(C) A overset(K_(4)Fe(CN)_(6))(to) underset(("Intense blue precipitate"))(D) overset(NaOH)(to) ("Red precipitate") A overset(Na_(2)HPO_(4))(to) underset(("Yellowish white precipitate"))(E) How many produces are correctly match? A=FeCl_(3), A=FeCl_(2), A=CuCl_(2), B=Fe(CN)_(3) B=K_(3)Fe(CN)_(6), C=K_(3)Fe(CN)_(6), C=K_(4)[Fe(CN)_(6)], D=Fe_(4)[Fe(CN)_(6)]_(3) D=Fe[Fe(CN)_(6)], E=FePO_(4) |
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Answer» `Fe(CN)_(3)+3KCNtoK_(3)[Fe(CN)_(6)]` `4FeCl_(3)+3K_(4)[Fe(CN)_(6)]toFe_(4)underset(D)([Fe(CN)_(6)]_(3))+12KCl` `Fe_(4)[Fe(CN)_(6)]_(3)+12NaOHto4Fe(OH)_(3)+3Na_(4)[Fe(CN)_(6)]` `FeCl_(3)+Na_(2)HPO_(4)tounderset(E)(FePO_(4))+2NaCl+HCl` |
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| 88205. |
A overset(H_2//"Raney Ni")to CH_3-CH_2-underset(H)underset(|)overset(H)overset(|)C-OH. A is ___________. |
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Answer» propanone |
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| 88206. |
A overset(Delta)underset(800^(@)C)rarr CH_(2)=C=O, Reactant 'A' in the reaction is |
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Answer» `CH_(3)CH_(2)CHO` |
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| 88207. |
A overset(4KOH.O_2)(to) underset("green")(2B ) + 2H_2O3B overset(4HCl)(to) underset("purple")(2C) + MnO_2+ 2H_2O2B overset(H_2O, KI)(to)2A + 2KOH + DIn the above sequence of reactions, A and D respectively, are : |
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Answer» `KIO_3` and `MnO_2` |
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| 88208. |
A overset(Cu//573K)larrCH_3 - CH_2OHoverset(Al_2O_3//Delta)toB. In this reaction A and B are respectively ................... . |
| Answer» Solution :Alkanal, Alkene | |
| 88209. |
A overset(" Na")(larr)CH_(3)CH_(2)Br overset("KOH alc.")(rarr)B. A and B are respectively |
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Answer» `C_(2)H_(4),C_(2)H_(6)` |
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| 88210. |
A overset( HBr) to B overset( AqKOH) toC overset( PDC) to CH_3 COCH_3 Identify the orgaic compounds A,B and C givenin the above sequence. |
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Answer» `CH_3 CHO, C_2H_2Br_2 and CH_3COOH` |
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| 88211. |
(a)Out ofAg_(2)SO_(4), CuF_(2), MgF_(2) and CuCl, whichcompound will be coloured and why ? (b) Explain :(i)CrO_(4)^(2-) is a strong oxidizing agent whileMnO_(4)^(2-) is not (ii) Zr and Hf have identical sizes. (iii) The lowestoxidation state of manganese is basic while the highestis acidic. (iv)Mn (II)shows maximumparamagnetic character amongst the divalent ions of the first transition series. |
| Answer» SOLUTION :(a) `CuF_(2)` (b) (i) Cr in`CrO_(4)^(2-)` is in the highest oxidation state, i.e., `+6` while Mn in `MnO_(4)^(2-)` is in oxidation state`+6` and its most STABLE oxidation state is`+7` (ii) due to lanthanoil contraction (iv)`Mn^(2+)` has `3d^(5) ` CONFIGURATION | |
| 88212. |
(A): Outermost electronic configuration of Pt is 5d^(9)s^(1) (R) : Pt in its ion attains pseudo inert gas configuration The correct answer is |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A) |
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| 88213. |
(a) Out of the following pairs, predict with reason which pair will allow greater conduction of electricity: (i) Silver wire at 30^(@)C or silver at 60^(@)C. (ii) 0.1 M CH_(3)COOH solution or 1 M CH_(3)COOH solution. (iii) KCl solution at 20^(@)C or KClsolution at 50^(@)C . Give two points of differences between electrochemical and electrolytic cells. |
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Answer» Solution :(a) (i) SILVER wire at `30^(@)C` or silver wire at `60^(@)C`. Silver wire at `30^(@)C` shows greater conduction of electricity. Conductivity of metals decreases with rise of temperature. This is because vibration of kernels increases with temperature. (ii) 0.1 M `CH_(3)COOH` solution or 1 M `CH_(3)COOH` solution. 0.1 M `CH_(3)COOH` shows greater conductivity. For weak ELECTROLYTES, conductivity increases with dilution because of increased ionisation and increase in the VOLUME of solution containing 1 mole of the electrolyte. (iii) KCl solution at `20^(@)C` or KCl solution at `50^(@)C`. KCl solution at `50^(@)C` will show greater conductance because ionic mobility increases with temperature. (b) Electrochemical cell and Electrolytic cell (i) Chemical energy of a reaction is converted into electrical energy in an electrochemical cell. Electrical energy is used to bring about a chemical change in electrolytic cell. (ii) An electrochemical cell consists of two parts, an anode where the oxidation takes place and a cathode where reduction takes place. The two parts are joined by a salt bridge. In an electrolyticcell, the substance in solution or molten state is taken in a vessel. Two electrodes a cathode and an anode of suitable material are dipped into it. On passing electricity, cations are deposited at cathode and anions are deposited at anode. |
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| 88214. |
(a). Out of which is an example of vinylichalide? (b). Out of whish is an example of benzylic halide? |
Answer» SOLUTION : .
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| 88215. |
(a) Out of Ag_(2)SO_(4),CuF_(2),MgF_(2) and CuCl, which compound will be coloured and why ? (b) Explain : (i) CrO_(4)^(2-) is a strong oxidising agent while MnO_(4)^(2-) is not (ii) Zr and Hr have identical sizes. (iii) The lowest oxidation state of manganese is basic while the highest is acidic. (iv) Mn (II) shows maximum paramagnetic character amongst the divalent ions of the first transition series. |
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Answer» Solution :(a) `CuF_(2)` is coloured due to vacant d-orbital and presence of unpaired electron. (B) (i) Cr in `CrO_(4)^(2-)` has oxidation state +6. It can reduce its oxidation state to +3 (in `Cr^(3+)`). THUS, it can act as oxidising agent. MN in `MnO_(4)^(2-)` has oxidation state +6. But its most stable oxidation state is +7. Therefore, it does not act as oxidising agent. (ii) Zr and Hf have identical sizes due to lanthanoid contraction. (III) Oxides of Mn in lowest oxidation state are ionic and form basic solution, whereas oxides in highest oxidation state are covalent and form acidic solution with water.  `Mn^(2+)` has the maximum number of unpaired electrons and, therefore, has the maximum paramagnetic character. |
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| 88216. |
(a) Out of t-butyl alcohol and n-butanol, which one will undergo acid catalysed dehydration faster and why? (b) Carry out the following conversions: (i) Phenol to Salicylaldehyde (ii) t-butylchloride to t-butyl ethyl ether (iii) Propene to Propanol |
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Answer» Solution :(a) t-butyl ALCOHOL will undergo acid CATALYSED dehydration faster. This is because the tertiary carbonium ion formed with t-butyl alcohol is more stable than primary carborium ion formed with n-butanol. `underset("t-Butyl CHLORIDE")(CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-Cl)overset(NaOH)tounderset("t-Butyl alcohol")(CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-OH)overset(Na)toCH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-ONa overset(CH_(3)CH_(2)Cl)tounderset("t-Butylethyl ether")(CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-OCH_(2)CH_(3))+NaCl` (b) (i) Phenol to Salicylaldehyde  (ii) t-butylchloride to t-butyl ethyl ether `CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-O^(-)Na^(+)+CH_(3)CH_(2)Cl tounderset("t-Butylethyl ether")(CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-O-CH_(2)CH_(3))` (iii) PROPENE to Propanol `underset("Propene")(CH_(3)-CH=CH_(2)) underset("Peroxide")overset("HBr")to CH_(3)-CH_(2)-CH_(2)Br overset(NaOH)to underset("Propanol")(CH_(3)CH_(2)CH_(2)OH)` |
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| 88217. |
(A) Orthorhombic crystal system has four Bravis lattices (R) Orthorhombic crystal system has equal sides and angles between faces. |
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Answer» Both (A) and (R) are TRUE and (R) is the correct explanation of (A) |
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| 88218. |
(A): Orthophosphoric acid can form two acidic salts and one normal salt (R): Orthophosphoric acid is a tribasic acid |
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Answer» Both (A) and (R ) are TRUE and (R ) is the correct EXPLANATION of (A) |
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| 88219. |
A organic compound (A) has pleasent odour, on boiling(A) with conc.H_2SO_4 at 443K produces colourless gas , which decolourises bromine water and Bayer's reagent. The original organic compound (A) is |
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Answer» `C_2H_5-Cl` |
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| 88220. |
(A) Order with respect to any reactant or product can be zero or positive but it is never negative. (R ) Rate does not decrease with increase in concentration of a reactant or product. |
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Answer» Both (A) and (R ) are TRUE and (R ) is the correct explanation of (A) |
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| 88221. |
(A) Ores like haematite, Iron pyrites can be concentrated by magnetic seperation method (R) Magneticseperation is used for ores in whichthe impurityor ore is magnetic |
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Answer» Both (A) and (R) are true and (R) is the CORRECT explanation of (A) |
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| 88222. |
(A) Order of recation is related to molecularity of reaction (R ) Molecularity of recation can be fractional. |
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Answer» Both (A) and (R ) are true and (R ) is the correct EXPLANATION of (A) |
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| 88223. |
A: Order of reaction is evaluated from the mechanism of a reaction R: Order of reaction can be zero |
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Answer» Both (A) and (R ) are true and (R ) is the CORRECT explanation of (A) |
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| 88224. |
(A) Order and molecularity are always equal (R ) Complex reactions take place in steps and the fastest step determines rate of reaction. |
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Answer» Both (A) and (R ) are true and (R ) is the CORRECT EXPLANATION of (A) |
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| 88225. |
(A): One mole of CrCl_3. 5NH_3 in aqueous solutions can give two moles of AgCl on addition with excess AgNO_3solution (R) : CrCl_3 . 5NH_3 has octahedral shape. |
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Answer» Both A & R are true, R is the correct explanation of A |
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| 88226. |
(a) One mole of an ideal gas expands isothermally and reversibly at 25^(@)C from a volume of 10 litres to a volume of 20 litres. (i) What is the change in entropy of the gas ? (ii) How much work is done by the gas ? (iii) What is q (surroundings) ? (iv) What is the change in the entropy of the surroundings ? (v) What is the change in the entropy of the system plus the surroundings ? (b) Also answer the questions (i) to (v) if the expansion of the gas occurs irreversibly by simply opening a stopcock and allowing the gas to rush into an evacuated bulb of 10-L volume. |
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Answer» Solution :(i) `DeltaS=2.303nR log. (V_(2))/(V_(1))=2.303xx1xx8.314xxlog.(20)/(10)=5.76J//K` (a) (II) `W_(rev)=-2.303nRTlog.(V_(2))/(V_(1))` `=2.303xx1xx8.314xx298xxlog.(20)/(10)=-1718J` (iii) For isothermal process, `DeltalI=0` and heat is absorbed by the gas, `q_(rev)=DeltalI-W=0-(-178)=1718J` `:.q_(rev)=1718J` (`:.` process is reversible) (IV) `DeltaS_(surr)=-(1718)/(298)=-5.76J//K` As entropy of the system increases by `5.76J`, the entropy of the SURROUNDING decreases by `5.76J`, since the process is carried out reversibly. (V) `DeltaS_(sys)+DeltaS_(surr)=0`.......for reversible process. (b) (i) `DeltaS=5.76J//K`, which is the same as above because S is a state function. `(ii) W=0` (`:. p_(ext)=0`) `(iii)` No heat is exchanged with the surroundings `(iv)DeltaS_(surr)=0` `(v)` The entropy of the system plus surrounding increases by `5.76 J//K` as we except entropy to increase in an irrevesible process. |
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| 88227. |
(A): One molar aqueous solution is always more concentrated than one molal aqueous solution. (R): The amount of solvent in 1M solution is less than in 1m solution . The correct answer is |
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Answer» Both (A) and (R) are true and (R) is the correct EXPLANATION of (A) |
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| 88228. |
A one molal solution is one that contains |
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Answer» 1 G. of the solute in `1000G` of solvent |
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| 88229. |
A one-litre vessel contained a gas at 27^(@)C. 6 g of charcoal was introduced into it. The pressure of the gas fell down from 700 mm to 400 mm. Calculate the volume of the gas (at S.T.P.) adsorbed per gram of charcoal. Density of charcoal sample used was 1.5g cm ^(-3). |
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Answer» Solution :VOLUME of carcoal presentin the eessel `=(6g)/(1.5g cm ^(-3))=4 cm` `THEREFORE` Volume of the gas initial at `27^(@)C,` 700 mm pressure `=100-4=996 cm^(3)` Volume of the gas at 400 cm and `27^(@)C` is again equal to the volume of the vessel EXCLUDING that of chrocol, i.e., `996 cm^(3).` Let US calculate equivalent volume of the gas at 700 mm at the same temperature `P_(1)V_(1)=P_(2)V_(2)i.e. 400 xx996=700xxV_(2)or V_(2)=569.1c c` `therefore` Volume of the gas adsorbed at `27^(@)C,` 700 mm pressure `=996-569.1=426.9cm^(3)` `therefore `Volume adsorbed pr gram of charcoal `=(426.9)/(6)cm^(3)G^(-1)=711cm^(3)g^(-1)` Converting this volume to S.T.P. we get `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2)),i.e., (700xx71.8)/(300)=(760xxV_(2))/(273)or V_(2)=59.6cm^(3)` |
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| 88230. |
A one litre flask is full of reddish brown bromine fumes.The intensity of brown colour of vapour will not decrease appreciably on adding to the flask some: |
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Answer» PIECES of marble |
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| 88231. |
A one litre flask is full of brown bromine vapour.The intensityof brown colour of vapour will not decreases appreciably on adding to the flask some |
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Answer» pieces of marble |
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| 88232. |
(A) on treatment with KCN gives a buff coloured precipitate which dissolves in excess of this reagent forming a compound (C). |
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Answer» |
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| 88233. |
(a) On mixing liquid X and Y, the volume of the resulting solution increases. What type ofdeviation from Raoult's law is shown by the resulting solution ? What change in temperaturewould you observe after mixing X and Y ? (b)How can the direction of osmosis be reversed ? Write one use of reverse osmosis. |
| Answer» Solution :(a) The resulting solution will SHOW positive DEVIATION from Raoult.s law. A LOWER temperaturewill be observed. (b) We can reverse the DIRECTION of osmosis by applying a pressure greater than the OSMOTICPRESSURE on the solution. Reverse osmosis is used in the purification of water for domesticand commercial purpose. | |
| 88234. |
(A) On heating with P_4O_10, carboxylic acids give anhydrides. (R) Carboxylic acids on reduction with LiA//H_4 give alcohols. |
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Answer» Both (A) and (R) are TRUE and (R) is the correct explanation of (A) |
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| 88235. |
A' on heating at 700^(@)C in air gives a white infusible amorphous powder (B) which is decomposed when heated in the currect of steam to give white powder 'C' and a gas 'D'.'D' turns red litmus blue and in aqueous solution , gives reddish brown ppt with K_(2)HgI_(4). compound 'C on strong heating gives 'E. 'D' is |
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Answer» `N_(2)` |
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| 88236. |
A' on heating at 700^(@)C in air gives a white infusible amorphous powder (B) which is decomposed when heated in the currect of steam to give white powder 'C' and a gas 'D'.'D' turns red litmus blue and in aqueous solution , gives reddish brown ppt with K_(2)HgI_(4). compound 'C on strong heating gives 'E. 'B' is |
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Answer» `BN` |
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| 88237. |
(A) On heating ferromagnetic or ferrimagnetic substances, they become paramagnetic (R) On heating randomsiation of magnetic domains occurs |
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Answer» Both (A) and (R) are true and (R) is the correct EXPLANATION of (A) |
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| 88238. |
A' on heating at 700^(@)C in air gives a white infusible amorphous powder (B) which is decomposed when heated in the currect of steam to give white powder 'C' and a gas 'D'.'D' turns red litmus blue and in aqueous solution , gives reddish brown ppt with K_(2)HgI_(4). compound 'C on strong heating gives 'E. 'C' is |
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Answer» `B_(2)O_(3)` [truns RED litmus BLUE because it is basic in NATURE] |
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| 88239. |
On dissolving 2.34g of non-electrolyte solute in 40g of benzene, the boilingpoint of solution was higher than benzene by 0.81K. Kb value for benzene is 2.53 K kgmol^(-1) .Calculate the molar mass of solute. [Molar mass of benzene is78 gmol^(-1)] |
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Answer» Solution :(a)Data: Mass of solute = `W_(2)` = 2.34 g, `K_(B) = 2.53 K.kg mol^(-1)`, Molar mass of solute = `M_(2)`= ? Mass of solvent benzene = WJ = 40 g Elevation in boiling point =0.81 K Formula: `M2=W_(2)/W_(1) 1000` Substituion: `M_(O_(2))=2.53xx2.34xx(1000)/(40)xx1/0.81` Answer: `M_(2) = 182.72 g mol^(-1)` (b) (II) Solubility of gas decreases with increase in temperature and INCREASES with increase in pressure. |
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| 88240. |
A' on heating at 700^(@)C in air gives a white infusible amorphous powder (B) which is decomposed when heated in the currect of steam to give white powder 'C' and a gas 'D'.'D' turns red litmus blue and in aqueous solution , gives reddish brown ppt with K_(2)HgI_(4). compound 'C on strong heating gives 'E. 'E' is |
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Answer» `B_(2)O_(3)` |
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| 88241. |
A' on heating at 700^(@)C in air gives a white infusible amorphous powder (B) which is decomposed when heated in the currect of steam to give white powder 'C' and a gas 'D'.'D' turns red litmus blue and in aqueous solution , gives reddish brown ppt with K_(2)HgI_(4). compound 'C on strong heating gives 'E. 'A' is |
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Answer» SOLUTION :`2B + 3O_(2) to B_(2)O_(3)` ( WHITE POWDER) `underset("('A')")(2B) + N_(2) to underset("('B')")(2BN) ` (White powder) |
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| 88242. |
(A) OF_(2) is named as oxygen difluoride (R ) In OF _(2), oxygen is less electronegative than fluorine |
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Answer» Both (A) and (R ) are true and (R ) is the CORRECT explanation of (A) |
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| 88243. |
(a) OF the ions Ag^(+), Co^(2+) and Ti^(4+), which onewill be coloured in aqueous solutions? ( Atomic nos. : Ag = 47 , Co = 27 ,T I = 22) (b) If each one of the above ionic species is in turn placed in a magnetic field, how will it response and why ? |
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Answer» Solution :(a) `Co^(2+)` will be colouredbecause it CONTAINS unpaired ELECTRONS. On the other hand , `Ag^(+)` has COMPLETELY filled 3d subshell and `Ti^(4+)` has empty 3d subshell, THEREFORE, they will be colourless. (b) `Co^(2+)` will be attracted by a magnetic field while`Ag^(+)` and `Ti^(4+)` ioniswill be repelled. |
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| 88244. |
(A) +O_(2)toX+Y+Z (Organic compound) Compound (A) is pure form does not give ppt. with AgNO_(3), solution. A mixture containing 70% of (A) and 30% of ether is used as an anaesthetic. Compound (X) and (Y) are oxides while (Z) is a pungen smelling gas. (X) is a neutral oxide which turns cobalt chloride paper pink. Compound (Y) turns lime water milky and produces an acidic solution with water. Compounds (A),(X),(Y) and (Z) respectivley will be |
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Answer» `CH_(4),H_(2)O,CO_(2),Cl_(2)` |
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| 88245. |
(A) o-Nitrophenol is less soluble in water than the m- and p-isomers of nitrophenol.(R) o-Nitrophenol exist as associated molecules. |
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Answer» Both (A) and (R) are true and (R) is the correct EXPLANATION of (A) |
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| 88246. |
(A) o-Phenol sulphonic acid on heating at 100^(@)C changes to p-phenol sulphonic acid. (R ) Sulphonation of phenol is a reversible process. |
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Answer» If both (A) and (R ) are CORRECT and (R ) is correct EXPLANATION of (A ). |
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| 88247. |
(A) O and p nitro phenols are separated by steam distillation(R) O-isomer is steam volatile due to intra molecular H-bond |
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Answer» Both (A) and (R) are true and (R) is the CORRECT explanation of (A) |
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| 88248. |
(A) O _(2) ^(-)ion is more stable than O _(2) ^(+) ion (R ) Negative ions are always more stable than positive inos |
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Answer» Both (A) and (R ) are true and (R ) is the CORRECT EXPLANATION of (A) |
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| 88249. |
(a) Number of orgsnic products obtained in more than 5% yield. (b )Number of moles of HI consumed. (c )Number of moles of I_(2) generated. ( d) Number of fraction which can be obtained on fractional distillation of organic product from mizture of products. Write answer of part a, b, c and d in the same order and present the four digit number as answer in OMR sheet. For example : If all these answers are 9 then fill 9999 in OMR sheet. |
Answer» 
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| 88250. |
(a) Number of ions which are regarded as ionic carbides amongst (C^(4-),C_(2)^(2-),C_(3)^(4-),C_(4)^(3-))=X (b) When oxalic acid is heated with conc. H_(2)SO_(4) it produces number of different gases amongst (CO,CO_(2),SO_(3))=Y ( c) Number of molecules having V-shaped structure amongst (SnCl_(2),PbCl_(2),COS)=Z Find X+Y+Z. |
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Answer» |
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