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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.
| 11551. |
The reagents needed to convert : |
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Answer» `KOH,Br_(2),LiAlH_(4)`<BR>`KOH,Br_(2),CH_(3)COCl` |
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| 11552. |
Which of the following is purified by zone refining ? |
| Answer» Answer :C | |
| 11553. |
To a 25 mL H_(2)O_(2) solution excess acidified solution of Kl was added. The iodine liberated20 ml of 0.3 N sodium thiosulphate solution. Use these data to choose the correct statements from the following : |
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Answer» The weight of `H_(2)O_(2)` present in 25 ml SOLUTION is 0.102 g |
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| 11554. |
The vapour pressure of a mixure of 1 mole of liquid 'A' and 3 mole of liquid 'B' is 550 mm of Hg. When one mole of liquid 'B' is further added, thevapour pressure of the solution increases by10 mm of Hg. The vapour pressure of pure liquid A and B are respectively in mm of Hg . |
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Answer» 400, 600 |
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| 11555. |
The standard oxidation potential of Ni//Ni^(+2) electrode is 0.236 V. If this electrode is combined with a hydrogen electrode in acid solution, at what pH of the solution will the measured emf be zero at 25^(@)C? Assume[Ni^(+2)] = 1M |
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Answer» 2 |
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| 11556. |
What are the common types of secondary structure of proteins? |
Answer» Solution :The conformation which the polypeptide chains assume as a result of hydrogen BONDING is called the secondary structure of the PROTEINS. The two types of secondary STRUCTURES are : cx HELIX and P-pleated sheet structure as shown in FIGURES below: 
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| 11557. |
Which of the following angle corresponds to sp^(2) hybridisation ? |
| Answer» Answer :B | |
| 11558. |
XeF_(2) has … struture |
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Answer» TRIANGULAR |
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| 11559. |
Write short note : Step growth polymerization. |
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Answer» Solution :This type of polymerization generally INVOLVES a repetitive condensation reaction between two different bi-functional or trifunctional mono-meric units. These POLYCONDENSATION reactions may result in the loss of some simple molecules as water, alcohol, HYDROGEN chloride, etc. and lead to the formation of high MOLECULAR mass condensation polymers. In these reactions, the product of each step is again a bi-functional species and the sequence of condensation goes on. Since, each step produces a distinct functionalized species and is independent of each other, this process is also called as step growth polymerization. The formation of terylene or DACRON by the interaction of ethylene glycol and terephthalic acid is an example of this type of polymerization. 
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| 11560. |
What are bio degradable polymers ? Give examples . |
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Answer» Solution :The materials that are readily decomposed by micro organisms in the environment are called biodegradable . Natural polymers degrade on their own after certain PERIOD of TIME but the synthetic polymers do not . The biopolymers which disintergrates by themselves in biological systems during a certain period of time by enzymatic hydrolysis and to some extent by OXIDATION are called biodegradable polymers . Examples : (i) POLYHYDROXY butyrate (PHB) . (ii) Polyhydroxy butyrate - co- `BETA`- hydroxyl valerate (PHBV) (iii) Polylactic acid (PLA) |
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| 11561. |
Which of the following acid produces violet colour with FeCl_(3) solution ? |
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Answer» Benzoic ACID |
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| 11562. |
which of the following electrilytes is most effective for the coagulation of Agl//Ag^(+) Sol? MgCl_(2).K_(2)SO_(4),K_(4)[Fe(CN)_(6)] |
| Answer» SOLUTION : `K_(4[Fe(CN)_(6)]` Is the most effective for coagulation ofAg/Ag SOL. | |
| 11563. |
Volume occupied by one molecule of water (density =1g//cm^(3)) is: |
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Answer» `3XX10^(-23)cm^(3)` `=2.98xx10^(-23)g` Volume of one molecule `=(M)/("DENSITY")=(2.98xx10^(-23))/(1)cm^(3(` `=3xx10^(-23)cm^(3)` |
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| 11564. |
Which of the following shows the correct reaction for nitrobenzene reduction ? |
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Answer» Nitrobenzene reacts with ZN dust and `NH_4Cl` to produce ANILINE |
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| 11565. |
Two platinum electrides are dipped in an aqueous solution of copper sulphate blue in colour. A current is passed through it. (a) What will happen on the two electrodes ? (b) What will happen to the colour of the solution ? (c ) Predict the nature of the solution left in the electrolytic cell. |
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Answer» Solution :(a) Copper will be DEPOSITED on the cathode i.e., the electrode connected to the negative terminal. Oxygen GAS `(O_(2))` will evolve at the anode. (B) The blue colour of the solution will slowly disappear and it will become colourless. (C ) The solution left in the ELECTROLYTIC cell will be acidic (aqueous solution of `H_(2)SO_(4)`). For details, consult text part. |
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| 11566. |
Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging. |
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Answer» Solution :* A secondary cell after use can be recharged by passing current through it in the opposite DIRECTION so that it can be used again. * On charging the battery the reaction is reversed and `PbSO_(4(S))` on ANODE and CATHODE is converted into Pb and `PbO_(2)` respectively. In recharging reaction, reduction is occurred near anode which is cathode in recharging. `PbSO_(4(S))+2e^(-) to Pb_((S))+SO_(4(aq))^(2-)`. . . (i) * (+) Cathode now becomes (+) anode and oxidation is CARRIED out. `PbSO_(4(S))+2H_(2)O_((l)) to PbO_(2(S))+SO_(4(aq))^(2-)+4H_((aq))^(+)+2e^(-)`. . (ii) * Over all reaction (i)+(ii) is as follows: `2PbSO_(4(S))+2H_(2)O to Pb_((S))+PbO_(2(S))+2H_(2)SO_(4(aq))` * So, on anode, `PbSO_(4)` is converted into pH and on cathode `PbSO_(4)` is converted into `PbO_(2)`. |
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| 11567. |
Which of the followingexhibits geometrical isomerism? |
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Answer» `C_(2)H_(5)Br` |
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| 11568. |
Which statement about Hg is correct? |
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Answer» HG is the only LIQUID metal |
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| 11569. |
Two pure liquids A & B which can from ideal solution have vapour pressures 300 torr & 800 torr . A mixture of the vapour of A & B for which the mole fraction of A is 0.25 is slowly compressed at temperature T. The molefraction of A in this first drop of condensate is : |
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Answer» 0.47 |
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| 11570. |
Two pure liquids A & B which can from ideal solution have vapour pressures 300 torr & 800 torr . A mixture of the vapour of A & B for which the mole fraction of A is 0.25 is slowly compressed at temperature T. The total pressure applied when only last bubble of vapour remains |
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Answer» 675 torr |
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| 11571. |
Two pure liquids A & B which can from ideal solution have vapour pressures 300 torr & 800 torr . A mixture of the vapour of A & B for which the mole fraction of A is 0.25 is slowly compressed at temperature T. The molefraction of B for a liquid solution whose normal boiling point is 'T' |
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Answer» 0.08 |
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| 11572. |
Two pure liquids A & B which can from ideal solution have vapour pressures 300 torr & 800 torr . A mixture of the vapour of A & B for which the mole fraction of A is 0.25 is slowly compressed at temperature T. The molefraction of B in the last track of vapour is : |
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Answer» 0.89 |
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| 11573. |
When a white crystalline compound X is heated with K_2Cr_2O_7 and concentrated H_2SO_4 a reddish brown gas A is evolved. On passing A into caustic soda solution, a yellow coloured solution B is obtained. Neutralizing the solution B with acetic acid on subsequent addition of lead acetate, a yellow precipitate C is formed. When X is heated with NaOH solution, a colourless gas is evolved and on passing this gas into K_2HgI_4 , solution, a reddish brown precipitate D is formed. Identify A, B, C, D and X. Write the equations of the reactions involved. |
Answer» SOLUTION : 
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| 11574. |
When chlorobenzene is heated with conc. NaOH solution at about 575 K under high pressure, the product is : |
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Answer» <P>PHENOL 
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| 11575. |
To which of the followingclass of organic compounds soap belongs |
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Answer» ESTERS |
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| 11576. |
what are thesalientfeature of Crystal FieldTheory(CFT) ? |
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Answer» Solution :Bethe and vanVleckdeveloped Crystal FieldTheory (CFT) TOEXPLAIN variousproperties of coordinationcompounds . Thesalientfeatureof CFT are as follows: (1) In a complex , the CENTRAL metal atom or ions is surrounded by various ligands whichareeithernegativelycharged ions `(F^(-),Cl^(-), CN^(-))` etc.ornetural molecules `(H_(2)O,NH_(3))` en etc.) and the mostelectronegativeatom in themor ionand thesurroundingligandsact as pointtowardscentralmetal ions . (2) The metalatom or ions and the surroundingligandsact as pointcharges involvingpurely electrosn attractionbetweenthem. (3) (i)Thecentralmetal ion has five ( n -1) d degenerateorbitalsnamely `d_(xy),d_(yz),d_((x^(2)-y^(2)) and d_(z^(2))` (II)Whenthe ligandsapproch themetalion, dueto repulsiveforces, the degeneracyof d-orbitalsis destroyedand they split into two groups of differentenergy `t_(2g)` and `e_(g)`orbitals. This effect is calledcrystal filed splittingwhichdependsuponthe geometryof thecomplex. (iii)The d-oribtalslying in thedirections of ligands are affected to agereater extentwhile thoselyingin betweenthe ligandsare affectedto a lessextent. (iv) Due torepulsion, the orbitalsalongthe axesof ligands acquirehigherenergywhilethoselying in betweentheligands acquirelessenergy . (v) Hencerepulsion by ligands GIVE tow sets of split up orbitalsofmetal ion withdifferentenergies.  (vi)Theenergy differencebetweentwo sets of d-orbitalsaftersplittingby ligands is calledcrystal field splittingenergy(CFSE) andrespesented by `Delta_(0)`or by arbitrary term`10D_(q)`. Thevalue of `Delta` or `10Delta_(q)`dependsupon thegeometryof the complex.   (4) Theelectrons of metalionoccupythe split d-orbitalsaccordingto Hund's ruleaufbauprincipleand thoseorbitalswithminimum repulsionand the firsthestawaysform theligands. (5) CFTdoes notaccountfor overlappingof orbitals of central metal ion and ligands , hencedoes notconsidercovalentnatureof thecomplex. (6) From thecrystal fieldstability ENERGY , the stability of thecomplexcan beknown. |
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| 11577. |
The reagents which cannot be used to distinguish benzyl chloride from chlorobenzene are |
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Answer» `Br_(2)//C Cl_(4)`<BR>Shaking with an aqueous solution of `AgNO_(3)` |
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| 11578. |
What will be the number of stereoisomers for complex [Ma_(2)b_(2)c_(2)]^(nt), if its two different monodentate ligends are replaced with one symmetrical bidentate ligand |
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Answer» 3 |
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| 11579. |
to dissolve 0.9 gm metal 100 ml of 1 N HCl is used. What is the equivalent wt. of metal |
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Answer» 7 `0.9/Exx1000=100xx1 IMPLIES E=9` |
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| 11580. |
Which of thefollowingfirstis morebasicthansecond ? |
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Answer»
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| 11581. |
What is the basic difference between the electronic configurations of transition and inner-transition elements? |
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Answer» Solution :GENERAL electronic CONFIGURATION of TRANSITION elements: [Noble gas ]`(n-1)d^(1-10) ns^(1-2)` General electronic configuration of inner-transition elements: `(n-2)f^(1-14)(n-1) d^(0-1) ns^(0-2)` Thus, in transition elements, last electron enters in (n-1) d- ORBITAL while in inner-transition elements, it enters in (n-2)f- orbital. |
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| 11582. |
What is the difference between multi-molecular and macro molecular colloids ? Give one example of each type. How are associated colloids different from these two types of colloids ? |
| Answer» | |
| 11583. |
The standard reduction potential at 25^@C of the reaction 2H_2O+2e^(-) leftrightarrow H_2 +2OH^(-) is -0.8277 volt. Calculate the equilibrium constant for the reaction 2H_2O=H_3O^+ +OH^(-) at 25^@C |
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Answer» Solution :Let the equilibrium constants for the following equations be as follows: `H_2O+H_2O LEFTRIGHTARROW H_3O^+ +OH^(-) K_1` .......(1) `H_3O^+ +e leftrightarrow 1/2H_2+H_2O K_2`......(2) Adding them we get `H_2O+ e leftrightarrow 1/2H_2 +OH^(-) K_3`........(3) We have now to calculate `K_1` Now for EQN(2) , `E_(H_3O^+)^@=0` `therefore K_2=1(because E^@=0.0591/1log K)`......(Eqn 2) For eqn. (2) `E^@=-0.8277` volt (given) `therefore E^@=0.0591/1 LOG K_3=-0.8277` `logK_3=-0.8277/0.0591, K_3=1 times 10^-14` Further as eqn (3) is the sum of eqns. (1) and (2) we have `K_3=K_1.K_2=K_1` `therefore K_1=1 times 10^-14` |
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| 11584. |
Write the name of different types of detergents. |
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Answer» SOLUTION :(1) ANIONIC DETERGENTS (2)Cationic detergents (3) Amphoteric detergents (4) Non-ionic detergents |
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| 11585. |
When a current of 0.75 A is passed through CuSO_(4) solution of 25 min, 0.369g of copper is deposited at the cathode. Calculate the atomic mass of copper. |
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Answer» SOLUTION :QUANTITY of electricity passed `=(0.75A)(25xx60s)=1125C` `Cu^(2+) to 2E^(-)toCu` Thus, 1 mole of Cu (i.e., gram atomic mass) is DEPOSITED by 2F, i.e, `2xx96500C` Now, `1125` C deposit Cu=0.369 g `2xx96500C` will deposit Cu=`(0.369)/(1125)xx2xx96500g=63.3g` hence, atomic mass of copper=63.3 |
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| 11586. |
Write IUPAC name of [PtBrCl(NO_(2))NH_(3)]^(-)or [PtCl(NH_(2)CH_(3))(NH_(3))_(2)]Clor [CoCl(NO_(2))(NH_(3))_(4)]Cl |
| Answer» SOLUTION :amminebromidochloridonitroplatinate (II), diamminechloridomethylamineplatinum (II) CHLORIDE tetraamminechloridonitrocobalt (III)chloride. | |
| 11587. |
Volume strength of a H_(2)O_(2) solution is 2.8. Normality of the solution is: |
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Answer» `0.1` `rArr N= (2.8)/(5.6)= (1)/(2)= 0.5` |
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| 11588. |
What is the product, when 1^(@) and 2^(@) alcohol is react with phosphorus Tribromide (PBr_(3)) ? |
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Answer» Bromo alcohol |
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| 11589. |
[T_i (H_2O)_6]^(3+ )is coloured ,while [Sc(H_2O)_6]^(3+) is colourless -explain |
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Answer» Solution :`Ti" in "[Ti(H_(2)O)^(6)]^(3+)" is in +3 oxidation state"` `Sc" in "[Sc(H_(2)O)_(6)]^(3+)" is in +3 oxidation state."` The outer electronic configuration of Sc, Ti and their trivalent ions are. `{:(SC,:,3D^(1)4S^(2)""Ti,:,3d^(2)4S^(2)),(SC^(3+),:,3d^(0)""Ti^(3+),:,3d^(1)):}` `Ti^(3+)` has one unpaired electron in 3d orbital and they undergoes d - d transition. This electron can be promoted to a higher ENERGY level by LIGHT absorption. Therefore `[Ti(H_(2)O)^(6)]^(3+)` is COLOURED. In the case of `[Sc(H_(2)O)_(6)]^(3+)`, there is no electron in 3d orbital of `Sc^(3+)`, hence there is no possibility of light absorbance. Therefore `[Sc(H_(2)O)_(6)]^(3+)` is colourless. |
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| 11590. |
Which of the following undergo SN reaction at a fastest rate than others |
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Answer» `CH_3CH_2-O-CH_2Cl` |
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| 11591. |
Write the reactions that take place during the manufacture of nitric acid by Ostwald's process. |
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Answer» SOLUTION :`4NH_(3)+5O_(2)overset("Pt/Rh")UNDERSET("500 K, 9 bar")rarr 4NO+6H_(2)O` `2NO+O_(2)hArr 2NO_(2)` `3NO_(2)+H_(2)Orarr 2HNO_(3)+NO` |
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| 11592. |
What is applied on wounds for making them free from microorganisms? |
| Answer» Solution :Tincture iodine | |
| 11593. |
Which of the following hydroxide is acidic? |
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Answer» `AL(OH)_(3)` `H_(3)BO_(3) or B(OH)_(3)+OH^(-) to B(OH)_(4)^(-)` |
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| 11594. |
Which is least soluble in water: |
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Answer» AgCl |
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| 11595. |
Which of the following species is not impected to be a ligand? |
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Answer» NO |
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| 11596. |
What is the main difference in the structure of natural rubber and gutta-percha ? |
| Answer» Solution :Both are POLYMERS of ISOPRENE. Whereas NATURAL rubber is cis (polyisoprene) while percha is TRANS (polyisoprene). | |
| 11597. |
Which is known as Philospher.s wool ? |
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Answer» HgO |
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| 11598. |
Which of the following statement about galvanic cell is incorrect |
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Answer» Anode is positive |
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| 11599. |
The vapour pressure of a 5% solution of a nonvolatile organic substance in waterat 373 K is 0.9935 xx 10^5Nm^(-1).Calculate the molecular mass of the solute (1 atm= 1.0132 xx 10^5 Nm^(-2)) |
| Answer» SOLUTION :0.0458 kg/mol | |
| 11600. |
The second law of thermodynamic introduced the concept of: |
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Answer» THIRD LAW of thermodynamics |
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