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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.
| 88351. |
A narrow spectrum antibiotic is active against _______ |
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Answer» gram positive or gram negative BACTERIA |
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| 88352. |
Define narrow spectrum antibiotics. |
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Answer» GRAM positive or gram NEGATIVE BACTERIA |
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| 88353. |
A narrow spectrum antibiotic is active against |
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Answer» gram POSITIVE or gram negative bacteria |
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| 88354. |
A narrow spectrum antibiotic is active against.. |
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Answer» gram positive or gram NEGATIVE bacteria. |
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| 88355. |
A narrow spectrum antibiotic is active against _________ |
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Answer» gram POSITIVE or gram NEGATIVE bacteria. |
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| 88356. |
A narrow spectrum antibiotic is active against ............ |
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Answer» GRAM positive or gram NEGATIVE bacteria |
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| 88357. |
A nanopeptide contains ……...peptide linkages. |
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Answer» A) 10 |
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| 88358. |
A nanopeptide contains ...........peptide linkages |
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Answer» 10 |
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| 88359. |
A nanopeptide contains how many peptide linkages ? |
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Answer» 10 |
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| 88360. |
A nanopeptide contain how many peptide bond |
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Answer» 7 |
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| 88361. |
A + NaNO_2 to N_2O + NaCl + 2H_2O in this reaction A can be |
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Answer» `H_2SO_4`.DIL |
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| 88362. |
A : NaN_(3) and Na_(3) N both are stable. R : Na when reacted with atmospheric nitrogen at different temperature forms. Stable NaN_(3) and Na_(3)N |
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Answer» If both ASSERTION & Reason are TRUE and the reason is the correct explanantion of the assertion, then mark (1). |
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| 88363. |
(a) Name three oxy-acids of phosphorus and draw their structures. (b) Discuss the anomalous behaviour of oxygen. |
Answer» Solution :(i) Hypophosphorus, (ii) Phosphorus acid, (m) Phosphoric acid :  (b) Anomalous behaviour of oxygen is due to : (b) It has small SIZE. (ii) It has high ionisation ENERGY. (iii) It has no vacant d-orbitals in its valence shell. Also the important points of difference are : (i) Oxygen exists as diatomic molecules whereas other elements are POLYATOMIC. (ii) Oxygen is gas whereas other members are solids at room temperature. (iii) Max. O.S. of O is + 2 witereas other elements show max. O.S. of +6. (iv) Oxygen is PARAMAGNETIC whereas other. elements are diamagnetic. |
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| 88364. |
a) Name the water insoluble component of starch. |
| Answer» SOLUTION :Amylopectin water INSOLUBLE COMPONENT of STARCH. | |
| 88365. |
(a) Name the starting material used in the industrial preparation of phenol. (b) Wriite complete reaction for the bromination of phenol in aqueous and non aqueous medium. (c) Explain why Lewis acid is not required in bromination of phenol. |
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Answer» Solution :(a) Cumene. <BR> (b) N/A (c) In bromination of benzene, Lewis acid is used to polarize `Br_(2)` to form the reactive electrophile, `Br^(+)`. In case of phenol, Lewis aciid is not REQUIRED because of O-atom of phenol itself polarizes the `Br_(2)` MOLECULE to form `Br^(+)` ions. further, +R-effect of OH group makes phenol highly activated towards electrophilic substitution reactions. |
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| 88366. |
(a) Name the reducing agent used in the extraction of zinc oxide. Give the equation. (b) What is the principle involved in zone refining of metals? |
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Answer» Solution :(a) Coke (carbon) `ZnO+CrarrZn+CO` (B) The IMPURITIES are more SOLUBLE in the MELT than in the solid STATE of the metal. |
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| 88367. |
a) Name the products X and Y? |
Answer» SOLUTION :Amides on heating with bromine and AQ NaOH gives `1^(@)`AMINES. This reaction is called Haffman.s bromamide reaction FOLLOWED by acetylation gives N-phenylethanamide. 
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| 88368. |
Name the phenomenon/process involved i) mixing of hydrated ferric oxide (+ve sol) and arsenious sulphide (-ve sol) ii) An impure sol is purified by removing dissolved particles using suitable membrane iii) Movement of dispersion medium is observed in an electric field. |
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Answer» Solution :i) mixing of hydrated ferric oxide (`+ve` SOL) and arsenious sulphide (`-ve` sol) is known as Mutual Cougulation. ii) An impure sol is purified by removing dissolved particles using suitable membrane is CALLED as DIALYSIS III) Movement of dispersion medium is observed in an electric field is called electro OSMOSIS. |
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| 88369. |
(a) Name the monomers used in the preparation of Nylon 6,6. (b) Explain Vulcanization of rubber. c) Give an example for biodegradable polymer. |
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Answer» Solution : NATURAL rubber BECOMES soft at high temperature and brittle at low temperature and show high water absorption capacity . In order to improve these physical properties a process of VULCANISATION ID CARRIED out . `Nylon- 2- Nylon - 6` . |
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| 88370. |
(a) Name the monomers of Nylon - 6,6.(b) How is Neoprene prepared? Given equation. (c)Give an example for thermoplastic polymer. |
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Answer» Solution :(a) Hexamethylene DIAMINE and adipic acid. (b) NEOPRENE is formed by the free radical POLYMERIZATION of CHLOROPRENE.  . (c ) Polystyrene,polyethylene. |
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| 88371. |
Name the monomers present in the following polymers. PVC |
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Answer» Solution :(a) (i) Vinylchloride or CHLOROETHENE (ii) Chloropene or 2-chloro 1, 3- butadiene (iii) Caprolactam (b) The process of heating a mixture of natural rubber with sulphur (and a appropriate additive), to in the physical property of rubber is CALLED Vulcanisation of rubber. (a) (i) Monomers of PVC is VINYL chloride `(CH_(2) = CH.Cl)`. (ii) Monomer of neoprene is chloroprene `(CH_(2)=CH-underset(Cl)underset(|)(C)=CH_(2))` (iii) Monomer of nylon -6 is caprolactum  (b) Vulcanisation of rubber : Natural rubber is soft and sticky. It stensile strength and elasticity is low. In order to give strength and elastisity, natural rubber is vulcanised to improve properties. The process is called vulcanisation of rubber. In this process, rubber is treated with sulphur or any suitable sulphur compound to modify its properties. The process gives cross-linkage, which provides mechanical strength and proper elasticity to the rubber. 
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| 88372. |
(a) Name the method of refining of nickel. (b) What is the role of cryolite in the extraction of aluminium ? (c ) What is the role of limestone in the extraction of iron from its oxides ? |
| Answer» Solution :(a) Mond's process (b) Lowers the melting point of alumina and makes it a good CONDUCTOR, (C ) It provides CaO (basci flux) upon HEATING which combines with ACIDIC impurities to FORM slag. | |
| 88373. |
(a) Name the halogen hydracid with highest bond energy. (b) When HI solution is kept exposed to air, it turns brown after a few days. Why ? |
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Answer» SOLUTION :(a) The hydracid (HF) having the smallest halogen (i.e., F) has the hightest bond energy. (b) Because of the LARGEST size of halogen, the H-I bond is the weakest. When exposed to air, it decomposes to liberate `I_(2)` which gives brown COLOUR to the solution. `4HI + underset(("Air"))(O_(2)) rarr 2 H_(2)O + underset("Brown")(2I_(2))` |
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| 88374. |
(a) Name one important ore of aluminium. Give its chemical composition. (b) How is copper extracted from low grade ore ? |
| Answer» SOLUTION :(a) BAUXITE, `(Al_(2)O,xH_(2)O)`. | |
| 88375. |
(a) Na_(2)B_(4)O_(7)+concentrated H_(2)SO_(4)+H_(2)O to underset((ii)"ignite")overset((i)C_(2)H_(5)OH)to (B) (B) is identify by the characteristic colour of the flame. Identify (A) and (B). (b) Complete the following reaction and identify the products formed.Na_(2)B_(4)O_(7) |
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Answer» Solution :(a) `Na_(2)B_(4)O_(7)+`concentrated `H_(2)SO_(4)+5H_(2)O to Na_(2)SO_(4)+4H_(3)BO_(3)` `H_(3)BO_(3)+3C_(2)H_(5)OH to B(OC_(2)H_(5))_(3)` -voltale (burn with green edged flame) + `3H_(2)O` (b) `Na_(2)B_(4)O_(7) underset((NaBO_(2))oversetDeltato(##RES_INO_CHM_XI_C06_S01_004_S01.png" width="80%"> |
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| 88376. |
A : Na and Li are stored under kerosene. R : Na and Li are soluble in kerosene. |
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Answer» If both Assertion & Reason are true and the reason is the correct explanantion of the assertion, then MARK (1). |
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| 88377. |
A : Na_(2) S_(2) O_3 on reaction with l_(2) gives Na_(2) S_(4) O_(6) . R : This reaction involves colour and electronic change both . |
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Answer» If both ASSERTION & Reason are TRUE and the reason is the correct explanation of the assertion , then mark (1) |
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| 88378. |
(A) N_(3)H exsist but not P_(3)H (R ) Nitrogen can form effective (P-d) pi bonds where as phosphorus cannot form such bonds |
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Answer» Both (A) and (R ) are TRUE and (R ) is the correct EXPLANATION of (A) |
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| 88379. |
(A) N_(2)O_(5) is more acidic than N_(2)O_(4) (R ) N_(2)O_(4) is a mixed anhydride where as N_(2)O_(5) is not |
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Answer» Both (A) and (R ) are true and (R ) is the correct explanation of (A) |
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| 88380. |
(A) N_(2)O is called laughing gas (R ) N_(2)O causes hysterical laugh. |
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Answer» Both (A) and (R ) are TRUE and (R ) is the correct explanation of (A) |
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| 88381. |
A^(+n) are arrangement on the corners of a cube. B^(-m) are occupied by all the octahedral voids and half of the tetrahedral voids. The formula of the compound is: |
| Answer» ANSWER :D | |
| 88382. |
(A) n-propyl alcohol is prepared from propene by the action of B_2H_6followed by oxidation with H_2O_2in basic medium(R) In Hydroboration oxidation reaction, rearranged products are not formed because it involves the formation of cyclic transition state |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT explanation of (A) |
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| 88383. |
A moving electron has 4.9 xx 10^(-25) joules of kinetic energy. Find out its de - Broglie wavelength (Given h = 6.626 xx 10^(-34) Js, m_e = 9.1 xx 10^(-31)kg). |
| Answer» SOLUTION :`7 XX 10^(-7) m` | |
| 88384. |
A moving electron has 4.55 xx 10^(-25) joules of kinetic energy. Calculate its wavelength (mass = 9.1 xx 10^(-31) kg and h = 6.626 xx 10^(-34) kg m^2 s^(-1) ). |
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Answer» Solution :Here we given kinetic ENERGY `1/2 mv^2 = 4.55 xx 10^(-25) J` `m = 9.1 xx 10^(-31) KG ` `h = 6.626 xx 10^(-34) kg m^2 s^(-1)` ` therefore 1/2 xx (9.1 xx 10^(-31) ) v^2 = 4.55 xx 10^(-25)` ` v^2= (4.55 xx 10^(-25) xx 2)/(9.1 xx 10^(-31)) = 10^6` ` v = 10^3 ms^(-1)` ` therefore lamda = (h)/(mv) = (6.626 xx 10^(-34))/((9.1 xx 10^(-31) ) xx 10^3)` ` = 7.25 xx 10^(-7) m` |
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| 88385. |
A moth repellent has the composition 49% C, 2.7%H and 48.3% Cl. Its molecular weight is 147 gm. Determine its molecular formula |
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Answer» |
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| 88386. |
(A) Most of the molecular reactions occur with moderate rates (R ) Molecular reactions involve bond rearrange-ments |
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Answer» Both (A) and (R ) are true and (R ) is the correct EXPLANATION of (A) |
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| 88387. |
(A): Morphine is an example for narcotic analgesic. (R) : Narcotic drugs have no addictive properties but is limited to mild aches and pains. |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT explanation of (A) |
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| 88388. |
A monotropic acid in 1.00 M solution is 0.01%ionised. What is the dissociation constant of the acid ? |
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Answer» `1.0xx10^(-3)` |
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| 88389. |
A monoprotic acid in a 0.1 M solution ionizes to 0.001%. Its ionisation constant is |
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Answer» `1.0 xx 10^(-3)` `HA hArr H^(+) + A^(-)` Ionisation constant = ? `alpha= 0.001 % = (0.001)/(100) = 10^(-5) RARR K = (alpha^(2))/(V) = ([10^(-5)]^(2))/(10) = 10^(-11)`. |
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| 88390. |
A monomer of a polymer on ozonolysis gives two moles of CH_2Oand one mol of CH_3- overset(O)overset(||)C-CHO. Write the structure of monomer and polymer and each step of reaction.Structure of monomer : CH_2O + O = overset(CH_3)overset(|)C-OCH_2 to CH_2= overset(CH_3)overset(|)C-CH=CH_2 Structure of polymer : |
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Answer» SOLUTION : STRUCTURE of monomer : `CH_2O + O = overset(CH_3)overset(|)C-OCH_2 to CH_2= overset(CH_3)overset(|)C-CH=CH_2` Structure of polymer : `[-CH_2-UNDERSET(H_3C)underset(|)C=underset(H)underset(|)C-CH_2-]_n` |
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| 88391. |
A monoprotic acid in 1.00 M solution is 0.01% ionized. The dissociation constant of this acid is |
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Answer» `1 xx 10^(-8)` `= 1 xx 10^(-8)`. |
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| 88392. |
A monobasic weak acid solution which is 0.002 M has pH value equal to 5, The percentage ionization value of the acid in the solution will be |
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Answer» 0.5 |
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| 88393. |
A monobasic weak acid solution has a molarity of 0.005 and pH of 5. What is the percentage ionization in this solution ? |
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Answer» `2.0` `K_(a)=([H^(+)][A^(-)])/([HA]), because [H^(+)]=10^(-pH)` `therefore [H^(+)]=10^(-5)`, and at equilibrium `[H^(+)]=[A^(-)]` `therefore K_(a)=(10^(-5)xx10^(-5))/(0.0015)=2xx10^(-8)` `alpha= sqrt((K_(a))/(005))=sqrt(4xx10^(-6))=2xx10^(-3)` PERCENTAGE IONIZATION = 0.2 |
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| 88394. |
A monobasic weak acid solution has a molarity of 0.005 and pH of 5. What is its percentage ionization in this solution |
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Answer» `2.0` `K_(a) = ([H^(+)][A^(-)])/([HA]) , pH = -log[H^(+)]` `[H^(+)] = 10^(-5) , [H^(+)] = [A^(-)]` `K_(a) = (10^(-5) xx 10^(-5))/(0.005) = 2 xx 10^(-8)` `ALPHA = sqrt((K_(a))/(C)) = 2 xx 10^(-3)` Percentage ionization = 0.2 |
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| 88395. |
A monobasic acid of phosphorus, which reduces HgCl_2 to black Hg is |
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Answer» hypophosphorus acid `H_(3)PO_(2)+4HgCl_(2)+2H_(2)O to 2Hg_(2)Cl_(2)+4HCl+H_(3)PO_(4)` `2HgCl_(2)+H_(3)PO_(2)+2H_(2)O to 4Hg+4HCl +H_(3)PO_(4)` |
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| 88396. |
A monoatomic ideal gas undergoes a process in which the ratio of P to V at any instant is constant and equal to unity. The molar heat capacity of gas is: |
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Answer» 1.5R |
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| 88397. |
A monoatomic gas lighter than air is |
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Answer» helium |
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| 88398. |
A monoacid organic base gave the following results on analysis 0.40g of the platinichlorie left 0.125g of Pt Calculate the molecular formula of the base |
| Answer» Solution :`(C_(7)H_(9)N)` | |
| 88399. |
A monoacid organic base gave the following data on analysis (a) 0.2790g of the base gave 0.7920g of CO_(2) and 0.1890g of H_(2)O (b) 0.1163g of the base gave 14mL of dry nitrogen at NTP (c ) 0.2980g of the platinichloride left 0.0975g of Pt Calculate the molecular formula of the base. |
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Answer» Solution :MOLES of `C=1 xx` moles of `CO_(2)= (0.792)/(44)= 0.018` Moles of `H= 2XX` moles of `H_(2)O =2 xx (0.1890)/(18)= 0.021` Moles of `N=2 xx` moles of `N_(2)=2 xx (14)/(22400)= 0.00125` Moles of N in 0.2790g of the base `=(0.00125)/(0.1163)xx 0.2790` = 0.003 `therefore` moles of `C: H: N= 0.018: 0.021: 0.003` `=18:21:3` `=6: 7:1` Hence, the empirical formula is `C_(6)H_(7)N or C_(6)H_(5)NH_(2)` Suppose the monoacid base is B `{:(B,+,H_(2)PtCl_(6),rarr,B_(2)H_(2)PtCl_(6),rarr,Pt),("Base",,"Acid",,"Chloroplatinate",,),(("Monoacid"),,,,("Platinichloride"),,),(,,,,0.2980g,,0.0975g):}` Applying POAC for Pt atoms: `1xx` moles of `B_(2)H_(2)PtCl_(6)`= moles of Pt in product `(0.2980)/("mol. wt of" B_(2)H_(2)PtCl_(6))=(0.0975)/(195)` `therefore` mol. wt of `B_(2)H_(2)PtCl_(6)=596` Molecular weight of B `=("mol. wt. of" B_(2)H_(2)PtCl_(6)-"mol. wt of" H_(2)PtCl_(6))/(2)= (596-410)/(2)= 93` Since empirical formula weight is also 93, therefore, the molecular formula of the base is `C_(6)H_(5)NH_(2)` |
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| 88400. |
(A): Mond's process is used to refine nickel metal (R) : With carbonmonoxide, nickel forms a neutral complex nickeltetracarbonyl. |
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Answer» Both A & R are TRUE, R is the correct explanation of A |
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