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 Multiple Choice QuestionsMultiple Choice Questions

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11.

Amount of oxalic acid present in a solution can be determined by its titration with KMnO4 solution in the presence of H2SO4. The titration gives unsatisfactory result when carried out in the presence of HCl, because HCl

  • gets oxidised by oxalic acid to chlorine

  • furnishes H+ions in addition to those from oxalic acid

  • reduces permanganate to Mn2+

  • reduces permanganate to Mn2+

313 Views

12.

The correct decreasing order of priority for the functional groups of organic compounds in the IUPAC system of nomenclature is

  • −COOH, −SO3H, −CONH2, −CHO

  • −SO3H, −COOH, −CONH2, −CHO

  • −CHO, −COOH, −SO3H, −CONH2

  • −CHO, −COOH, −SO3H, −CONH2

188 Views

13.

CH3Br + Nu- → CH3 -Nu +Br-
The decreasing order of the rate of the above reaction with nucleophiles (Nu–) A to D is
[Nu–=
(A) PhO
(B) AcO
(C) HO
(D) CH3O
]

  • D > C > A > B

  • D > C > B > A

  • A > B > C > D

  • A > B > C > D

379 Views

14.

The increasing order of stability of the following free radicals is

  • left parenthesis CH subscript 3 right parenthesis subscript 2 space straight C with • on top straight H space less than thin space left parenthesis CH subscript 3 right parenthesis subscript 3 straight C with • on top space less than thin space left parenthesis straight C subscript 6 straight H subscript 5 right parenthesis subscript 2 straight C with • on top straight H space less than thin space left parenthesis straight C subscript 6 straight H subscript 5 right parenthesis subscript 3 straight C with • on top
  • left parenthesis straight C subscript 6 straight H subscript 5 right parenthesis subscript 3 straight C with • on top space less than thin space left parenthesis straight C subscript 6 straight H subscript 5 right parenthesis subscript 2 straight C with • on top straight H space less than space left parenthesis CH subscript 3 right parenthesis subscript 3 straight C with • on top space less than thin space left parenthesis CH subscript 3 right parenthesis subscript 2 space straight C with • on top straight H
  • left parenthesis straight C subscript 6 straight H subscript 5 right parenthesis subscript 2 straight C with • on top straight H space less than space left parenthesis CH subscript 3 right parenthesis subscript 3 straight C with • on top space less than thin space left parenthesis CH subscript 3 right parenthesis subscript 3 straight C with • on top space less than space left parenthesis straight C subscript 6 straight H subscript 5 right parenthesis subscript 2 straight C with • on top straight H
  • left parenthesis straight C subscript 6 straight H subscript 5 right parenthesis subscript 2 straight C with • on top straight H space less than space left parenthesis CH subscript 3 right parenthesis subscript 3 straight C with • on top space less than thin space left parenthesis CH subscript 3 right parenthesis subscript 3 straight C with • on top space less than space left parenthesis straight C subscript 6 straight H subscript 5 right parenthesis subscript 2 straight C with • on top straight H
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15.

Due to the presence of an unpaired electron, free radicals are:

  • Chemically reactive

  • Chemically inactive

  • Anions

  • Anions

192 Views

16.

The decreasing order of nucleophilicity among the nucleophiles

  • (a), (b), (c), (d)

  • (d), (c), (b), (a)

  • (b), (c), (a), (d)

  • (b), (c), (a), (d)

322 Views

17.

Of the five isomeric hexanes, the isomer which can give two monochlorinated compounds is

  • n-hexane

  • 2, 3-dimethylbutane

  • 2,2-dimethylbutane

  • 2,2-dimethylbutane

402 Views

18.

An organic compound having molecular mass 60 is found to contain C = 20%, H = 6.67% and N = 46.67% while rest is oxygen. On heating, it gives NH3 along with a solid residue. The solid residue give violet colour with alkaline copper sulphate solution. The compound is

  • CH3NCO

  • CH3CONH2

  • (NH2)2CO

  • (NH2)2CO

298 Views

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19.

The ammonia evolved from the treatment of 0.30 g of an organic compound for the estimation of nitrogen was passed in 100 mL of 0.1 M sulphuric acid. The excess of acid required 20 mL of 0.5 M sodium hydroxide solution hydroxide solutio for complete neutralization. The organic compound is

  • acetamide

  • thiourea

  • urea

  • urea

444 Views

20.

When metal  ‘M’ is treated with NaOH, a white gelatinous precipitate ‘X’ is obtained, which is soluble in excess of NaOH. Compound ‘X’ when heated strongly gives an oxide which is used in chromatography as an adsorbent. The metal ‘M’ is

  • Fe

  • Zn

  • Ca

  • Al


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