Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

 Multiple Choice QuestionsMultiple Choice Questions

1.

Four successive members of the first-row transition elements are listed below with atomic numbers. Which one of them is expected to have the highest straight E subscript straight M to the power of 3 plus end exponent divided by straight M to the power of 2 plus end exponent end subscript superscript straight o  value?

  • Cr (Z =24)

  • Mn(Z =25)

  • Fe (Z = 26)

  • Co (Z= 27)


D.

Co (Z= 27)

530 Views

2.

Given, 

straight E subscript Cr to the power of 3 plus end exponent divided by Cr end subscript superscript straight o space equals space minus space 0.74 space straight V semicolon
straight E subscript MnO subscript 4 superscript minus divided by Mn to the power of 2 plus end exponent end subscript superscript straight o space equals space 1.51 space straight V
straight E subscript Cr subscript 2 straight O subscript 7 superscript 2 minus end superscript divided by Cr to the power of 3 plus end exponent end subscript superscript straight o space equals space 1.33 space straight V semicolon
straight E subscript Cl divided by Cl to the power of negative 1 end exponent end subscript superscript straight o space equals space 1.36 space straight V
Based on the data given above, strongest oxidising agent will be

  • Cl

  • Cr3+

  • Mn2+

  • MnO4-


D.

MnO4-

Higher the SRP, better is an oxidising agent, among the givenstraight E subscript MnO subscript 4 superscript minus divided by Mn to the power of 2 plus end exponent end subscript superscript straight o is highest, hence, MnO4- is a strongest oxidising agent.

265 Views

3.

Two Faraday of electricity is passed through a solution of CuSO4. The mass of copper deposited at the cathode is: (at. mass of Cu = 63.5 amu)

  • 0 g

  • 63.5 g

  • 2 g

  • 127 g


B.

63.5 g

Atomic mass of Cu = 63.5 u
Valency of the metal Z= 2
We have,
CuSO4 → Cu2+ + SO42-
Cu2+ + 2e-  →  Cu

1 mol   2 mol       1 mol = 63.5 g
670 Views

4.

How many litres of water must be added to 1 L to an aqueous solution of HCl with a pH of 1 create an aqueous solution with PH of 2?

  • 0.1 L

  • 0.9 L 

  • 2.0 L 

  • 9.0 L


D.

9.0 L

Initial pH = 1, i.e. [H+] = 0.1 mole/litre
New pH = 2, i.e. [H+] = 0.01 mole/litre
In case of dilution: M1V1 = M2V2
0.1 ×1 =0.01 × V2
V2 = 10 litre.
A volume of water added = 9 litres.

201 Views

5.

The degree of dissociation (α) of a weak electrolyte ,AxBy is related to van't Hoff factor (i) by the expression

  • straight alpha space equals space fraction numerator 1 minus 1 over denominator left parenthesis straight x plus straight y minus 1 right parenthesis end fraction
  • straight alpha space equals space fraction numerator 1 minus 1 over denominator left parenthesis straight x plus straight y plus 1 right parenthesis end fraction
  • straight alpha space equals space fraction numerator straight x plus straight y minus 1 over denominator straight i minus 1 end fraction
  • straight alpha space equals space fraction numerator straight x plus straight y plus 1 over denominator straight i minus 1 end fraction

A.

straight alpha space equals space fraction numerator 1 minus 1 over denominator left parenthesis straight x plus straight y minus 1 right parenthesis end fraction
straight A subscript straight x straight B subscript straight y space rightwards harpoon over leftwards harpoon space space xA to the power of straight y plus end exponent space plus space yB to the power of straight x minus end exponent
left parenthesis 1 minus straight alpha right parenthesis space space space space space xα space space space space space space space space space space yα
straight i space equals space straight n space left parenthesis straight A subscript straight x straight B subscript straight y right parenthesis space plus space straight n left parenthesis straight A to the power of straight y plus end exponent right parenthesis plus straight n left parenthesis straight B to the power of straight x minus end exponent right parenthesis
space equals space 1 minus straight alpha space plus xα space plus yα
space equals 1 space plus straight alpha space left parenthesis straight x plus straight y minus 1 right parenthesis
straight alpha space equals space fraction numerator straight i minus 1 over denominator left parenthesis straight x plus straight y minus 1 right parenthesis end fraction
665 Views

6.

Galvanization is applying a coating of:

  • Cr

  • Cu

  • Zn

  • Pb


C.

Zn

Zinc metal is the most stable metal to cover iron surfaces. The process of coating the iron surface by zinc is called galvanization.

267 Views

7.

Which among the following metals is employed to provide cathodic protection to iron?

  • Zinc

  • Nickel

  • Tin

  • Lead


A.

Zinc

Cathodic protection of iron involves using another more reactive metal or a sacrificial anode. Since, in electrochemical series, Zn is placed above Fe, so it is used for such action.


8.

The reduction potential of hydrogen half-cell will be negative if 

  • p(H2) = 1 atm and [H+] = 2.0 M

  • p(H2) = 1 atm and [H+] = 1.0 M

  • p(H2) = 2 atm and [H+] = 1.0 M

  • p(H2) = 2 atm and [H+] = 2.0 M


C.

p(H2) = 2 atm and [H+] = 1.0 M

2 straight H to the power of plus left parenthesis aq right parenthesis space plus space 2 straight e to the power of minus space rightwards arrow with space on top space straight H subscript 2 space left parenthesis straight g right parenthesis
straight E subscript red space equals space straight E subscript red superscript 0 minus fraction numerator 0.0591 over denominator straight n end fraction space log space fraction numerator straight P subscript straight H subscript 2 end subscript over denominator left parenthesis straight H to the power of plus right parenthesis squared end fraction
straight E subscript red space equals space 0 space minus fraction numerator 0.0591 over denominator 2 end fraction space log space fraction numerator 2 over denominator left parenthesis 1 right parenthesis squared end fraction
straight E subscript red space equals space minus fraction numerator 0.0591 over denominator 2 end fraction space log space 2
Thus correct answer is c.
429 Views

9.

The equivalent conductance of NaCl at concentration C at infinite dilution are λC and λ respectively. The correct relationship between λC and λ  is given as (where the constant B is positive)

  • λC = λ +(B)C

  • λC = λ -(B)C

  • λC = λ -(B)square root of straight C

  • λC = λ +(B)square root of straight C


C.

λC = λ -(B)square root of straight C

343 Views

10.

Given below are the half-cell reactions
Mn2+ + 2e- → Mn; Eo = - 1.18 eV
2(Mn3+ + e- →Mn2+); Eo = +1.51 eV
The Eo for 3Mn2+ → Mn + 2Mn3+ will be

  • -2.69 V; the reaction will not occur

  • -2.69 V; the reaction will occure

  • -0.33 V; the reaction will not occur

  • -0.33 V; the reaction will occur


A.

-2.69 V; the reaction will not occur

Standard element potential of reaction [ Eo] can be calculated as
Eocell = ER-EP
where ER = SRP of reactant
EP = SRP of product
If Eocell = +ve, then the reaction is spontaneous otherwise non-spontaneous.
Mn to the power of 3 plus end exponent space rightwards arrow with straight E subscript 1 superscript straight o space equals space 1.51 space straight V on top space Mn to the power of 2 plus end exponent
Mn to the power of 2 plus end exponent space rightwards arrow with straight E subscript 2 superscript straight o space equals space minus space 1.18 space straight V on top space Mn
therefore, For Mn2+ disproportionation
Eo = - 1.51 V - 1.18 V = - 2.69 V < 0

568 Views