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 Multiple Choice QuestionsMultiple Choice Questions

401.

If the molar conductance values of Ca2+ and Cl- at infinite dilution are respectively 118.88 X 10-4 m2 Ωho mol-1 and 77.33 X 10-4m2 Ωho mol-1 then that of CaCl2 is ( in m2 Ωho mol-1): 

  • 118.88 X 10-4

  • 154.66  X 10-4

  • 273.54  X 10-4

  • 196.21 X 10-4


402.

The ionic conductance of Ba2+ and Cl- are respectively 127 and 76 ohm-1 cm2 at infinite dilution. The equivalent conductance (in ohm-1cm2) of BaCl2 at infinite dilution will be

  • 139.5

  • 203

  • 279

  • 101.5


403.

The reduction electrode potential, E of 0.1 M solution of M+ ions (ERP = -2.36 V) is

  • -4.82 V

  • -2.41 V

  • +2.41 V

  • None of these


404.

The amount of silver deposited on passing 2 F of electricity through aqueous solution of AgNO3

  • 54g

  • 108g

  • 216g

  • 324g


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405.

EMF of hydrogen electrode in term of pH is (at 1 atm pressure).

  • EH2=RTF×pH

  • EH2=RTF·1pH

  • EH2=2.303 RTFpH

  • EH2=- 0.591 pH


406.

For the cell reaction, 

2Ce4+ + Co → 2Ce3+ + Co3+; Ecell° is 1.89 V. If ECo2+/ Co is -0.28V, what is the value of ECe4+ / Ce3+°?

  • 0.28 V

  • 1.61 V

  • 2.17 V

  • 5.29 V


407.

The equivalent conductance of silver nitrate solution at 250°C for an infinite dilution was found to be 133.3 Ω-1 cm2 equiv-1. The transport number of Ag+ ions in very dilute solution of AgNO3 is 0.464. Equivalent conductances of Ag+ and NO3- (in Ω-1 cm2 equiv-1) at infinite dilution are respectively.

  • 195.2; 133.3

  • 61.9; 71.4

  • 71.4; 61.9

  • 133.3; 195.2


408.

The standard reduction potential for Mg2+ /Mg is - 2.37 V and for Cu2+ /Cu is 0.337. The Ecell° for the following reaction is

Mg + Cu2+ → Mg2+ + Cu

  • +2.03 V

  • -2.03 V

  • -2.7 V

  • +2.7 V


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409.

Cu+ (aq) is unstable in solution and undergoes simultaneous oxidation and reduction, according to the reaction

                    2Cu+(aq)  Cu+(aq) + Cu(s)

choose correct E° for above reaction if

ECu2+/Cu°= 0.34 V     andECu2+/Cu+° = 0.15 V

  • -0.38 V

  • + 0.49 V

  • + 0.38 V

  • -0.19 V


C.

+ 0.38 V

From given data ( From G°= -nE°F)(i) Cu(s) Cu2+(aq) + 2e-     G1°= -2 ×(-0.34)× F(ii) Cu2+(aq) + e- Cu+(aq)  G2°= -1 × (0.15) × FOn additionCu(s)  Cu+(aq) + e-,          G3° = -1 × E°× EG3°=G1° ×G2°        =(-2 ×-0.34× F)+ (-1 x 0.15 × F)        = + 0.68 F-0.15 F         = 0.53 For E° = 0.53 VReaction,2Cu+(aq)   Cu2+ (aq) + Cu(s)     E°=?So, Cu+(aq)   Cu (s),                    E° = + 0.53 VCu+(aq)   Cu2+ (aq) + e-           E°= - 0.15V2Cu+(aq)   Cu2+ (aq) + Cu(s)     E°= +0.38V


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410.

The number of electrons required to reduce 4.5 × 10-5 of Al is

  • 1.03 × 1018

  • 3.01 × 1018

  • 4.95 × 1026

  • 7.31 × 1020


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