The degree of dissociation (α) of a weak electrolyte ,AxBy is r

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341.

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)

  • Fe (Z = 26)

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342.

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 

  • 2.0 L 

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343.

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+] = 1.0 M

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344.

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 minus 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
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345.

The Gibbs energy for the decomposition of Al2O3 at 500°C is as follows
2 over 3 space Al subscript 2 straight O subscript 3 space rightwards arrow space 4 over 3 space Al space plus space straight O subscript 2 comma space increment subscript straight r space straight G space equals space plus 966 space kJ space mol to the power of negative 1 end exponent
The potential difference needed for electrolytic reduction of Al2O3 at 500°C is at least

  • 4.5 V

  • 3.0 V

  • 2.5 V

  • 2.5 V

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346.

Given,

straight E subscript Cl subscript 2 divided by Cl to the power of minus end subscript superscript 0 space equals 1.36 space comma
straight E subscript Cr to the power of 3 plus end exponent divided by Cr end subscript superscript 0 space equals space minus 0.74 space straight V
straight E subscript Cr subscript 2 straight O subscript 7 superscript 2 minus end superscript end subscript superscript 0 space equals space 1.33 space straight V comma
straight E subscript MnO subscript 4 superscript minus divided by Mn to the power of 2 plus end exponent end subscript superscript 0 space equals space 1.51 space straight V
Among the following, the strongest reducing agent is

  • Cr

  • Mn2+

  • Cr3+

  • Cr3+

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347.

Given: straight E subscript Fe to the power of 3 plus end exponent divided by Fe end subscript superscript 0 space equals space minus 0.036 straight V comma space straight E subscript Fe to the power of 2 plus end exponent divided by Fe end subscript superscript 0 space equals space minus 0.439 straight V The value of standard electrode potential for the change Fe subscript left parenthesis aq right parenthesis end subscript superscript 3 plus end superscript space plus space straight e to the power of minus space rightwards arrow Fe to the power of 2 plus end exponent space left parenthesis aq right parenthesis will be

  • -0.072

  • 0.385 V

  • 0.770 V

  • 0.770 V

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348.

On the basis of the following thermochemical data: (∆f G° H(aq)+=0)
H2O(l) → H+(aq) + OH(aq); ∆H =57.32kJ
H2(g) + 1/2O2(g) → H2O(l); ∆H = –286.20 kJ
The value of enthalpy of formation of OHion at 25°C is

  •  –22.88 kJ

  • –228.88 kJ

  • +228.88 kJ

  • +228.88 kJ

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349. Given space 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 0.72 space straight V comma space straight E subscript Fe to the power of 2 plus end exponent divided by Fe end subscript superscript straight o space equals space minus space 0.42 space straight V. space The space potential space for space the space cell
Cr vertical line Cr to the power of 3 plus end exponent space left parenthesis 0.1 right parenthesis vertical line vertical line Fe to the power of 2 plus end exponent space left parenthesis 0.01 space straight M right parenthesis vertical line Fe space is
  • 0.26

  • 0.399

  • 0-399

  • 0-399

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350.

The cell, Zn|Zn2+ (1M)|| Cu2+ (1M|Cu(Eocell = 1.10 V), was allowed to be completely discharged at 298 K. THe relative concentration of Zn2+ to Cu2+ open square brackets fraction numerator open parentheses Zn to the power of 2 plus end exponent close parentheses over denominator left parenthesis Cu to the power of 2 plus end exponent right parenthesis end fraction close square brackets space is

  • antilog (24.08)

  • 37.3

  • 1037.3

  • 1037.3

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