The number of moles of KMnO4 reduced by one mole of KI in al

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 Multiple Choice QuestionsMultiple Choice Questions

281.

The number average molar mass and mass average molar mass of a polymer are respectively 30,000 and 40,000. The
polydispersity index (PDI) of the polymer is 

  • -1

  • 0

  • > 1

  • < 1


282.

0.001 mole of [Co(NH3)5(NO3) (SO4)] was passed through a cation exchanger and the acid coming out of it required 20mL of 0.1 M NaOH for neutralisation. Thus, the complex is

  • [Co(NH3)5(NO3)] SO4

  • [Co(NH3)5(SO4)]NO3

  • [Co(NH3)5] NO3.SO4

  • None of the above


283.

In order to prepare one litre 1N solution of KMnO4, how many grams of KMnO4 are required, if the solution to be used in acid medium for oxidation?

  • 128g

  • 41.75 g

  • 31.60 g

  • 62.34 g


284.

The osmotic pressure of blood is 8.21 atm at 37°C. How much glucose would be used for an injection that is at the same osmotic pressure as blood?

  • 22.17 gL-1

  • 58.14 gL-1

  • 61.26 gL-1

  • 75.43 gL-1


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285.

A solution made by dissolving 40 g NaOH in 1000 g of H2O is

  • 1 molar

  • 1 normal

  • 1 molal

  • None of these


286.

0.1 mol HCl is equal to

  • 3.65 g

  • 36.5 g

  • 18 g

  • 1.8 g


287.

A 0.5 g/L solution of glucose is found to be isotonic with a 2.5 g/L solution of an organic compound. What will be the molecular weight of that organic compound ?

  • 300

  • 600

  • 900

  • 200


288.

The pH of a 0.001 M solution of HCl is

  • 0

  • 3

  • 5

  • 10


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289.

The mass of carbon anode consumed (giving only carbondioxide) in the production of 270 kg of aluminium metal from bauxite by the Hall process is (Atomic mass Al= 27)

  • 180 kg

  • 270 kg

  • 540 kg

  • 90 kg


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290.

The number of moles of KMnO4 reduced by one mole of KI in alkaline medium is

  • one-fifth

  • five

  • one

  • two


D.

two

In alkaline solution, KMnO4 is reduced to MnO2 (colourless).

           2KMnO4 + 2H2O  2MnO2 + 2KOH          + 3[O]Ki + 3[O]     KIO3                                                             2KMnO4 + 2H2O + KI  2MnO2 + 2KOH + KIO3

Hence, two moles of KMnO4 are reduced by one mole of KI.


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