Advertisements
Advertisements
प्रश्न
Choose A, B, C or D to match the descriptions (i) to (v) below . Some alphabets may be repeated.
(A) Non-electrolyte, (B) Strong electrolyte, (C) Weak electrolyte, (D) Metallic conductor
(i) Molten ionic compound
(ii) Carbon tetrachloride
(iii) An aluminium wire
(iv) A solution containing solvent molecules, solute molecules and ions formed by the dissociation of solute molecules
(v) A sugar solution with sugar molecules and water molecules.
उत्तर
(i) Molten ionic compound - Strong electrolyte
(ii) Carbon tetrachloride- Non-electrolyte
(iii) An aluminium wire- Metallic conductor
(iv) A solution containing solvent molecules, solute molecules and ions formed by the dissociation of solute molecules- weak electrolyte
(v) A sugar solution with sugar molecules and water molecules- Non-electrolyte
APPEARS IN
संबंधित प्रश्न
Powdered sodium chloride (common salt) does not conduct an electric current, but it does so when ______ or when ______.
Give reason for the following:
In the electrolysis of acidified water, dilute sulphuric acid is preferred to dilute nitric acid for acidification.
Give reason for the following:
Ammonia is unionised in the gaseous state but in the aqueous solution, it is a weak electrolyte.
Write down the word or phrase from the bracket that will correctly fill in the blank in the following sentence:
We can expect that pure water ________________ (will / will not) normally conduct electricity.
Write the difference between with examples:
A strong electrolyte and a weak electrolyte
Fill in the black.
The metal plate through which ____________ leaves from an electrolyte is called ____________ .It has ______________ of electrons.
Choose the correct answer from the option given below:
Electrolysis of acidulated water is used in the production of
Why is it necessary to add acid to water before proceeding with electrolysis of 'water'?
Identify the following reactions as either oxidation or reduction : O + 2e- → O-2
Identify the following reactions as either oxidation or reduction : Fe+3 + e- → Fe+2
