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An aqueous solution of salt 'R' when treated with dil HCl, a colourless gas is given out. The gas so evolved when passed through acidified KMnO4 decolourises KMnO4 solution. The salt 'R' is
  • a)
    Na2CO3
  • b)
    NaClO3
  • c)
    NaNO2
  • d)
    Na2SO3
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
An aqueous solution of salt 'R' when treated with dil HCl, a colourle...
Answer:

Introduction:
In this question, we are given an aqueous solution of an unknown salt 'R' which gives out a colourless gas when treated with dilute hydrochloric acid (HCl). When this gas is passed through acidified potassium permanganate (KMnO4) solution, the KMnO4 solution gets decolourized. We need to identify the salt 'R' from the given options.

Explanation:
To determine the identity of the salt 'R', let's analyze the given information step by step.

Step 1: Reaction with Dilute HCl
When the aqueous solution of salt 'R' is treated with dilute hydrochloric acid (HCl), a colourless gas is evolved. This indicates the presence of a gas that is not soluble in water.

Step 2: Reaction with Acidified KMnO4
The evolved gas is then passed through acidified potassium permanganate (KMnO4) solution. This reaction is a test for the reducing property of the gas.

If the gas is a reducing agent, it will react with the acidified KMnO4 and reduce the Mn(VII) ions to Mn(II) ions. As a result, the purple color of KMnO4 solution will get decolourized.

Conclusion:
Based on the given information, we can conclude that the salt 'R' is sodium sulfite (Na2SO3). When sodium sulfite reacts with HCl, it produces sulfur dioxide (SO2) gas, which is a reducing agent. This gas then reacts with acidified KMnO4 and reduces the Mn(VII) ions to Mn(II) ions, leading to decolorization of the KMnO4 solution.

Therefore, the correct option is 'D' - Na2SO3.
Free Test
Community Answer
An aqueous solution of salt 'R' when treated with dil HCl, a colourle...
Ince 'R' gives a colourless gas on reaction with dil HCl, so it contains CO32- or SO32- as anion (i.e., CO2 or SO2 is evolved).
Since the gas decolourises acidified KMnO4 solution so it is SO2 and thus the anion present is SO32- i.e., the salt 'R' is Na2SO3 .
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An aqueous solution of salt 'R' when treated with dil HCl, a colourless gas is given out. The gas so evolved when passed through acidified KMnO4 decolourises KMnO4 solution. The salt 'R' isa)Na2CO3b)NaClO3c)NaNO2d)Na2SO3Correct answer is option 'D'. Can you explain this answer?
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