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A colourless water-soluble compound on s...

A colourless water-soluble compound on strong heating liberates a brown colored gas and leaves a yellow residue that turns white on cooling. An aqueous solution of the original solid gives a white precipitate with `(NH_(4))_(2)S`. The original solid is :

A

`Zn(NO_(3))_(2)`

B

`Ca(NO_(3))_(2)`

C

`Al(NO_(3))_(3)`

D

`NaNO_(3)`

Text Solution

AI Generated Solution

The correct Answer is:
To determine the original solid based on the provided clues, we can follow these steps: ### Step 1: Analyze the clues provided The compound is colorless and water-soluble. Upon strong heating, it liberates a brown gas and leaves a yellow residue that turns white on cooling. Additionally, it forms a white precipitate when treated with ammonium sulfide. ### Step 2: Identify the brown gas The brown gas that is liberated upon heating is likely nitrogen dioxide (NO₂). This is a common gas released during the thermal decomposition of certain nitrogen-containing compounds. ### Step 3: Identify the yellow residue The yellow residue that turns white upon cooling suggests the presence of zinc oxide (ZnO). Zinc oxide appears yellow when hot due to its ability to absorb blue light, and it turns white upon cooling. ### Step 4: Relate the observations to a specific compound Given that we have identified the brown gas as nitrogen dioxide and the yellow residue as zinc oxide, we can deduce that the original solid is likely zinc nitrate (Zn(NO₃)₂). When zinc nitrate is heated, it decomposes to produce nitrogen dioxide, oxygen, and zinc oxide. ### Step 5: Confirm with the ammonium sulfide test When zinc nitrate is dissolved in water and treated with ammonium sulfide ((NH₄)₂S), it forms zinc sulfide (ZnS), which is a white precipitate. This confirms our identification of the original solid. ### Conclusion The original solid is **zinc nitrate (Zn(NO₃)₂)**. ---
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