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When metal ‘M’ is treated with NaOH, a w...

When metal ‘M’ is treated with NaOH, a white gelatinous precipitate ‘X’ is obtained, which is soluble in excess of NaOH. Compound ‘X’ when heated strongly gives an oxide which is used in chromatography as an adsorbent. The metal ‘M’ is

A

Fe

B

Zn

C

Ca

D

Al

Text Solution

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The correct Answer is:
To solve the problem step by step, we can follow this reasoning: ### Step 1: Identify the Reaction When metal ‘M’ reacts with sodium hydroxide (NaOH), it produces a white gelatinous precipitate ‘X’. This indicates that metal ‘M’ is likely to be a metal that forms a hydroxide which is insoluble in water but soluble in excess NaOH. **Hint:** Look for metals that form amphoteric hydroxides. ### Step 2: Analyze the Precipitate The white gelatinous precipitate ‘X’ is soluble in excess NaOH. This characteristic is typical of amphoteric hydroxides, which can dissolve in both acids and bases. **Hint:** Recall that aluminum hydroxide (Al(OH)₃) is an amphoteric hydroxide. ### Step 3: Consider the Heating of Precipitate The problem states that when compound ‘X’ is heated strongly, it gives an oxide that is used as an adsorbent in chromatography. This suggests that the oxide formed from ‘X’ is significant in chromatography. **Hint:** Think about common metal oxides used in chromatography. ### Step 4: Identify the Metal Given that aluminum hydroxide is a white gelatinous precipitate that is soluble in excess NaOH and that heating it produces aluminum oxide (Al₂O₃), which is indeed used as an adsorbent in chromatography, we can conclude that the metal ‘M’ is aluminum. **Hint:** Remember that Al₂O₃ is commonly used in chromatography. ### Step 5: Conclusion Thus, the metal ‘M’ is aluminum. **Final Answer:** The metal ‘M’ is aluminum. ---
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