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:

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answer is C.

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Detailed Solution

$\begin{array}{l} Al + 3 H_{2} O \underset{(NaOH}{⟶} Al (OH)_{3} ↓ + 3 / 2 H_{2} (g) \\ White gelatinous ppt. \\ Al (OH)_{3} \underset{excess of NaOH}{⟶} Na [Al (OH)_{4}] \\ Soluble \\ 2 Al (OH)_{3} \overset{Δ}{(X)} {Al}_{2} O_{3} + 3 H_{2} O \end{array}$
Al₂O₃ is used as adsorbent in chromatography. Thus, metal 'M' is Al.

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