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--
OPTION A CALCIUM
OPTION B ALUMINIUM
OPTION C IRON
OPTION D ZINC
JEE MAINS QUESTION 2018
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PAPER CODE C QUESTION NUMBER 11 CHEMISTRY
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Answer → Option B. Aluminium.
Explanation ⇒
When Aluminum (or Metal M) is treated with the Sodium Hydroxide, Aluminium Hydroxide which is a Gelatinous white precipitate is formed.
Reaction is ⇒
2Al + 6H₂O ------(NaOH)---→ 2Al(OH)₃↓ + 3H₂↑
Al(OH)₃ (or X) is the Gelatinous white precipitate, which will be soluble in the excess of the NaOH.
When Aluminium hydroxide (or X) is further made to reacts with the Sodium hydroxide, it forms the Clear Solution of Sodium Aluminate.
Reaction is ⇒
Al(OH)₃ + NaOH ---------→ Na[Al(OH)₄] (Sodium Aluminate).
When Aluminium Hydroxide (or Compound X) is Heated Strongly it forms the Aluminium Oxide.
2Al(OH)₃ -----Δ----→ Al₂O₃ + 3H₂O
This Oxide of Aluminum is used in the Chromatography as an Adsorbent.
[ Note ⇒ Aluminum also forms the Gelatinous Precipitate and is soluble in the excess of the sodium hydroxide but the Zinc oxide is not used in the chromatography. Thus, it can be correct answer. ]
Calcium Forms the White Precipitate and Iron of Valency 2 Forms Dirty Green precipitate and Iron of Valency 3 form the Reddish Brown Precipitate and they both are insoluble in the Excess of the NaOH Thus, these option also can not be correct.
Hence, the Answer is Aluminium (or Metal M).
Hope it helps.
Answer → Option B. Aluminium.
Explanation ⇒
When Aluminum (or Metal M) is treated with the Sodium Hydroxide, Aluminium Hydroxide which is a Gelatinous white precipitate is formed.
Reaction is ⇒
2Al + 6H₂O ------(NaOH)---→ 2Al(OH)₃↓ + 3H₂↑
Al(OH)₃ (or X) is the Gelatinous white precipitate, which will be soluble in the excess of the NaOH.
When Aluminium hydroxide (or X) is further made to reacts with the Sodium hydroxide, it forms the Clear Solution of Sodium Aluminate.
Reaction is ⇒
Al(OH)₃ + NaOH ---------→ Na[Al(OH)₄] (Sodium Aluminate).
When Aluminium Hydroxide (or Compound X) is Heated Strongly it forms the Aluminium Oxide.
2Al(OH)₃ -----Δ----→ Al₂O₃ + 3H₂O
This Oxide of Aluminum is used in the Chromatography as an Adsorbent.
[ Note ⇒ Aluminum also forms the Gelatinous Precipitate and is soluble in the excess of the sodium hydroxide but the Zinc oxide is not used in the chromatography. Thus, it can be correct answer. ]
Calcium Forms the White Precipitate and Iron of Valency 2 Forms Dirty Green precipitate and Iron of Valency 3 form the Reddish Brown Precipitate and they both are insoluble in the Excess of the NaOH Thus, these option also can not be correct.
Hence, the Answer is Aluminium (or Metal M).
Hope it helps.
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