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Question

A soft silver-white metal 'A' burns with a golden-yellow flame to give a yellow power 'B,' which on treatment with water gives a clear solution 'C'. The solution placed on Al attacks this metal, liberating a gas 'D' forming a water-soluble compound 'E'. The metal 'A' dissolves in liquid NH3 to form a deep-blue solution which is a good conductor and an excellent reducing agent. Identify 'C':

A
KOH
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B
Al(OH)3
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C
NaOH
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D
LiOH
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Solution

The correct option is C NaOH
The intense yellow flame points to sodium metal (A). It burns in air to give a yellow powder.
Purely owing to the colour, (B) is Na2O2 and not Na2O which is white.

2Na2O2+2H2O4NaOH+O2

Al is attacked by NaOH and thus the gas (D) is H2:

2Al+2NaOH+2H2O2NaAlO2+3H2

The question only asks you to identify (C) and nothing else.
You can disregard any information not helpful in identifying it.

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