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 $N{H}_3$ to form a deep-blue solution which is a good conductor and an excellent reducing agent. Identify 'C':

• A. $KOH$
• B. $Al(OH{)}_3$
• C. $NaOH$
• D. $LiOH$

Answer 1

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 $N{a}_2{O}_2$ and not $N{a}_{2}O$ which is white.

$2N{a}_2{O}_2+2H_{2}O⟶4NaOH+O_2$

Al is attacked by $NaOH$ and thus the gas (D) is $H_2$:

$2Al+2NaOH+2H_{2}O⟶2NaAl{O}_2+3H_2$

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