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When a metal X is treated with cold water, it gives a basic solution Y with molecular formula XOH (molecular $mass = 40$ ) and liberates a gas Z which easily catches fire. X, Y and Z respectively are

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X: Na, Y: NaOH, Z:

H_{2} O

X: K, Y: KOH, Z:

H_{2}

X: Ca, Y:

Ca (OH)

, Z:

H_{2}

X: Na, Y: NaOH Z:

H_{2}

answer is B.

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Detailed Solution

$\overset{}{\underset{(X)}{4 Na (s)}} + O_{2} (g) \to 2 {Na}_{2} O (s)$

${Na}_{2} O + H_{2} O \to \underset{\underset{(Y)}{(Mol .mass = 40)}}{2 NaOH} + H_{2} (Z) ↑$

Therefore X: Na, Y: NaOH Z: $H_{2}$

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