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The volume of $HCl$ , containing 73 gL^-1,required to completely neutralize $NaOH$ obtained by reacting 0.69 g of metallic sodium with water, is ____________ ml. ( Nearest integer)
(Given: Molar Masses of Na, Cl, O, H are 23,35.5,16 and 1g mol^-1 respectively).

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answer is 15.

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Detailed Solution

$\begin{array}{l} \underset{1 mol}{Na} + H_{2} O ⟶ \underset{1 mol}{NaOH} + \frac{1}{2} H_{2} \\ n_{Na} = \frac{0.69}{23} = 0.03 \\ n_{NaOH} formed = 0.03 \end{array}$

Milli equivalent of NaOH = milli equivalent of HCl

$\begin{aligned} 0.03 \times 10^{3} = 2 \times V (ml) \\ V = 15 ml \end{aligned}$

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