The volume of HCl , containing 73g L ^ - 1 , required to completely...

Mole of $NaOH = \dfrac{0 . 69}{23} = 3 \times {10}^{- 2}$

$2Na + 2{H}_{2}O ⟶ 2NaOH + {H}_{2}$

Number of moles of $Na$ = Number of moles of $NaOH$

$NaOH + HCl \to NaCl + {H}_{2}O$

Moles of $HCl$ = moles of $NaOH$

$⇒ ( \text{Molarity} \times V{)}_{HCl} =$Number of moles of $NaOH$

$⇒ \left(\dfrac{73}{36 . 5 \times 1}\right) \times V = 0 . 03$

$⇒ V = 15 \times {10}^{- 3}Lit$

$⇒ V = 15mL$