Question

The volume of $HCl$, containing $73 \,g \, L ^{-1}$, required to completely neutralise $NaOH$ 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 $1 \,g \,mol ^{-1}$ respectively)

Answer 1

Mole of $Na =\frac{0.69}{23}=3 \times 10^{-2}$
$Na + H _2 O \longrightarrow NaOH +\frac{1}{2} H _2$
By using POAC
Moles of $NaOH =3 \times 10^{-2}$
$NaOH$ reacts with $HCl$
No. of equivalent of $NaOH = No$. of equivalent of $HCl$
$3 \times 10^{-2} \times 1=\frac{73}{36.5} \times V (\text { in } L ) \times 1$
$V =1.5 \times 10^{-2} L$
Volume of $HCl =15\, ml$.