Question

The amount of volume (in ml) of a solution of hydrochloric acid containing $73$ g of acid per litre that would be required for the exact neutralization of sodium hydroxide obtained by allowing $0.46$ g of metalic sodium to act upon water is :
$(Cl=35.5,Na=23.0,O=16)$

Answer 1

The reactions are as follows:

$Na+H_{2}O\to NaOH+\frac{1}{2}{H}_2$
$HCl+NaOH\to NaCl+H_{2}O$

Meq. of $Na$ $=$ Meq. of $NaOH=$ Meq. of $HCl$

$\frac{0.46}{23}\times 1000=\frac{73}{36.5}\times V$ ($N_{HCl}=\frac{73}{36.5}$ and Meq. $=N\times V$)

$⟹V=10$ mL

Hence, the volume is $10$ mL.