Question

$1g$ of the complex $\left[Cr\right(H_{2}O{)}_{5}Cl]C{l}_2.H_{2}O$ was passed through a cation exchanger to produce $HCl$ . The acid liberated was diluted to $1$ litre . What is normality of this acid solution .

Answer 1

Mol. wt. of $\left[Cr\right(H_{2}O{)}_{5}Cl]C{l}_2.H_{2}O=266.5$

$\therefore$ Moles of complex = $\frac{1}{266.5}$

Note : $1$ mole of $\left[Cr\right(H_{2}O{)}_{5}Cl]C{l}_2.H_{2}O$ will give $2$ mole $C{l}^{-}$ ions

Thus , mole of $HCl$ formed = $\frac{2\times 1}{266.5}$

$\therefore$ $N_{HCl}=\frac{2\times 1}{266.5\times 1}=0.0075N$