Question

The volume (in ml) of $0.1$ M $AgN{O}_3$ required for complete precipitation of chloride ions present in $30ml$ of $0.01M$ solution of $\left[Cr\right(H_{2}O{)}_{5}Cl]C{l}_2$, as silver chloride, is close to:

• A. $6$
• B. $8$
• C. $9$
• D. $7$

Answer 1

The correct option is A $6$

Number of ionisable $C{l}^{-}$ in $\left[Cr\right(H_{2}O{)}_{5}Cl]C{l}_2$ is $2$.

$\therefore$ Millimoles of $C{l}^{-}=30\times 0.01\times 2=0.6$

$\therefore$ Millimoles of $A{g}^{+}$ required $=0.6$

Now, $\frac{0.6}{V\left(inml\right)}=0.1$

or, $V=6$ ml.

So, the correct option is $A$