Question

The volume (in mL) of 1.0 M $AgN{O}_3$ required for the complete precipitation of chloride ions present in 30 mL of 0.01M solution of $\left[Cr\right(H_{2}O{)}_6]C{l}_3$, as silver chloride is:

• A. 0.9
• B. 1.0
• C. 1.8
• D. 2.0

Answer 1

The correct option is A 0.9

$\left[Cr\right(H_{2}O{)}_6]C{l}_3+3AgN{O}_3\to [Cr\left(H_{2}O{)}_6\right](N{O}_3{)}_3+3AgCl$
Because $\left[Cr\right(H_{2}O{)}_6]C{l}_3$ has 3 ionisable $C{l}^{-}$ ions.
We know that, $M_1{V}_1=M_2{V}_2$
$M_{1}and{M}_2,V_{1}and{V}_2$ are concentration and volume of $\left[Cr\right(H_{2}O{)}_6]C{l}_{3}andAgN{O}_{3}respectively.$

Millimoles of $C{l}^{-}$ = 30 x 0.01 M x 3 = 0.9 M
So, volume of $A{g}^{+}$ required =0.9 x 1 = 0.9 mL