A coordination compound $CrC{l}_3.4H_{2}O$ precipitates silver chloride when treated with silver nitrate. The molar conductance of its solution corresponds to a total of two ions. Write structural formula of the compound and name it.

A molar conductance equivalent to two ions suggests that complex is made up of two ions i.e., one complex ion and one ionizable counter ion.

Since the compound $CrC{l}_3.4H_{2}O$ precipitates silver chloride $\left(AgCl\right)$ when treated with silver nitrate, one $C{l}^{-}$ should be present outside the coordination sphere as counter ion and the remaining atoms/ions should be the part of coordination sphere. Thus, the formula of the compound should be:

$\left[Cr\right(H_{2}O{)}_{4}C{l}_2]Cl$

It ionizes as:
$\left[Cr\right(H_{2}O{)}_{4}C{l}_2]Cl\to [Cr(H_{2}O{)}_{4}C{l}_2{]}^{+}+C{l}^{-}$

And its reaction with $AgN{O}_3$ will be:

$\left[Cr\right(H_{2}O{)}_{4}C{l}_2]Cl+AgN{O}_3\to [Cr\left(H_{2}O{)}_{4}C{l}_2\right]N{O}_3+AgCl\downarrow$

Thus, the formula of the complex is $\left[Cr\right(H_{2}O{)}_{4}C{l}_2]Cl$ and its IUPAC name is tetraaquadichloridochromium(III) chloride.