Explanation

(d)
mmol of complex =30$\times$0.01 =0.3 Also, 1 mole of complex $\left\lbrack Cr\left( H_{2}O \right)_{5}Cl \right\rbrack{Cl}_{2}$ gives only two moles of chloride ion when dissolved in solution.
$\left\lbrack Cr\left( H_{2}O \right)_{5}Cl \right\rbrack{Cl}_{2} \longrightarrow \left\lbrack Cr\left( H_{2}O \right)_{5}Cl \right\rbrack^{2 +} + \ 2\ {Cl}^{-}$
⇒mmol of $Cl^{-}$ ion produced from its 0.3 mmol =0.6
Hence, 0.6 mmol of $Ag^{+}$ would be required for precipitation.
⇒ 0.60 mmol of $Ag^{+}$=0.1 M $\times$V (in mL)
⇒ V=6 mL