The volume (in mL) of 0.1 `M AgNO_(3)` required to completely precipitat the chloride ions present in 30 mL of 0.01 M of `[Cr(H_(2)O)_(5)Cl]Cl_(2)`, as silver chloride is close to :
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The correct Answer is:
6
`underset(n_(1)=2)(2AgNO_(3))+[Cr(H_(2)O)_(5)Cl]Cl_(2)to [Cr(H_(2)O)_(5)Cl](NO_(3))_(2)+2AgCl` `(M_(1)V_(1))/(n_(1))=(M_(2)V_(2))/(n_(2))` `(0.1xxV)/(2)=(0.01xx30)/(1)` V=6 mL
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