Calculate the volume of 0.1M `AgNO_(3)` solution required for complete precipitation of chloride ions present in 30 mL of 0.01M solution of `[Cr(H_(2)O)_(6)]Cl_(3)`.

Explanation

Millimoles of `[Cr(H_(2)O)_(6)]Cl_(3)`
=`"volume (in mL)"xx"molarity"=30xx0.01=0.3`
`therefore" Millimoles of "Cl^(-)" in "`
`[Cr(H_(2)O)_(6)]Cl_(3)=3xx" millimoles of "[Cr(H_(2)O)]Cl_(3)`
`=3xx0.3=0.9`
So, millimoles of `Ag^(+)` ion required = millimoles of `Cl^(-)` ion present in`[Cr(H_(2)O)_(6)]Cl_(3)=0.9`
But millimoles of `Ag^(+)` ion = volume of `AgNO_(3)` solution `"(in mL)"xx"molarity of "AgNO_(3)" solution"`
`therefore" 0.9 = volume of "AgNO_(3)" solution "`
Hence volume of `AgNO_(3)` solution (in mL) `=(0.9)/(0.1)=9`