A solution containing 0.319 g of complex `CrCl_(3).6H_(2)O` was passed through cation exchanger and the solution given out was neutralised by 28.5 ml of 0.125 m NaOH. The correct formula of the complex will be: [ molecular weight of complex =266.5]

Options

A. `[CrCl(H_(2)O)_(5)]Cl_(2).H_(2)O`
B. `[Cr(H_(2)O)_(6)]Cl_(3)`
C. `[CrCl_(2)(H_(2)O)_(4)]Cl.2H_(2)O`
D. All are correct

Answer

B

Explanation

`Cl^(-)=HCl=NaOH`
`nCl^(-) +nH^(+) to nHCl `
Thus 1 mol of complex will form n mol of HCl
1 mole of complex`=(0.319)/(266.5)=0.0012`, mole of NaOH use `=(28.5xx0.125)/(1000)=0.0036`
So 0.0012 moel of complex =0.0036 mole of NaOH=0.0036 mole of HCl
1 mole of complex `=(0.0036)/(0.0012)=`3 mole of HCl
`:.` n=3
So complex is `[Cr(H_(2)O)_(6)]Cl_(3)`.