When any solution passess through a cation exchange resin that is in acidic form, H ion of the resin is replaced by cations of the solution. A solution containing 0.319 g of an isomer with molecular formula `CrCl_(3).6H_(2)O` is passed through a cation exchange resin in acidic form. The eluted solution requires 19 `cm^(3)` fo 0.125 N NaOH. The isomer is

Options

A. triaquatrichlorochrominum (III) chloride trihydrate
B. hexaaquachromium (III) chloride
C. pentaaquamonochlorochromium (III) chloride monohydrate
D. tetraaquadichlorochromium (III) chloride dihydrate

Answer

C

Explanation

`CrCl_(3).6H_(2)Oto` diluted solution requires 19 mm and 0.125N NaOH=2.375 m mole `OH^(-)`
`0.319/266.5 = 0.00119`
`CrCl_(3).6H_(2)O to 0.00119` mole,
`[Cr(H_(2)O)_(5)Cl]Cl_(2).H_(2)Oto 2Cl^(-)to 2H^(+)`
`2H^(+)` requires `to 2OH^(-)`
`0.00119 xx 2` moles of `Cl^(-1)=nH^(+)`
`1000 xx 0.00238`=2.38 moles