Question

Question #c7336

Answer 1

Taking atomic masses of constituent elements of the complex as

`Crto52g"/"mol`

`Cl to35.5g"/"mol`

`O to16g"/"mol`

`H to1g"/"mol`

We have formula mass of the complex

`H_12O_6Cl_3Cr->12xx1+6xx16+3xx35.5+1xx52=266.5g"/"mol`

On heating with conc `H_2SO_4` ,the complex looses 6.75% of its original weight.Actually conc `H_2SO_4` dehydrates it.

So loss of total mass of water from formula mass is

`=266.5xx6.75%g~~18g`

Molar mass of water `18g"/"mol`

Hence number of moles of water released `1mol`. So one moecule of water was not in complex sphere.

Hence the formula of the complex as monohydrate should be

`[Cr(H_2O)_5Cl]Cl_2 * H_2O`

And its actual dehydrated formula is

`color(magenta)([Cr(H_2O)_5Cl]Cl_2)`