Question #c7336

1 Answer
Aug 13, 2017

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) #