# Question c7336

Aug 13, 2017

Taking atomic masses of constituent elements of the complex as

$C r \to 52 g \text{/} m o l$

$C l \to 35.5 g \text{/} m o l$

$O \to 16 g \text{/} m o l$

$H \to 1 g \text{/} m o l$

We have formula mass of the complex

${H}_{12} {O}_{6} C {l}_{3} C r \to 12 \times 1 + 6 \times 16 + 3 \times 35.5 + 1 \times 52 = 266.5 g \text{/} m o l$

On heating with conc ${H}_{2} S {O}_{4}$ ,the complex looses 6.75% of its original weight.Actually conc ${H}_{2} S {O}_{4}$ dehydrates it.

So loss of total mass of water from formula mass is

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

Molar mass of water $18 g \text{/} m o l$

Hence number of moles of water released $1 m o l$. So one moecule of water was not in complex sphere.

Hence the formula of the complex as monohydrate should be

$\left[C r {\left({H}_{2} O\right)}_{5} C l\right] C {l}_{2} \cdot {H}_{2} O$

And its actual dehydrated formula is

$\textcolor{m a \ge n t a}{\left[C r {\left({H}_{2} O\right)}_{5} C l\right] C {l}_{2}}$