Question

The theoretical percentage weight loss when $\left[Cr\right(H_{2}O{)}_{4}C{I}_2]CI.2H_{2}O$ is kept over conc. $H_{2}S{O}_4$ in a closed desicator is:

• A. $13.50\%$
• B. $6.75\%$
• C. $18\%$
• D. $36\%$

Answer 1

The correct option is A $13.50\%$

Molecular mass of $\left[Cr\right(H_{2}O{)}_{4}C{l}_2]Cl.2H_{2}O$ is$=52+35.5\times 3+6\times 18=266.5$g

When $\left[Cr\right(H_{2}O{)}_{4}C{l}_2]Cl.2H_{2}O$ is kept oper conc.$H_{2}S{O}_4$ 2 water molecules separated.

Mass of two water molecules$=2\times 18=36$

%weight loss$=\frac{36}{267}\times 100=13.5$%