Question

The elevation of boiling point of $0.10m$ aqueous $CrC{1}_3.xN{H}_3$ solution is two times that of $0.05m$ aqueous $CaC{l}_2,$ solution. The value of $x$ is [Assume $100\%$ ionisation of the complex and $CaC{l}_2$, coordination number of $Cris6$, and that all $N{H}_3$ molecules are present inside the coordination sphere]
(JEE MAIN 2020)

Answer 1

Molality of $CaC{l}_2$ solution$=0.05m$ $\Delta T_b=i{K}_{b}m=3\times K_b\times 0.05=0.15K_b$ Moltality of $CrC{l}_3.xN{H}_3=0.10m$ $\Delta T_b^{\prime }=i{K}_b\times 0.10;\Delta T_b^{\prime }=2\Delta T_b$ $i{K}_b\times 0.10=2\times 0.15\times K_b\Rightarrow i=3$ Since, Co-ordination number of $Cris6$. $\therefore \text{the complex is}\left[Cr\right(N{H}_3{)}_{5}Cl)C{l}_2$ $\therefore x=5$