Question

The elevation of boiling point of 0.10 m aqueous \[{\rm{CrC}}{{\rm{l}}_{\rm{3}}}{\rm{xN}}{{\rm{H}}_{\rm{3}}}\] solution is two times that of 0.05m aqueous \[{\rm{CaC}}{{\rm{l}}_{\rm{2}}}\] solution. The value of x is ……
[Assume 100% ionisation of the complex and \[{\rm{CaC}}{{\rm{l}}_{\rm{2}}}\], coordination number of Cr as 6, and that all \[{\rm{N}}{{\rm{H}}_{\rm{3}}}\] molecules are present inside the coordination complex ]

Answer 1

Hint: Carbon exists in a wide number of allotropic forms and diamond is one of them. It is the purest form of carbon. It is a huge three-dimensional polymer containing a large number of \[s{p^3}\] hybridised carbon atoms organised in a tetrahedral structure.

Formula used: The formula of elevation of boiling point is,
\[\Delta {T_b}\, = \,\,i{K_b}m\]
Here, i is van’t Hoff factor, \[{K_b}\] is molal elevation constant, m is molality of the solution.

Complete Step by Step Solution:
Let’s first calculate i for \[{\rm{CaC}}{{\rm{l}}_{\rm{2}}}\] . Calcium Chloride undergoes dissociation into one Calcium ion and two chloride ions.

So, the number of ions formed is three. So, i for \[{\rm{CaC}}{{\rm{l}}_{\rm{2}}}\] is 3.
Now, we find out the \[\Delta {T_b}\] of \[{\rm{CaC}}{{\rm{l}}_{\rm{2}}}\] and \[{\rm{CrC}}{{\rm{l}}_{\rm{3}}}{\rm{xN}}{{\rm{H}}_{\rm{3}}}\] using the above formula.
For \[{\rm{CaC}}{{\rm{l}}_{\rm{2}}}\], m=0.05 and i=3
\[\Delta {T_b}_{_1}\, = \,\,3 \times {K_b} \times 0.05\] ---- (1)

For, \[{\rm{CrC}}{{\rm{l}}_{\rm{3}}}{\rm{xN}}{{\rm{H}}_{\rm{3}}}\], m=0.10
\[\Delta {T_{{b_2}}}\, = \,\,i \times {K_b} \times 0.10\] ---- (2)
Given that, the \[\Delta {T_b}\] of \[{\rm{CrC}}{{\rm{l}}_{\rm{3}}}{\rm{xN}}{{\rm{H}}_{\rm{3}}}\] solution is twice of \[{\rm{CaC}}{{\rm{l}}_{\rm{2}}}\] solution.
So,
\[ \Rightarrow \,\,i \times {K_b} \times 0.10 = \,\,2\,\left( {3 \times {K_b} \times 0.05} \right)\]
\[ \Rightarrow \,\,i \times {K_b} \times 0.10 = \,\,6 \times {K_b} \times 0.05\]
\[ \Rightarrow \,\,i = \,\,\dfrac{{6 \times 0.05}}{{0.10}}\]
\[ \Rightarrow i = 3\]
Therefore, van’t Hoff factor for
\[{\rm{CrC}}{{\rm{l}}_{\rm{3}}}{\rm{xN}}{{\rm{H}}_{\rm{3}}}\]is 3, that means, it dissociates into three ions.
As given, the dissociation of \[{\rm{CrC}}{{\rm{l}}_{\rm{3}}}{\rm{xN}}{{\rm{H}}_{\rm{3}}}\]is 100% and all ammonia molecules are inside the complex, so the chemical equation is,
$CrCl_3xNH_3 \rightarrow [Cr(NH_3)_xCl]^{2+} + 2Cl^{-}$

Also, given that the coordination number of Cr is 6, that means, the Cr is bonded to 6 atoms, so, the number of ammonia molecules is 5.
Hence, the value of x is 5. Therefore, the compound is \[{\rm{CrC}}{{\rm{l}}_{\rm{3}}}{\rm{5N}}{{\rm{H}}_{\rm{3}}}\].

Note: The phenomenon of elevation of boiling point states that addition of a compound to a solvent results in the increase of boiling point than the pure solvent.