The elevation of boiling point of $$0.10\,\text{m}$$ aqueous $$\text{CrCl}_3 \cdot x\text{NH}_3$$ solution is two times that of $$0.05\,\text{m}$$ aqueous $$\text{CaCl}_2$$ solution. The value of $$x$$ is............ [Assume 100% ionisation of the complex and $$\text{CaCl}_2$$, coordination number of Cr as 6, and that all $$\text{NH}_3$$ molecules are present inside the coordination sphere]

For boiling‐point elevation we always use the relation

$$\Delta T_b = i\,K_b\,m,$$

where $$\Delta T_b$$ is the rise in boiling point, $$i$$ is the van’t Hoff factor, $$K_b$$ is the molal elevation constant of the solvent (same for both solutions because the solvent is water), and $$m$$ is the molality of the solution.

We are told that

$$\Delta T_b(\text{CrCl}_3\cdot x\text{NH}_3,\;0.10\,\text{m}) = 2\,\Delta T_b(\text{CaCl}_2,\;0.05\,\text{m}).$$

Substituting the formula for each $$\Delta T_b$$ we have

$$i_{\text{complex}}\;K_b\;(0.10) = 2\;\bigl[i_{\text{CaCl}_2}\;K_b\;(0.05)\bigr].$$

Because $$K_b$$ is common to both solutions, it cancels out. The given condition of 100 % ionisation means that

$$i_{\text{CaCl}_2}=3$$

(one $$\text{Ca}^{2+}$$ ion and two $$\text{Cl}^-$$ ions).

So we get

$$i_{\text{complex}}\,(0.10)=2\,(3)\,(0.05).$$

Multiplying on the right-hand side first,

$$2\times3\times0.05 = 0.30.$$

Now dividing both sides by 0.10,

$$i_{\text{complex}} = \frac{0.30}{0.10} = 3.$$

Thus the van’t Hoff factor of the chromium complex must be $$3$$, which means the complex salt produces three ions per formula unit in water.

Next we interpret the formula $$\text{CrCl}_3\cdot x\text{NH}_3$$ on the basis of coordination chemistry. Chromium(III) has a coordination number of 6, and every $$\text{NH}_3$$ molecule is inside the coordination sphere. Let us assume that inside the square brackets the complex is

$$[\text{Cr}(\text{NH}_3)_x\text{Cl}_y]^{(3-y)+},$$

where $$y$$ chloride ions are also coordinated. The remaining $$(3-y)$$ chloride ions are outside the bracket as counter-ions. Therefore the complete neutral compound is

$$[\text{Cr}(\text{NH}_3)_x\text{Cl}_y]\;\text{Cl}_{\,3-y}.$$

Because the total number of ligands inside the bracket must be six,

$$x + y = 6.$$

When this salt dissolves, 100 % ionisation gives

$$[\text{Cr}(\text{NH}_3)_x\text{Cl}_y]^{(3-y)+} + (3-y)\,\text{Cl}^-.$$

The number of particles (ions) produced is therefore

$$1 + (3-y) = 4 - y.$$

This count must equal the required van’t Hoff factor:

$$4 - y = i_{\text{complex}} = 3.$$

So

$$y = 4 - 3 = 1.$$

Finally, substituting $$y = 1$$ into $$x + y = 6$$ gives

$$x = 6 - 1 = 5.$$

Thus the complex is $$[\text{Cr}(\text{NH}_3)_5\text{Cl}]\;\text{Cl}_2.$$ So, the answer is $$5$$.