Question

For $[\text{CrC}{{\text{l}}_{3}}\text{.xN}{{\text{H}}_{3}}]$, elevation in boiling point of one molal solution is double of one molal solution of glucose; hence x is, if complex is 100% ionised :

Answer 1

Hint: An attempt to this question can be made by understanding the colligative property, elevation in boiling point. Determine the elevation in boiling point of glucose and equate it to the elevation in boiling point of the complex mentioned in the question. Based on this you can determine the number of ions after dissociation and thus determine the value of x.

Complete answer:
Colligative properties are the properties of solutions that depend on the ratio of the number of solute particles to the number of solvent molecules in a solution, and not on the nature of the chemical species present.
The number ratio can be related to the various units for concentration of a solution, for example, molarity, molality, normality etc.
The colligative properties are:
Relative lowering of vapor pressure
Elevation of boiling point
Depression of freezing point
Osmotic pressure
When a non-volatile solute is dissolved in a liquid solvent, the boiling point temperature of the solution becomes greater than the boiling point of the pure liquid solvent.
The formula to find elevation in boiling point is :
$\Delta {{T}_{b}}\text{ = i }\text{. }{{\text{K}}_{b}}\text{ }\text{. m}$
Where,
$\Delta {{T}_{b}}$ is the difference in freezing point
i is the Van't Hoff factor
${{K}_{b}}$ is the ebullioscopic constant
m is the molality of solution
It is given to us that 1 molal solution of glucose has half the elevation of 1 molal solution of $[\text{CrC}{{\text{l}}_{3}}\text{.xN}{{\text{H}}_{3}}]$.
Elevation in boiling point of 1 molal solution of glucose can be written as,
$\Delta {{T}_{g}}\text{ = i }\text{. }{{\text{K}}_{b}}\text{ }\text{. m}$
$\Delta {{T}_{g}}\text{ = 1 }\text{. }{{\text{K}}_{b}}\text{ }\text{. 1}$
Elevation in boiling point of 1 molal solution of $[\text{CrC}{{\text{l}}_{3}}\text{.xN}{{\text{H}}_{3}}]$ can be written as,
$\Delta {{T}_{b}}\text{ = i }\text{. }{{\text{K}}_{b}}\text{ }\text{. m}$
$\Delta {{T}_{b}}\text{ = i }\text{. }{{\text{K}}_{b}}\text{ }\text{. 1}$
Equating the two equations keeping in regard the multiplying factor,
$\Delta {{T}_{b}}\text{ = 2 }\Delta {{T}_{g}}$
This gives the value of i as 2.
Thus 2 moles of ions are dissociated for every mole of $[\text{CrC}{{\text{l}}_{3}}\text{.xN}{{\text{H}}_{3}}]$. This means that one chlorine atom is outside the inner complex. But we know that the coordination number of chromium ions is 6. Thus, there must be 4 ammonia molecules in the inner complex.

Therefore, the value of x is 4 and the correct answer is option (A).

Note:
Van't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as per its mass. The value of the Van't Hoff factor is greater than 1 for dissociation of ionic compounds. The factor is used when the solute undergoes dissociation or association in the solvent.