JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Solutions

The elevation of boiling point of 0.10 m aqueous

CrCl_{3} \times NH_{3}

solution is two times that of 0.05 m aqueous

CaCl_{2}

solution. The value of x is ____
[Assume

100%

ionisation of the complex and

CaCl_{2}

, coordination number of Cr as 6, and that all

NH_{3}

molecules are present inside the coordination sphere]

Answer: 5

Solution

First fix the reference:

CaCl X_{2}

0.05 m

ionizes completely into three ions (

i = 3

), giving an effective molality of

3 \times 0.05 = 0.15 m

for boiling-point elevation. The complex shows twice this elevation, so its effective molality is

0.30 m

; since the complex itself is

0.10 m

, its van't Hoff factor is

i = \frac{0.30}{0.10} = 3

, meaning it dissociates into three ions. With chromium of coordination number

6

and all the ammonia bound inside the sphere, the formula giving three ions is

[Cr (NH X_{3}) X_{5} Cl] Cl X_{2}

, which releases

[Cr (NH X_{3}) X_{5} Cl] X^{2 +}

and two

Cl X^{-}

. Counting the ammonia ligands inside the sphere gives

x = 5

JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Integer answer