Question

Which of the complex in its 1 molal solution with 100% ionisation will show a boiling point of ${102.08}^o$? $(K_B$ for water $=0.52Kmolalit{y}^{-1})$

• A. $F{e}_4\left[Fe\right(CN{)}_6]$
• B. $K_3\left[Fe\right(CN{)}_6]$
• C. $K_4\left[Fe\right(CN{)}_6]$
• D. none of these

Answer 1

The correct option is B $K_3\left[Fe\right(CN{)}_6]$

Option (B) is correct.
$\Delta T=K_b\times molality\times i$

$\therefore i=\frac{2.08}{0.52}=4$

Also, $i=1-\alpha +x\alpha +y\alpha (\because \alpha =1)$

$\therefore x+y=4$

$i.e.,3K^{+}$ and $Fe(CN{)}_6^{-3}$