Question

For a given value of the degree of dissociation ($\alpha$), which of the following have correct van’t Hoff factor?

• A. $NaCl;i=2+\alpha$
• B. $Ca(N{O}_3{)}_2;i=1+2\alpha$
• C. $K_4\left[\right(Fe\left(CN{)}_6\right];i=1+4\alpha$
• D. $(N{H}_4{)}_{3}P{O}_4;i=3+\alpha$

Answer 1

Answer: (B), (C)

$Ca(N{O}_3{)}_2;i=1+2\alpha$
$K_4\left[\right(Fe\left(CN{)}_6\right];i=1+4\alpha$

The expression to determine van't Hoff factor of a solution is given by:

$i=1+(n-1)\alpha$

$n$= number of ions formed after dissociation

$\alpha$= Degree of dissociation

where, $i$= van't Hoff factor

A. For $NaCl$; $n=2$ so $i=1+\alpha$

B. For $Ca(N{O}_3{)}_2$; $n=3$ so $i=1+2\alpha$

C. For $K_4\left[Fe\right(CN{)}_6]$; $n=5$ so $i=1+4\alpha$

D. For $(N{H}_4{)}_{3}P{O}_4$; $n=4$ so $i=1+3\alpha$

So, option B and C are correct.