Answer:
2 is the van’t Hoff factor for NaCl if it completely dissociates in water.
Explanation:
Formula used :
[tex]\alpha =\frac{i-1}{n-1}[/tex]
[tex]\alpha [/tex] = Degree of dissocation.
n = number of ions produced by dissociation of ionic compound
i = van't Hoff factor
We have:
[tex]NaCl(aq)\rightarrow Na^+(aq)+Cl^-(aq)[/tex]
n = 2. [tex]\alpha = 100\%=1[/tex]
[tex]1=\frac{i-1}{2-1}[/tex]
i = 2
2 is the van’t Hoff factor for NaCl if it completely dissociates in water.
