Chemistry - Liquid Solution Question with Solution | TestHub

The van't Hoff factor ($i$) is calculated as $i=1+(n-1)\alpha$, where $n$ is the number of ions produced per formula unit and $\alpha$ is the degree of dissociation.

A. For K₄[Fe(CN)₆], $n=5$. $i=1+(5-1)(0.50)=1+4(0.50)=3.0$.

For Mohr salt (ammonium iron(II) sulfate hexahydrate, (NH₄)₂Fe(SO₄)₂·6H₂O), $n=5$ (2NH₄⁺, Fe²⁺, 2SO₄²⁻). $i=1+(5-1)(0.80)=1+4(0.80)=4.2$.

Values are not identical.

B. For NaCl, $n=2$. $i=1+(2-1)(0.80)=1+0.80=1.8$.

For BaCl₂, $n=3$. $i=1+(3-1)(0.40)=1+2(0.40)=1.8$.

Values are identical.

C. For Na₃PO₄, $n=4$. $i=1+(4-1)(0.60)=1+3(0.60)=2.8$.

For K₄[Fe(CN)₆], $n=5$. $i=1+(5-1)(0.45)=1+4(0.45)=2.8$.

Values are identical.

D. For NaNO₃, $n=2$. $i=1+(2-1)(0.90)=1+0.90=1.9$.

For FeCl₃, $n=4$. $i=1+(4-1)(0.30)=1+3(0.30)=1.9$.

Values are identical.

Therefore, options B, C, and D have identical van't Hoff factors.