Question

For $1\mathrm{molal}$ aqueous solution of the following compounds, which one will show the highest freezing point?

• A. $\left[\mathrm{Co}{\left({\mathrm{H}}_2\mathrm{O}\right)}_5\mathrm{Cl}\right]{\mathrm{Cl}}_2.{\mathrm{H}}_2\mathrm{O}$
• B. $\left[\mathrm{Co}{\left({\mathrm{H}}_2\mathrm{O}\right)}_4{\mathrm{Cl}}_2\right]\mathrm{Cl}.2{\mathrm{H}}_2\mathrm{O}$
• C. $\left[\mathrm{Co}{\left({\mathrm{H}}_2\mathrm{O}\right)}_3{\mathrm{Cl}}_3\right].3{\mathrm{H}}_2\mathrm{O}$
• D. $\left[\mathrm{Co}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]{\mathrm{Cl}}_3$

Answer 1

The correct option is C

$\left[\mathrm{Co}{\left({\mathrm{H}}_2\mathrm{O}\right)}_3{\mathrm{Cl}}_3\right].3{\mathrm{H}}_2\mathrm{O}$

Answer (C)

The solution which shows maximum freezing point must have minimum number of solute particles.

Explanation for correct option:

$\left[\mathrm{Co}{\left({\mathrm{H}}_2\mathrm{O}\right)}_3{\mathrm{Cl}}_3\right].3{\mathrm{H}}_2\mathrm{O}->\left[\mathrm{Co}{\left({\mathrm{H}}_2\mathrm{O}\right)}_3{\mathrm{Cl}}_3\right].3{\mathrm{H}}_2\mathrm{O}],\mathrm{i}=1$

Explanation for incorrect option:

(A) $\left[\mathrm{Co}{\left({\mathrm{H}}_2\mathrm{O}\right)}_5\mathrm{Cl}\right]{\mathrm{Cl}}_2.{\mathrm{H}}_2\mathrm{O}->{\left[\mathrm{Co}{\left({\mathrm{H}}_2\mathrm{O}\right)}_5\mathrm{Cl}\right]}^{2+}+2{\mathrm{Cl}}^{-},\mathrm{i}=3$

(B) $\left[\mathrm{Co}{\left({\mathrm{H}}_2\mathrm{O}\right)}_4{\mathrm{Cl}}_2\right]\mathrm{Cl}.2{\mathrm{H}}_2\mathrm{O}->{\left[\mathrm{Co}{\left({\mathrm{H}}_2\mathrm{O}\right)}_4{\mathrm{Cl}}_2\right]}^{+}+{\mathrm{Cl}}^{-},\mathrm{i}=2$

(D)$\left[\mathrm{Co}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\mathrm{Cl}\right]{\mathrm{Cl}}_3->{\left[\mathrm{Co}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{3+}+3{\mathrm{Cl}}^{-},\mathrm{i}=4$

So, solution $1\mathrm{molal}\left[\mathrm{Co}\right({\mathrm{H}}_2{\mathrm{O}}_3\left){\mathrm{Cl}}_3\right].3{\mathrm{H}}_2\mathrm{O}$ will have minimum number of particles in an aqueous state.

Hence option (C) is correct option.