A specifically designed aqueous solution of an ionic coordination compound is recorded to freeze at — Solutions and Colligative Properties Chemistry Question
Question
A specifically designed $0.0020\text{ m}$ aqueous solution of an ionic coordination compound $[Co(NH_3)_5(NO_2)]Cl_n$ is recorded to freeze at exactly $-0.00744^\circ\text{C}$. Assuming complete $100\%$ ionization of the outer sphere and given that $K_f$ for water is $1.86^\circ\text{C m}^{-1}$, what is the total number of moles of ions produced when exactly one mole of this complex dissolves in water?
💡 Solution & Explanation
The depression in freezing point formula featuring the van't Hoff factor is $\Delta T_f = i \times K_f \times m$. Substituting the measured values gives $0.00744 = i \times 1.86 \times 0.0020$. Multiplying the constants yields $0.00744 = i \times 0.00372$. Solving for $i$ gives $i = \frac{0.00744}{0.00372} = 2$. Since $100\%$ ionization is assumed, $i = n$. Therefore, exactly 2 ions are produced per molecule (one complex cation and one $Cl^-$ anion, meaning $n=1$ in the formula).