Find the Vant Hoff factor of a mixture of two moles of KI with 1 mole of HgI₂ in a solution of water.

To determine the Van 't Hoff factor (i) for a mixture of potassium iodide (KI) and mercury(II) iodide (HgI2) in water, we first need to understand what the Van 't Hoff factor represents. It indicates the number of particles into which a solute dissociates in solution. Let's break down the components of your mixture and calculate the Van 't Hoff factor step by step.

Understanding the Components

In your mixture, we have:

• 2 moles of KI: Potassium iodide dissociates completely in water into potassium ions (K⁺) and iodide ions (I^-). Therefore, 1 mole of KI produces 2 moles of particles.
• 1 mole of HgI₂: Mercury(II) iodide, when dissolved, dissociates into one mercury ion (Hg²⁺) and two iodide ions (I^-). Thus, 1 mole of HgI₂ produces 3 moles of particles.

Calculating the Total Number of Particles

Now, let’s calculate the total number of particles produced by each component:

• From 2 moles of KI:
  • 2 moles of KI × 2 particles/mole = 4 particles
• From 1 mole of HgI₂:
  • 1 mole of HgI₂ × 3 particles/mole = 3 particles

Summing Up the Total Particles

Now, we can add the particles from both solutes:

• Total particles = 4 (from KI) + 3 (from HgI₂) = 7 particles

Determining the Van 't Hoff Factor

The Van 't Hoff factor (i) for the entire solution can be calculated by taking the total number of particles produced and dividing it by the total number of moles of solute present:

• Total moles of solute = 2 moles of KI + 1 mole of HgI₂ = 3 moles
• Van 't Hoff factor (i) = Total particles / Total moles of solute = 7 particles / 3 moles = 2.33

Final Thoughts

In summary, the Van 't Hoff factor for the mixture of 2 moles of KI and 1 mole of HgI₂ in water is approximately 2.33. This factor is crucial in colligative properties calculations, as it helps us understand how the presence of solutes affects properties like boiling point elevation and freezing point depression.