Question

5) The following are both water soluble, find the number of moles of the particles which will be released into solution:

a. Calcium chloride is a water soluble ionic compound, if 0.075 mol CaCl₂ are dissolved in water, how many moles of particles will be released into solution?

b. Acetone is a water soluble molecular compound, if 0.134mol C₃H₆O is dissolved in water how many moles of particles will be released into solution?

Answer 1

The chemical formula for Calcium Chloride is CaCl₂, it a white, ionic and water soluble substance which on dissociation in water furnishes 3 ions i.e., 1 Ca²⁺, and two Cl^- ions.

The reaction for dissociation of Calcium Chloride in water can be written as:

CaCl₂ (s)  -----------------> Ca²⁺ (aq) + 2Cl^- (aq)

From the above reaction it is clear that 1 mole of CaCl₂ release 1 mole of Calcium ions and two moles of chloride ions, thus, in total, 1 mole of CaCl₂ liberates 3 mole of ions into the solution.

1 mole of CaCl₂ releases = 3 mole of ions

0.075 mol of CaCl₂ would release = 3 (0.075 mol) of ions or particles

Hence, 0.075 mol of CaCl₂ release 0.225 mol of particles.

b. Acetone is a well-known solvent used in many organic reactions, it is a transparent, covalent liquid with low boiling point. It dissolves in water readily by forming hydrogen bond, however being covalent in nature it does not dissociates in water, thus it does not liberate any ion in the solution.

CH₃COCH₃ (l) --------------> CH₃COCH₃ (aq)

From the above equation it is clear that the number of particle before and after dissolution remains same, thus, after dissolution of acetone in water number of particles would remain 0.134 mol