An aqueous solution is prepared by dissolving 0.5 moles of MgCl ₂ in 342 g of water. If the vapour pressure of pure water at the given temperature is 760 mm Hg and the vapour pressure of the solution is 722 mm Hg , the percentage dissociation of MgCl ₂ in the solution is _____.

Step-by-step solution

Moles of water ( n_ solvent ) = 342 18 = 19 moles Moles of MgCl ₂ ( n_ solute ) = 0.5 moles Using Raoult's law for relative lowering of vapour pressure: p^ - p_s p_s = i n_ solute n_ solvent 760 - 722 722 = i 0.5 19 38 722 = i 0.5 19 1 19 = i 0.5 19 i 0.5 = 1 i = 2 For MgCl ₂ , dissociation yields 3 ions ( Mg ²⁺ and 2 Cl ^- ), so n = 3 . Van 't Hoff factor i = 1 + (n - 1) 2 = 1 + (3 - 1) 2 = 1 + 2 = 0.5 Percentage dissociation = 0.5 100 = 50 % Answer: 50