12.25 ~g of CH ₃ CH ₂ CHClCOOH is added to 250 ~g of water to make a solution. If the dissociation constant of above acid is 1.44 10⁻³ , the depression in freezing point of water in ^ C is (K_f . for water is .1.86 ~K ~kg ~mol ⁻¹ )

Step-by-step solution

Step I Calculation of degree of dissociation Molar concentration of solution aligned & = Mass of solute Molar mass of solute Mass of solvent 1000 & = 12.25 ~g 1000 122.5 ~g ~mol ⁻¹ 250 ~g & aligned & Molecular weight of solute ( CH ₃ CH ₂ CHClCOOH ) = & =122.5 ~g / mol ] aligned & =0.40 ~m aligned If is the degree of dissociation of CH ₃ CH ₂ CHClCOOH , then aligned & K_a= C ^2 (1- ) C ^2 & = K_a C & = 1.44 10⁻³ 0.4 = 36 10⁻⁴ =0.06 aligned Step II Calculation of van't-Hoff factor CH ₃ CH ₂ CHClCOOH CH ₃ CH ₂ CHClCO