A solution of ( 1.25 ~g ) of ' ( P ) ' in ( 50 ~g ) of water lowers freezing point by ( 0.3 ^ C ). Molar mass of ' ( P ) ' is 94. ( K_ f ( water ) =1.86 Kg mol ⁻¹ ). The degree of association of ' ( P ) ' in water is

Step-by-step solution

Given, w_ B =1.25 ~g w_ A = mass of solvent =50 ~g T_ f =0.3^ C Molecular mass of P =94K_ f =1.86 ~K ~kg ~mol ⁻¹ T_ f =0-(-0.3)=0.3^ C M_ B = K_ f w_ B 1000 T_ f w_ A M_ B = 1.86 1.25 1000 0.3 50 =155 Now, van't Hoff factor, i= Normal molar mass Observed molar mass = 94 155 =0.6064 Initial moles Moles after association 1- a / 2 (If a is the degree of association) Degree of association ( )= n(1-i) n-1 = 2 (1-0.6064) 1 =78.7 %