Q. 334.1( 22 Votes )

19.5 g of CH

Answer :

Given- w1 = 500g


W2 = 19.5g


Kf = 1.86 K kg mol-1


Molar mass of CH2FCOOH = 12 + 2 + 19 + 12 + 16 + 16 + 1


= 78 g mol-1


The depression in freezing point is calculated by,


ff (where, m is the molality)





f (calculated) = 0.93


To find out the vant Hoff’s factor, we use the formula,





CH2FCOOH CH2FCOO- + H +


To find out the degree of dissociation α, we use




With degree of dissociation, known to us, we can easily calculate the dissociation constant, Ka using the formula,





Thus, the vant Hoff’s factor is 1.07 an the dissociation constant is 2.634x10-3


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :

A statement of asChemistry - Exemplar

Match the laws giChemistry - Exemplar