1.

Van't hoff factor for a dilute aqueous solution of hcn is 1.00002 .the percent degree of dissociation of the acid is​

Answer»

ANSWER:

DEGREE of DISSOCIATION (a) = I -1/n-1

a = 1.00002 - 1/2-1

a = 0.0002



Discussion

No Comment Found

Related InterviewSolutions