Question

30. A 0.01 m aqueous solution of Aici, freezes at - 0.068 °C. Calculate the
percentage of dissociation. [Given : Kc for Water = 1.86 K kg mol-1]​

Answer 1

Answer:

given kc=1.86

T=0--0.068=0.068

t=i*kc*m

0.068=i*1.86*0.01

i=3.65

degree of dissociation=$\frac{i-1}{n-1}$

where n=2

degree of dissociation=3.65-1

                                     =2.65=26.5%

Explanation:

Answer 2

Answer:

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Explanation: