Question

5 The molecular weight of NaCl determined by
studying freezing point depression of its 0.5%
aqueous solution is 30. The apparent degree of
dissociation of NaCl is​

Answer 1

Answer:

0.95.

Explanation:

We know the formulae of freezing point which is ΔTf = i*Kf*m in which the T(f) being the freezing point of the mixture, van't Hoff factor is i and the depression constant is kf with the molecular mass as m.  

Since, we know that the van't Hoff factor i is the given molar mass/ the observed molar mass which on substituting from the question we will get that 58.5/30 = 1.95(where 58.5 sis the molar mass of NaCl). If the degree of dissociation is taken as a. So, 1+a will be 1.95 which on solving we will get a=0.95.

Answer 2

Answer: 0.95

Explanation:

Nacl= na+ + cl-

1- a. a. a

i = 1-a+2a

i=1+a

a= i-1

i = normal molecular mass /observed molecular mass

= 58.5/30

= 1.90

a = 1.90-1

a=0.90

Apparent degree of dissociation of Nacl is 0.90