Question

25. 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
(1) 0.95
(2) 0.45
(3) 0.60
(4) 0.35​

Answer 1

answer : option (1) 0.95

explanation : observed molecular weight of NaCl = 30

actual molecular weight of NaCl = 58.5

using formula,

i = actual molecular weight/observed molecular weight

= 58.5/30 = 1.95 ......(1)

see dissociation of NaCl,

NaCl ⇔ Na^+ + Cl^-

at eq 1 - α α α

i = 1 - α + α + α = 1 + α

from equation (1),

1.95 = 1 + α ⇒α = 0.95

hence, degree of dissociation of NaCl is 0.95

hence, option (1) is correct choice.

Answer 2

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