The mol. mass odf NaCl determined by studying freezing point depression of its 0.5% aq. sol. is 30 . The degree of dissociation of NaCl will be?

Observed Molar Mass of NaCl = 30

Actual Molar Mass of NaCl = 58.5 g/mol

Van't Hoff Factor, i = Normal Molar Mass of NaCl / Observed Molar Mass of NaCl 

i = 58.5 / 30 = 1.95

NaCl ionizes as : 

NaCl ⇔ Na+ + Cl-

Let α be the degree of dissociation, then

Initial moles

1

0

0

Moles after dissociation

1- α

α 

α

Total number of moles after dissociation = 1 - α + α + α = 1 + α

We know that , i = Observed moles of solute / Normal moles of solute 

i = (1 + α) / 1

(1 + α) / 1 = 1.95

1 + α = 1.95

α = 1.95 – 1 = 0.95

Therefore, degree of dissociation of 0.5% aqueous solution of NaCl = 0.95 which means that at this concentration NaCl is 95% dissociated.

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