a 1.7 % solution of AGNO3 is isotonic with 3.4% solution of glucose.calculate the degree of ddissociation of AGNO3.{88 %}

Asked by PratibhaPandey | 11th May, 2014, 06:47: PM

Expert Answer:

Let us first calculate observed molar mass of AgNO3.

P(AgNO3) = P(glucose)

and P = w RT

MV

Osmotic pressure of AgNO3

P(AgNO3) = 1.7 RT

MX0.1

P(glucose) = 3.4 RT

MX0.1

1.7 RT = 3.4 RT

MX0.1 180 X0.1

M = 180 X 1.7 = 90

3.4

Normal molar mass of AgNO3 = 169

i = Normal molar mass = 169 = 1.88

Observed molar mass 90

AgNO3 dissociates as:

AgNO3 Ag+ + NO3-

If a is the degree of dissociation

Initial moles 1 0 0

Moles after dissociation 1-α α α

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

i=1-α + α + α = 1 + α

i=Moles of solute after dissociation = 1+α

Now, 1+α = 1.88

1

α = 1.88 - 1 = 0.88 = 88%

Answered by Vaibhav Chavan | 12th May, 2014, 06:26: PM