CBSE Class 12-science Answered
a 1.7 % solution of AGNO3 is isotonic with 3.4% solution of glucose.calculate the degree of ddissociation of AGNO3.{88 %}
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%