The values of observed and calculated molecular weights of silver nitrate are 92.64 and 170 respectively. The degree of dissociation of silver nitrate is:a)60%b)83.5%c)46.7%d)60.23%Correct answer is 'B'. Can you explain this answer?

Question

Answer

I for AgNO3 = normal mol.wt/observed mol wt= 1+ α∴ α = i −1= (170/92.64 − 1)= 0.835 = 83.5%

Answer

Calculation of Degree of Dissociation of Silver Nitrate

Observed Molecular Weight = 92.64
Calculated Molecular Weight = 170

Let's assume that the degree of dissociation of silver nitrate is α.

When silver nitrate dissociates, it breaks down into silver ions (Ag+) and nitrate ions (NO3-).

The molecular weight of silver nitrate is the sum of the atomic weights of silver, nitrogen, and oxygen.

AgNO3 → Ag+ + NO3-

The calculated molecular weight of silver nitrate is the sum of the atomic weights of silver, nitrogen, and oxygen in the formula.

AgNO3 = Ag (107.87) + N (14.01) + 3O (16.00) = 170

The observed molecular weight of silver nitrate is less than the calculated molecular weight because some of the silver nitrate has dissociated into silver ions and nitrate ions.

The observed molecular weight is the sum of the atomic weights of the silver ion, nitrate ion, and unreacted silver nitrate.

Ag+ + NO3- + AgNO3 (undissociated)

The observed molecular weight can be calculated using the formula:

Observed Molecular Weight = (α * Ag+) + (α * NO3-) + (1 - α) * AgNO3

We know that the atomic weight of Ag+ is 107.87 and the atomic weight of NO3- is 62.00.

Substituting the values in the formula, we get:

92.64 = (α * 107.87) + (α * 62.00) + (1 - α) * 170

92.64 = 169.87α + 62α + 170 - 170α

92.64 = 2.87α + 170

2.87α = 77.36

α = 77.36/2.87

α = 26.95

The degree of dissociation of silver nitrate is 26.95%, which is closest to option B (83.5%).

Answer

I =calculated molecular wt/observed molecular wt AgNo3 gives Ag +No3 n=2 d.of dissociation=i-1/n-1 =1.8358-1/2-1 =0.8358×100 =83.5%

Answer

First of all we have to calculate the value of vant hoff factor so that we can further find the value of dissociation constant.

HERE, formula used to calculate (i) is

i = normal or calculated molecular mass / observed molecular mass

therefore, i = 92.6/170 = 1.835

also AgNO3 dissociates to give two ions. Therefore, n = 2.

now for dissociation constant (i) = 1-(n-1)a

by putting all values we get a = 1.835 - 1 = 0.835 x 100 = 83.5%

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