The values of observed and calculated molecular weights of silver nitrate are 92.64 and 170 respectively. The degree of dissociation of silver nitrate will be

Question

The values of observed and calculated molecular weights of silver nitrate are 92.64 and 170 respectively. The degree of dissociation of silver nitrate will be (A) 60 % (B) 83.5 % (C) 46.7 % (D) 60.23 %

Tardigrade Admin · Accepted Answer

i=( text Normal mol. mass / text Observed mol. mass ) =(170/92.64)=1.835 α=(i-1/n-1) ; where, n=2 for AgNO 3 α=(1.835-1/2-1) =0.835=83.5 %

Q. The values of observed and calculated molecular weights of silver nitrate are 92.64 and 170 respectively. The degree of dissociation of silver nitrate will be

$60%$

$83.5%$

$46.7%$

$60.23%$

$i = \frac{Normal mol. mass}{Observed mol. mass}$

$= \frac{170}{92.64} = 1.835$

$α = \frac{i - 1}{n - 1};$ where, $n = 2$ for $A g N O_{3}$

$α = \frac{1.835 - 1}{2 - 1}$

$= 0.835 = 83.5%$

Q. The values of observed and calculated molecular weights of silver nitrate are 92.64 and 170 respectively. The degree of dissociation of silver nitrate will be

60%

83.5%

46.7%

60.23%

i= Observed mol. mass Normal mol. mass ​ =92.64170​=1.835 α=n−1i−1​; where, n=2 for AgNO3​ α=2−11.835−1​ =0.835=83.5%

$60%$

$83.5%$

$46.7%$

$60.23%$

$i = \frac{Normal mol. mass}{Observed mol. mass}$

$= \frac{170}{92.64} = 1.835$

$α = \frac{i - 1}{n - 1};$ where, $n = 2$ for $A g N O_{3}$

$α = \frac{1.835 - 1}{2 - 1}$

$= 0.835 = 83.5%$