Question

The value of observed and the calculated molecular weight of silver nitrate are 92.64 and 170 respectively degree of dissociation of the silver nitrate will be

Answer 1

Answer: Degree of dissociation of the silver nitrate will be 83.50 %.

Explanation:

$AgNO_3(aq)\rightarrow Ag^+(aq)+NO_3^-(aq)$

The observed molecular mass of $AgNO_3$=92.64 g/mol

The calculated molecular mass of $AgNO_3$=170 g/mol

i = Van't Hoff factor

$i=\frac{\text{calculated molecular mass}}{\text{observed molecular mass}}$

$i=\frac{170 g/mol}{92.64 g/mol}=1.8350$

n = 2

$\alpha =\frac{(i-1)}{(n-1)}$

$\alpha$ = degree of dissociation

n =  number ions produced by single formula unit.

$\alpha =0.8350=83.50 \%$

Degree of dissociation of the silver nitrate will be 83.50 %.

Answer 2

Explanation:

AgNO

3

(aq)→Ag

+

(aq)+NO

3

−

(aq)

The observed molecular mass of AgNO_3AgNO

3

=92.64 g/mol

The calculated molecular mass of AgNO_3AgNO

3

=170 g/mol

i = Van't Hoff factor

i=\frac{\text{calculated molecular mass}}{\text{observed molecular mass}}i=

observed molecular mass

calculated molecular mass

i=\frac{170 g/mol}{92.64 g/mol}=1.8350i=

92.64g/mol

170g/mol

=1.8350

n = 2

\alpha =\frac{(i-1)}{(n-1)}α=

(n−1)

(i−1)

\alphaα = degree of dissociation

n = number ions produced by single formula unit.

\alpha =0.8350=83.50 \%α=0.8350=83.50%

Degree of dissociation of the silver nitrate will be 83.50 %.

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