The value of observed and calculate molecular weight of silver nitrate are 92.64 and 170 respectively, the degree of dissociation of silver nitrate is

60%

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83.5%

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46.7%

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60.23%

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Solution

The correct option is B 83.5%
$i$ for $A g N O_{3} = \frac{n o r m a l m o l w t}{o b s e r v e d m o l w t}$
$= 1 + α$
$α = i - 1$
$= 17092.64 - 1$
$= 0.835 = 83.5 %$

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Similar questions

Q. The value of observed and calculate molecular weights of silver nitrate are

92.64

and

170

respectively.

The degree of dissociation of silver nitrate will be :

Q. The value of the observed and calculated molecular weight of silver nitrate is 92.64 and 170, respectively. The degree of dissociation of silver nitrate is:

Sodium chloride (

N a C l

) reacts with silver nitrate (

A g N O_{3}

) to produce silver chloride (

A g C l

) and sodium nitrate (

N a N O_{3}

). The weight of silver nitrate reacts with 1 mol of sodium chloride is _________ .

10 g NaCl solution is mixed with 17 g of silver nitrate solution. Calculate the weight of silver chloride precipitated.
AgNO₃ + NaCl $\overset{}{\to}$ AgCl + NaNO₃ [Ag = 108]

Q. Calculate the number of coulombs required to deposit

50 g

of silver at cathode from silver nitrate solution.
(Atomic mass of silver

= 108

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The value of observed and calculate molecular weight of silver nitrate are 92.64 and 170 respectively, the degree of dissociation of silver nitrate is

The correct option is B 83.5%i for AgNO3=normal mol wtobserved mol wt=1+αα=i−1=17092.64−1=0.835=83.5%

The correct option is B 83.5%
$i$ for $A g N O_{3} = \frac{n o r m a l m o l w t}{o b s e r v e d m o l w t}$
$= 1 + α$
$α = i - 1$
$= 17092.64 - 1$
$= 0.835 = 83.5 %$