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:

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 $A g N O_{3} = \frac{normal mol.wt}{observed mol wt}$

$= 1 + α$

$α = i - 1$

$= 17092.64 - 1$

$= 0.835 = 83.5$ %

So, the correct option is $B$

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Similar questions

Q. 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

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]

Silver nitrate is 85% by mass. To prepare 125ml of 0.05molar silver nitrate solution , silver nitrate sample required is?

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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:

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%So, the correct option is B

The correct option is B 83.5%
$i$ for $A g N O_{3} = \frac{normal mol.wt}{observed mol wt}$

$= 1 + α$

$α = i - 1$

$= 17092.64 - 1$

$= 0.835 = 83.5$ %

So, the correct option is $B$