The molar conductivity of a 0.5 mol/dm3 solution of AgNO3 with electrolytic conductivity of 5.76 x 10-3 S cm-1 at 298 K is

  • Option 1)

    2.88 S cm2/mol

  • Option 2)

    11.52 S cm2/mol

  • Option 3)

    0.086 S cm2/mol

  • Option 4)

    28.8 S cm2/mol

 

Answers (1)

None

 

Formula of Molar Conductivity -

 \Lambda m (S\:cm^{2}mol^{-1})=\frac{\kappa (S\:cm^{-1})}{1000Lm^{-3}\times molarity(mol\:L^{-1})}

-

 

 \Lambda _{m} = \frac{K \times 1000}{Molarity (M)}

\frac{5.76 \times 10^{-3} Scm^{-1} \times 1000}{0.5 moI cm^{-3}}

=11.52 S cm ^{2}/moI

 


Option 1)

2.88 S cm2/mol

This is incorrect option

Option 2)

11.52 S cm2/mol

This is correct option

Option 3)

0.086 S cm2/mol

This is incorrect option

Option 4)

28.8 S cm2/mol

This is incorrect option

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