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0.1 "mole" AgNO(3) is added in 250 mL of...

0.1 "mole" `AgNO_(3)` is added in 250 mL of saturated solution of AgCl at `25^(@)C` without changing volume. Given: `K_(sp)` of `AgCl=1.0xx10^(-10)M^(2)`
Ionic conductance of `Ag^(+)"ion"=60Omega^(-1)cm^(2) "mol"^(-1)`
Ionic conductance of `Cl^(-)` ion =`75Omega^(-1)cm^(2) "mol"^(-1)`
Ionic conductance of `NO_(3)^(-) "ion"=75 Omega^(-1)`
`[Cl^(-)]`in the final solution is equal to :

A

`10^(9)M`

B

`2.5xx10^(-10)M`

C

`10^(-5)M`

D

`2.5xx10^(-7)M`

Text Solution

Verified by Experts

The correct Answer is:
B
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0.1 "mole" AgNO_(3) is added in 250 mL of saturated solution of AgCl at 25^(@)C without changing volume. Given: K_(sp) of AgCl=1.0xx10^(-10)M^(2) Ionic conductance of Ag^(+)"ion"=60Omega^(-1)cm^(2) "mol"^(-1) Ionic conductance of Cl^(-) ion = 75Omega^(-1)cm^(2) "mol"^(-1) Ionic conductance of NO_(3)^(-) "ion"=75 Omega^(-1) The conductivity of solution is:

0.1 "mole" AgNO_(3) is added in 250 mL of saturated solution of AgCl at 25^(@)C without changing volume. Given: K_(sp) of AgCl=1.0xx10^(-10)M^(2) Ionic conductance of Ag^(+)"ion"=60Omega^(-1)cm^(2) "mol"^(-1) Ionic conductance of Cl^(-) ion = 75Omega^(-1)cm^(2) "mol"^(-1) Ionic conductance of NO_(3)^(-) "ion"=75 Omega^(-1) If the solution is electrolysed using inert electrodes the expected electrode products are:

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