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For a saturated solution of AgCl at 25^(...

For a saturated solution of `AgCl` at `25^(@)C,k=3.4xx10^(-6)ohm^(-1)cm^(-1)` and that of `H_(2)O(l)` used is `2.02xx10^(-6)ohm^(-1)cm^(-1). lambda_(m)^(@)` for `AgCl` is `138ohm^(-1)cm^(2)mol^(-1)` then the solubility of AgCl in moles per liter will be

A

`10^(-5)`

B

`10^(-10)`

C

`10^(-14)`

D

`10^(-16)`

Text Solution

Verified by Experts

The correct Answer is:
A
`lambda^(@)=(100xx(k_("sol")-k_(H_(2)O)))/(S)`
`138=(100xx(3.4xx2.02)xx10^(-6))/(S)`
`S=1xx10^(-5)M` Ans.
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