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_(

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 138 `ohm^(-1) cm^(2) "mol"^(-1)` then the solubility of AgCl in "mole"s per litre will be:
A. `10^(-5)`
B. `10^(-10)`
C. `10^(-14)`
D. `10^(-16)`

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

Salim Ahmed Kawsar

Correct Answer - A
`lambda^@=1000xx(K_(Sol)-k_(H_2O))/S`
`138=(1000xx(3.4-2.02)xx10^(-6))/S`
`S=1xx10^(-5)`

4 views Answered 2023-02-05 15:29

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