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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

Text Solution

Verified by Experts

The correct Answer is:
9
For `AgCl` molarity `=` normality
`k=3.41xx10^(-6)-1.60xx10^(-6)=1.81xx10^(-6)Omega^(-1)cm^(-1)`
For saturated solution of sparingly soluble salt,
`lamda_(V)=lamda^(oo)` and solubility `=` normality
`:. lamda_(V)=lamda^(oo)=(1000xxk)/s`
`:.s=(1000xxk)/(lamda^(oo))=(1000xx1.81xx10^(-6))/138.3=1.31xx10^(-5)`
Now,
`K_(sp(AgCl))=s^(2)=(1.31xx10^(-5))^(2)=0.176xx10^(-9)`
`:. x=9`
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