For a saturated solution of `AgCl` at `25^(@)C`, specific conductance is `3.41xx10^(-6)ohm^(-1) cm^(-1)` and that of water used for preparing the solution is `1.60xx10^(-6)ohm^(-1)cm^(-1)`. If `lamda_(AgCl)^(oo)=138.3ohm^(-1)cm^(2)eq^(-1)`, the `K_(sp)` of `AgCl` is `0.17xx10^(-x)M^(2)`. The value of `x` is :

Answer

9

Explanation

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`