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At 298 K, The specific conductivity of a...

At 298 K, The specific conductivity of a saturated solution of silver chloride in water is `2.30xx10^(-5)Scm^(-1)`. Calculate its solubility in `gL^(-1)` at 298 K. Given `lambda_(m)^(@)(Ag^(+))" and "lambda_(m)^(@)(Cl^(-))` are 61.9 and 76.3 S `cm^(2)mol^(-1)` respectively.

Text Solution

Verified by Experts

The correct Answer is:
`2.382xx10^(-3) gL^(-1)`
`Lambda_(m(AgCl))^(oo)=Lambda_(m(Ag^(+)))^(oo)+Lambda_(m(Cl^(-)))^(oo)=(61.9+76.3)" S "mol^(-1)cm^(2)=138.2" S " mol"^(-1)cm^(2)`
Solubility (in mol `L^(-1)`) `=(kxx1000" cm"^(3)L^(-1))/(Lambda_(m)^(oo))=((2.3xx10^(-5)" S "cm^(-1))xx(1000 cm^(3)L^(-1)))/((138.2" S mol"^(-1)"cm"^(2)))`
`=0.0166xx10^(-2) mol L^(-1)`
Solubility (in g`L^(-1)`) `=(0.0166xx10^(-2)" mol "L^(-1))xx("Molar mass of" AgCl)`
`=(0.0166xx10^(-2)" mol "L^(-1))xx(143.5" g mo"l^(-1))=2.382xx10^(-2)" g "L^(-1)`.
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