1.

The specific conductance of a saturated solution of AgCl at25^(@) Cis 2.28 xx 10^(-6) S cm^(-1) . If solubility ofAgCl" is " 2.35 xx 10^(-x) gL^(-1) ( lambda_(AgCl)^(@) = 138.3 S cm^(2) mol^(-1)) then x is

Answer»


Solution :`S = M = ( K xx 1000)/(^^_M) = (2.28 xx 10^(-6) xx 10^(3))/(138.3) , S = 0.016486 xx 10^(-3)`
`=1.6486 xx 10^(-2) xx 10^(-3) = 1.65 xx 10^(-5) = " mole/L" = 2.36 xx 10^(-3)g//L`


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