1.

Solubility of AgCl at 20^(@)C is 1.435 xx 10^(-3) gm per litre. The solubility product of AgCl is

Answer»

`1 xx 10^(-5)`
`1 xx 10^(-10)`
`1.435 xx 10^(-5)`
`108 xx 10^(-3)`

Solution :`S = 1.435 xx 10^(-3) g//l, (1.435 xx 10^(-3))/(143.5) = 10^(-5) m`
`K_(SP) = S xx S = 10^(-10)`.


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