1.

The solubility of AgCl in 0.1M NaCI is (K_(sp) " of AgCl" = 1.2 xx 10^(-10))

Answer»

`0.1 M `
` 1.2 XX 10 ^(-5) `
` 1.095 xx 10 ^(-5)`
` 1.2 xx 10 ^(-9) `

SOLUTION :`AGCL hArr AG^(+) +Cl^(-) `
` "" s "" 0.1`
`KSP= 5 xx 0.1 rArr S =1.2 xx 10 ^(-9) `


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