Determine the molar solubility of agi in pure water. ksp (agi) = 8.51 × 10-17.

When the salt AgI dissolves in water it dissociates as follows;
AgI ---> Ag⁺ + I⁻
molar solubility is the number of moles that are dissolved in 1 L of solution.
If molar solubility of AgI is x, then molar solubility of Ag⁺ is x and I⁻ is x.
the formula for solubility product constant - ksp of AgI is given below 
ksp = [Ag⁺][I⁻]
ksp = (x)(x)
ksp = 8.51 x 10⁻¹⁷ 
therefore,
x² = 8.51 x 10⁻¹⁷
x = 9.22 x 10⁻⁹
since molar solubility of AgI is x, then molar solubility of AgI is 9.22 x 10⁻⁹ M

Answers are 1,2,4 its correct on edgenuity.

H tricky one let me think and get back to u