What minimum concentration of clâ is required to begin to precipitate pbcl2? for pbcl2, ksp=1.17ã10â5?

Missing in your question : [Pb2+] = 0.085 M

So from this balanced equation:
PbCl2  ↔ Pb+2   +   2Cl-

when Ksp = [Pb+2][Cl]^2
when we have Ksp = 1.17x10^-5 & [Pb+2] = 0.085 M so by substitution:
1.17x10^-5 = 0.085 [Cl-]^2

∴[Cl]^2 = (1.17x10^-5) / 0.085 = 1.38x10^-4
∴[Cl-] = √0.0002
          = 0.0117 M
So 0.0117 M is the minimum concentration requires precipitating PbCl2

Molarity is moles of solute over liters of solution. idk if that is what you are talking about, but you can convert moles of solute to grams of solute using dimensional analyzes by using molar mass found on the periodic table.

Rate law for this reaction israte of the reaction = k[a]²[b].second order in a  means concentration of a in power of two, and  first order in b means concentration of b in power of one.