Given that 25.0 mL of 0.100 M lead(II) nitrate reacts with 37.5 mL of lithium iodide solution

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Given that 25.0 mL of 0.100 M lead(II) nitrate reacts with 37.5 mL of lithium iodide solution according to the equation 

Pb(NO3)2(aq) + 2 LiI(aq) → PbI2(s) + 2 LiNO3(aq)

(a) What is the molarity of the LiI solution?

(b) What is the mass of PbI2 precipitate?

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