Answer to Question #114248 in General Chemistry for Sophia

Question

Excess aqueous NaI was added to 50 mL of
aqueous CuNO3 solution, and 19.81 g of CuI
precipitate was formed. What was the molar concentration of CuNO3 in the original
solution?
Answer in units of M

Answer 1

V(CuNO₃) = 50 mL;

m(CuI) = 19.81 g;

M(CuI) = 190.45 g/mol;

CuNO₃ + NaI => CuI + NaNO₃

n=m/M;

n(CuI)= 19.81 g / (190.45 g/mol) = 0.104 moles;

According to the reaction: n(CuNO₃) = n(CuI) = 0.104 moles;

Molar concentration C = n/V;

C(CuNO₃) = 0.104 mol/50 mL= 0.00208 mol/mL = 2.08 mol/L;

Answer: 2.08 M