Answer to Question #349072 in General Chemistry for baylee

Question

How many grams of CuCl2·2H2O are needed to prepare 0.500 L of solution containing 0.250 M Cl-(aq)?

Answer 1

C_M (Cl^-) = 0,25 M

V = 0,5 L

M_r (CuCl₂ • 2H₂O) = 171 g/mol

n(Cl^-) = C_M * V = 0,25 * 0,5 = 0,125 mol Cl^-

CuCl₂ • 2H₂O => Cu²⁺ + 2 Cl^- + 2H₂O

1mol CuCl₂ • 2H₂O => 2mol Cl^-

X => 0,125mol Cl^-

X = 0,125 * 1 / 2 = 0,0625 mol

m = n * M_r

m(CuCl₂ • 2H₂O) = 0,0625 * 171 = 10,6875 g

Answer: 10,6875 g CuCl₂ • 2H₂O