find the number of millimoles of solute in:
a. 2.00 L of 2.76 x 10-3 M KMnO4
b. 250 mL of a solution that contains 4.20 ppm of CuSO4
a) CM = n / V
n = CM x V = 2.76 x 10-3 x 2.00 = 5.52 x 10-3 mol = 5.52 mmol
b) ppm = mg/L
m (CuSO4) = 4.20 x 0.25 = 1.05 mg = 0.00105 g
n = m / M
M (CuSO4) = 159.6 g/mol
n (CuSO4) = 0.00105 / 159.6 = 0.00000658 mol = 0.00000000658 mmol
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