A commercial sample of a protective colloid is only 80% pure by mass (the impurity has no protective action). 1.25 mg of the sample just protects 25 mL of a gold sol against coagulation by NaCl. What mass of the pure colloid, in mg, is needed to protect 1.5 L of the same gold sol? (nearest 0.1 mg)

Numerical answer type, answer in mg, tolerance 0.5.

Answer: 60 mg

Pure colloid in 1.25 mg of sample

= 1.25 \times 0.80 = 1.00

mg, which protects 25 mL. The gold number (per 10 mL) is

1.00 \times 10/25 = 0.40

mg. For 1500 mL the mass is

0.40 \times 1500/10 = 60.0

mg.