Tincture of iodine used as an antiseptic contains 2.6 g of iodine per 100 mL of the alcohol-water solution. A bottle holds 85 mL of it. Taking I = 127, calculate the amount of iodine ( $I X_{2}$ ) in the bottle in millimoles, to two decimal places.

Numerical answer type, answer in mmol, tolerance 0.01.

Answer: 8.7 mmol

Mass of iodine =

2.6 \times 85/100 = 2.21 g

. Molar mass of

I X_{2}

2 \times 127 = 254

g/mol. Moles =

2.21/254 = 8.70 \times 1 0^{- 3}

mol, i.e. 8.70 mmol.