Question

Prepare a solution of 1M iodine in 250mls of water. Molar 126.2g/mol

Answer 1

M(I₂) = 252.4 g/mol

Proportion for 1 M solution:

1 L – 252.4 g

0.250 L – x g

$x = \frac{0.250 \times 252.4}{1} = 63.1 \; g$

You need 63.1 g of I₂, but in real laboratory practice I₂ will not dissolve in pure water without KI.