Answer
The necessary mass of potassium iodide is equal to 232 g.
Work Step by Step
$5.00 \times 10^{2} mL$ = $500$ $mL$ = 0.500L
1. Find the number of moles:
$Concentration(M) = \frac{n(mol)}{V(L)}$
$2.80 = \frac{n(mol)}{0.500}$
$2.80 * 0.500 = n(mol)$
$1.40\ moles = n(mol)$
2. Determine the molar mass of this compound (KI):
39.1* 1 + 126.9* 1 = 166 g/mol
3. Calculate the mass
$mm(g/mol) = \frac{mass(g)}{n(mol)}$
$mm(g/mol) * n(mol) = mass(g)$
$ 166 * 1.40 = mass(g)$
$232 = mass(g)$
