What is the molarity if 6664 grams of boron iodide is dissolved in 5.25 liters of water?

3.24 mol/L

Explanation:

Given that:

mass of Boron triiodide = 6664 grams

molar mass of BI_3 = 391.52 g/mol

Recall that:

number of moles = mass/molar mass

∴

number of moles = 6664 g /391.52 g/mol

number of moles = 17.02 mol

Also;

Molarity = moles for solute/liter for solution

= 17.02 mol/5.25 L

= 3.24 mol/L