what's the mass of barium chloride needed to make 275mL of a 0.280 mol/L solution.

Answer 1

Molar mass BaCl₂ = 208.2 g/mole

Converting 275 ml to liters, we have 275 ml x 1 L/1000 ml = 0.275 L

0.280 mol/L x 0.275 L = 0.077 moles BaCl₂ needed

0.077 moles x 208.2 g/mol = 16.0 g BaCl₂ needed (to 3 significant figures)