What is the mass in grams of BaCl2 that is needed to prepare 200 mL of a 0.500 M solution

= 20.82 g of BaCl2

Explanation:

Given,

Volume = 200 mL

Molarity = 0.500 M

Therefore;

Moles = molarity * volume

= 0.2 L * 0.5 M

= 0.1 mole

But; molar mass of BaCl2 is 208.236 g/mole

Therefore; 0.1 mole of BaCl2 will be equivalent to;

= 208.236 g/mol x 0.1 mol

= 20.82 g

Therefore, the mass of BaCl2 in grams required is 20.82 g