Question

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

Answer 1

Answer:

= 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