Question

How many grams of BaCl2 are required to make 1.50L of a 0.60M solution?

Answer 1

M=n.M
We have C=0.60M, V=150L, M[BaCl2]=208g/mol
—>n=CxV = 0.60x1.50=0.9mol
So m= 0.9x208=187.2gram

Answer 2

Answer:

187.407g

Explanation:

Moles of solute = Molarity × Liters of solution

Multiply 0.60 M by 1.50L:

0.60 mol/ 1L×1.50L=0.9 mol

To obtain the mass of solute, we will need to the molar mass of BaCl2, which is 208.23g/mol:

Finally, multiply the number of moles by208.23g /mol

0.9mol×208.23g/1mol =187.407g

Boom, you have a mass of:

187.407g