To prepare 1.00 L of 14g BaCl2, what Molarity would it be?

The molarity would be 0,067M.

Explanation:

The molarity corresponds to a concentration measurement that expresses the moles of solute in 1 liter of solution. To calculate the molarity we must first calculate the weight of 1 mole of BaCl2, obtaining the atomic weights of Ba and Cl from the periodic table:

Weight 1 mol BaCl2= Weight Ba + 2 x(Weight Cl)= 137, 33 g + 2 x35, 45 g=

Weight 1 mol BaCl2= 208, 23 g/mol

208, 23 g1 mol BaCl2

14 g         x= (14 g   x 1 mol BaCl2)/208, 23 g=0,067 mol NaCl2

We have 0,0632 mol of BaCl2 in 1,00 Liter of solution; thus the solution is 0,067 M.

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