How much water should be added to 16.6 g CaCl2 (s) (molar mass 110.9 g/mol) to prepare a solution such that the concentration of Cl– (aq) is 0.60 M? Assume that the volume of CaCl2 is negligible.
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n(CaCl2)=16.6/110.9 = 0.1496844 mol = 0.15 mol of Cl2, which is 0,30 mol of Cl
(because Cl2--> 2 Cl )
0.60 M = 0.60 mol/liter
0,30 mol / ? L = 0.6
so 0.30/0.6 = 0.50 L of water
(because Cl2--> 2 Cl )
0.60 M = 0.60 mol/liter
0,30 mol / ? L = 0.6
so 0.30/0.6 = 0.50 L of water
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That's not the way it works. You dissolve the solute in water, and then dilute the resulting solution to the final volume. You don't determine the amount of water needed. That's for computing molality. The question is flawed, even with the statement about the negligible volume of CaCl2.
We can compute the volume of the solution, but that may not be the actual amount of water that is added. I certainly wouldn't advocate that to my students. Again, it's a poor question.
The molar mass of CaCl2 is NOT 110.9 g/mol. Again, poor question design. The molar mass, to three decimal places, is 110.986 g/mol, which rounds to a tenth as 111.0 g/mol. And if we started with atomic weights to the nearest tenth, we would get 111.1 g/mol, not 110.9.
16.6 g CaCl2 x (1 mol CaCl2 / 111.0 g) x (2 mol Cl / 1 mol CaCl2) x (1.00 L / 0.60 mol Cl) = 0.50 L or 500 mL of SOLUTION (not added water).
We can compute the volume of the solution, but that may not be the actual amount of water that is added. I certainly wouldn't advocate that to my students. Again, it's a poor question.
The molar mass of CaCl2 is NOT 110.9 g/mol. Again, poor question design. The molar mass, to three decimal places, is 110.986 g/mol, which rounds to a tenth as 111.0 g/mol. And if we started with atomic weights to the nearest tenth, we would get 111.1 g/mol, not 110.9.
16.6 g CaCl2 x (1 mol CaCl2 / 111.0 g) x (2 mol Cl / 1 mol CaCl2) x (1.00 L / 0.60 mol Cl) = 0.50 L or 500 mL of SOLUTION (not added water).