how many grams of CaCl2 would be required to be added to water to prepare 1 liter (1000 ml) of a 1.72% (w/v) aqueous solution of CaCl2?

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  1. 1.72% w/v means 1.72 g CaCl2 in 100 mL of solution so that would be 10x that for 1 L of solution or 17.2 g CaCl2. The answer to the problem, therefore, is 17.2 g CaCl2 but it makes a difference how the solution is prepared. You can NOT add 17.2 g CaCl2 to a liter (1000 mL) of water because that will be more than 1000 mL OF SOLUTION. You prepare it correctly by adding 17.2 g CaCl2 to a 1000 mL volumetric flask, adding some water, swirl until all of the CaCl2 is dissolved, then make the final volume to 1000 mL (by adding water to the mark on the flask).

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