Achemist wants to make 6.0 l of a 0.260 m cacl2 solution. part a what mass of cacl2 (in g) should the chemist use?

In order to solve this problem the formula for molarity has to be revisited. The formula for molarity is M=(mole/volume). We are given the volume which is 6.0 L and the molarity of the solution of CaCl2 is 0.260 M. So, rearranging the molarity formula we can solve for the amount of moles in the 6.0 L solution of CaCl2 (0.260M * 6.0L = 1.56 mol CaCl2). Now using stoichiometry the we can find the grams of CaCl2 needed to make this solution. (1.56mol CaCl2) * (110.984 g / 1 mol CaCl2) = 173.135 g CaCl2 . So with the stoichiometric calculation we found that 173.153 g of CaCl2 are needed. Then following significant figures we can conclude that we need 173. g CaCl2.

While unproven, it uses facts to support part of a conclusion. it is still factual and has predictive power, however.