How many grams of CaCl2 are needed to prepare 1.0 L of a 1.0M solution?
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Molarity (M) can be described as the number of moles of a substance present in a given litre of solution.
Molarity = Moles/Litre
So, a 1.0M solution of CaCl2 would contain 1 mole of CaCl2 dissolved in 1 litre of solution. This is the concentration of the particular solution.
We've made the comparison between molarity and moles, so let's now look at the relationship between the number of moles of a substance and it's weight in grams. This obviously depends on the atoms or molecules we are dealing with - 1 mole of NaCl, for example, would not weigh the same amount as CaCl2, as their molecular weights would be different.
A good point to remember is that 1 mole of a substance (atom, ion, molecule) will be that substances molecular weight in grams. For example, 1 mole of NaCl (sodium chloride) will weigh 58.442468 g. I got this from the molecular weight of the compound, which is the sum of the atomic weights of the individual atoms that it composes of. If we were to dissolve 58.443468 g of NaCl in 1 litre of a solvent, we would form a 1M solution.
Returning to our example, if you're able to find the molecular weight of CaCl2, you can then find the number of grams in 1 mole of the substance, which is the number of grams that we will use to form a 1M solution in 1 litre of solvent. Hope that helps! :)