What mass of calcium chloride, CaCl2, is needed to prepare 2.85 L of a 1.56 M solution?

Question

What mass of calcium chloride, CaCl₂, is needed to prepare 2.85 L of a 1.56 M solution? 

316 grams

203 grams

493 grams

25.0 grams

Answer 1

First, find the molar mass of CaCl₂:

On the periodic table, we can see that the molar mass of Ca is 40.08 g/mol and the molar mass of Cl is 35.45 g/mol. There are 1 Ca and 2 Cl's in CaCl₂ so 40.08 + 2(35.45) = 110.98 g/mol of CaCl₂

Next, use the fact that M = mol/L to solve for n (the number of moles):

1.56 M = n mol/2.85 L --> multiply both sides by 2.85 L to isolate n --> n = 4.446 moles of CaCl₂

We want grams so we need to multiply g/mol by mol to cancel the moles and get grams:

110.98 g/mol * 4.446 moles = 493 g of CaCl₂

[Side note: some chemistry teachers want you to use sig figs rules after every step, while other teachers want you to round only at the end. From the answer choices, your teacher is doing the latter (if we had rounded before, we would have gotten 4.45 moles, which would have made us get 494 g as a final answer).]