# How many grams of CaCl_2 would be required to produce a 3.5 M solution with a volume of 2.0 L?

Aug 28, 2017

${\text{600 g CaCl}}_{2}$ would be required to make $\text{2 L}$ of a $\text{3.5 M}$ solution.

#### Explanation:

"Molarity"=("moles of solute")/("liters of solution")

A $\text{3.5 M CaCl"_2}$ solution contains $\text{3.5 moles}$ $\text{CaCl"_2"/L solution}$.

We need to convert moles ${\text{CaCl}}_{2}$ to grams. We do this by multiplying moles ${\text{CaCl}}_{2}$ by its molar mass $\left(\text{110.978 g/mol}\right)$.

3.5color(red)cancel(color(black)("mol CaCl"_2))xx(110.978"g CaCl"_2)/(1color(red)cancel(color(black)("mol CaCl"_2)))="300 g CaCl"_2" rounded to two significant figures

Since $\text{300 g CaCl"_2}$ are needed to make $\text{1 L}$ of a $\text{3.5 M}$ solution, ${\text{600 g CaCl}}_{2}$ are needed to make $\text{2 L}$ of a $\text{3.5 M}$ solution.