Question

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

Answer 1

`"600 g CaCl"_2` would be required to make `"2 L"` of a `"3.5 M"` solution.

Explanation:

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

A `"3.5 M CaCl"_2"` solution contains `"3.5 moles"` `"CaCl"_2"/L solution"`.

We need to convert moles `"CaCl"_2` to grams. We do this by multiplying moles `"CaCl"_2` by its molar mass `("110.978 g/mol")`.

`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 `"300 g CaCl"_2"` are needed to make `"1 L"` of a `"3.5 M"` solution, `"600 g CaCl"_2` are needed to make `"2 L"` of a `"3.5 M"` solution.