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

1 Answer
Meave60
Aug 28, 2017

"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.

Related questions
See all questions in Molarity
Impact of this question
71115 views around the world
You can reuse this answer
Creative Commons License