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.

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