Question #322e0

The first thing that you need to do here is to figure out the number of moles of sodium chloride that is present in #"0.707 L"# of #"0.22 M"# sodium chloride solution.

As you know, the molarity of the solution tells you the number of moles of sodium chloride, the solute, present in exactly #"1 L"# of this solution.

In your case, you know that

#"0.22 M " implies " 0.22 moles NaCl for every 1 L of solution"#

This implies that your solution will contain

#0.707 color(red)(cancel(color(black)("L solution"))) * "0.22 moles NaCl"/(1color(red)(cancel(color(black)("L solution")))) = "1.5554 moles NaCl"#

To convert the number of moles of sodium chloride to grams, use the compound's molar mass, which tells you the mass of exactly #1# mole of sodium chloride.

#"58.44 g/mol " implies " 58.44 g for every 1 mole of NaCl"#

You will end up with

#1.5554 color(red)(cancel(color(black)("moles NaCl"))) * "58.44 g"/(1color(red)(cancel(color(black)("mole NaCl")))) = color(darkgreen)(ul(color(black)("9.1 g")))#

The answer is rounded to two sig figs, the number of sig figs you have for the molarity of the solution.