# Question 322e0

Oct 20, 2017

$\text{9.1 g}$

#### Explanation:

The first thing that you need to do here is to figure out the number of moles of sodium chloride that is present in $\text{0.707 L}$ of $\text{0.22 M}$ 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 $\text{1 L}$ of this solution.

In your case, you know that

$\text{0.22 M " implies " 0.22 moles NaCl for every 1 L of solution}$

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.

$\text{58.44 g/mol " implies " 58.44 g for every 1 mole of NaCl}$

You will end up with

$1.5554 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles NaCl"))) * "58.44 g"/(1color(red)(cancel(color(black)("mole NaCl")))) = color(darkgreen)(ul(color(black)("9.1 g}}}}$

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