# Question 9a2cf

Nov 17, 2015

$\text{0.154 M}$

#### Explanation:

As you know, molarity is defined as moles of solute, which in your case is sodium chloride, $\text{NaCl}$, divided by liters of solution.

Now, a weight by volume percent concentration is defined as mass of solute per volume of solution. In this case, a $\text{0.90%w/v}$ saline solution will contain $\text{0.9 g}$ of sodium chloride for every $\text{100 mL}$ of solution.

Now, since molarity is determined by liters of solution, it's easier to take a $\text{1.00-L}$ sample of this saline solution and figure out how many moles of sodium chloride it contains.

Since you know that $\text{1.00 L}$ of saline solution contains $\text{9.000 g}$ of sodium chloride, use the compound's molar mass to determine how many moles you have

9.000color(red)(cancel(color(black)("g NaCl"))) * "1 mole NaCl"/(58.44color(red)(cancel(color(black)("g NaCl")))) = "0.154 moles NaCl"

This means that the sample's molarity will be

$c = \frac{n}{V}$

c = "0.154 moles"/"1.00 L" = color(green)("0.154 M")#

I'll leave the answer rounded to three sig figs, despite the fact that you only have two sig figs for the percent concentration of the solution.

SIDE NOTE The result must be exactly the same regardless of the volume sample you pick for the solution. I recommend using different volume samples to test this out.