# Question 95762

Oct 7, 2017

${\text{0.2 mol L}}^{- 1}$

#### Explanation:

My guess is that you're referring to the mass of exactly $1$ mole of sodium hydroxide, $\text{NaOH}$.

The idea here is that you need to convert the number of grams of sodium hydroxide present in your solution to moles by using the compound's molar mass.

Sodium hydroxide has a molar mass of approximately ${\text{40 g mol}}^{- 1}$, which means that $1$ mole of sodium hydroxide has a mass of $\text{40 g}$.

This means that your sample contains

4 color(red)(cancel(color(black)("g"))) * "1 mole NaOH"/(40color(red)(cancel(color(black)("g")))) = "0.1 moles NaOH"

Now, the molarity of the solution tells you the number of moles of solute present in exactly $\text{1 L} = {10}^{3}$ $\text{mL}$ of solution. Since you know how many moles of solute are present in $\text{500 mL}$ of solution, you can say that ${10}^{3} \textcolor{w h i t e}{.} \text{mL" = "1 L}$ of solution will contain

10^3 color(red)(cancel(color(black)("mL solution"))) * "0.1 moles NaOH"/(500color(red)(cancel(color(black)("mL solution")))) = "0.2 moles NaOH"#

Therefore, you can say that the solution has a molarity of

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{molarity = 0.2 mol L}}^{- 1}}}}$

The answer is rounded to one significant figure.