# Question 1c29c

Oct 6, 2017

${\text{0.38 mol L}}^{- 1}$

#### Explanation:

A solution's molar concentration, or molarity, tells you the number of moles of solute, which in your case is sodium hydroxide, $\text{NaOH}$, present in exactly $\text{1 L}$ of the solution.

You know that your solution has a concentration of ${\text{15 g L}}^{- 1}$ sodium hydroxide, which means that $\text{1 L}$ of this solution contains $\text{15 g}$ of sodium hydroxide.

Now, in order to convert the number of grams of sodium hydroxide present in $\text{1 L}$ of this solution to moles, you need to use the molar mass of sodium hydroxide.

The problem provides you with the molar masses of its constituent elements

• ${\text{Na" = "23 g mol}}^{- 1}$
• ${\text{O" = "16 g mol}}^{- 1}$
• ${\text{H" = "1 g mol}}^{- 1}$

to find the molar mass of the compound. Since $1$ mole of sodium hydroxide contains $1$ mole of sodium, $1$ mole of oxygen, and $1$ mole of hydrogen, you can say that the molar mass of sodium hydroxide is equal to

${\text{23 g mol"^(-1) + "16 g mol"^(-1) + "1 g mol"^(-1) = "40 g mol}}^{- 1}$

This tells you that $1$ mole of sodium hydroxide has a mass of $\text{40 g}$.

You can thus say that the number of moles of sodium hydroxide present in $\text{1 L}$ of solution will be

15 color(red)(cancel(color(black)("g"))) * "1 mole NaOH"/(40color(red)(cancel(color(black)("g")))) = "0.375 moles NaOH"#

Therefore, the molarity of the solution will be

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{molarity = 0.38 mol L}}^{- 1}}}}$

I'll leave the answer rounded to two sig figs, the number of sig figs you have for the concentration in grams per liter.