# Question 9d6d5

May 20, 2016

${\text{3.1 mol L}}^{- 1}$

#### Explanation:

The idea here is that the volume of the solution will decrease, but that the number of moles of sodium hydroxide, $\text{NaOH}$, will remain unchanged.

Since molarity is defined as number of moles of solute per liter of solution, decreasing the volume of the solution while keeping the number of moles of solute constant will increase the solution's concentration.

That happens because you have the same number of moles of solute in a smaller volume of solution.

The new volume of the solution will be

${V}_{\text{new" = "495 mL" - "16 mL" = "479 mL}}$

The initial solution contained

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{c = {n}_{\text{solute"/V_"solution" implies n_"solute" = c * V_"solution}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

n_"NaOH" = "3.0 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(495 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(blue)("volume in liters"))

${n}_{\text{NaOH" = "1.485 moles NaOH}}$

The final solution will thus have a molarity of

c_"final" = "1.485 moles"/(479 * 10^(-3)"L") = color(green)(|bar(ul(color(white)(a)"3.1 mol L"^(-1)color(white)(a/a)|)))#

The answer is rounded to two sig figs.

As predicted, the molarity of the solution increased as a result of the decrease in volume.