# Question 1f4c5

Oct 18, 2017

${\text{4.00 mol L}}^{- 1}$

#### Explanation:

For starters, you know that the density of pure water is approximately equal to ${\text{1.00 g mL}}^{- 1}$. By comparison, this aqueous solution has a density of ${\text{1.25 g mL}}^{- 1}$.

The idea here is that the difference between the mass of $\text{1 mL}$ of pure water and the mass of $\text{1 mL}$ of this solution will represent the mass of the solute.

For $\text{1 mL}$ of this solution, you have

overbrace("1.25 g")^(color(blue)("mass of solution")) - overbrace("1.00 g")^(color(blue)("mass of solvent")) = overbrace("0.25 g")^(color(blue)("mass of solvent"))

Use the molar mass of the solvent to find the number of moles present in $\text{1 mL}$ of this solution.

0.25 color(red)(cancel(color(black)("g"))) * "1 mole solute"/(62.5color(red)(cancel(color(black)("g")))) = "0.00400 moles solute"

As you know, the molarity of the solution tells you the number of moles of solute present in ${10}^{3} \textcolor{w h i t e}{.} \text{mL" = "1 L}$ of the solution.

Since you know that $\text{1 mL}$ contains $0.00400$ moles of solute, you can say that ${10}^{3}$ $\text{mL}$ will contain

10^3 color(red)(cancel(color(black)("mL solution"))) * "0.00400 moles solute"/(1color(red)(cancel(color(black)("mL solution")))) = "4.00 moles solute"#

You can thus say that the molarity of the solution is equal to

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{molarity = 4.00 mol L}}^{- 1}}}}$

The answer is rounded to three sig figs.