# Question e8b23

Feb 22, 2017

Here's how you can do that.

#### Explanation:

The first thing to do here is to figure out the density of the solution by using its specific gravity.

The specific gravity of a substance is defined as the density of that substance divided by the density of water at ${4}^{\circ} \text{C}$, the temperature at which the density of water is maximum.

color(blue)(ul(color(black)("SG" = rho_"substance"/rho_ ("water at 4"^@"C"))))

You can approximate the density of water at ${4}^{\circ} \text{C}$ to

rho_ ("water at 4"^@"C") = "1.00 g mL"^(-1)

This means that the density of your cinnamaldehyde solution will be equal to

${\rho}_{\text{cinnamaldehyde" = 1.05 * "1.00 g mL}}^{- 1}$

${\rho}_{\text{cinnamaldehyde" = "1.05 g mL}}^{- 1}$

Now, molarity is defined as the number of moles of solute present in $\text{1.0 L}$ of solution.

At this point, your strategy will be to pick a $\text{1.0-L}$ sample of this solution and figure out how many moles of cinnamaldehyde, your solute, it contains.

Calculate the mass of the solution first by using its density

1.0 color(red)(cancel(color(black)("L"))) * (10^3color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * "1.05 g"/(1color(red)(cancel(color(black)("mL solution")))) = "1050 g"

This solution is 98% cinnamaldehyde by mass, which basically means that you get $\text{98 g}$ of cinnamaldehyde for every $\text{100 g}$ of solution.

In your case, the sample will contain

1050 color(red)(cancel(color(black)("g solution"))) * "98 g solute"/(100color(red)(cancel(color(black)("g solution")))) = "1029 g solute"

To convert this to moles of cinnamaldehyde, use the compound's molar mass

1029 color(red)(cancel(color(black)("g"))) * "1 mole cinnamaldehyde"/(132.16color(red)(cancel(color(black)("g")))) = "7.786 moles cinnamaldehyde"#

Since you know that the solution contains $7.786$ moles of solute in $\text{1.0 L}$ of solution, the volume of the sample, you can say that its molarity will be

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{molarity = 7.8 mol L}}^{- 1}}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the solution's percent concentration.