# Question #e8b23

##### 1 Answer

#### Answer:

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 *maximum*.

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

You can approximate the density of water at

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

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

#rho_"cinnamaldehyde" = 1.05 * "1.00 g mL"^(-1)#

#rho_ "cinnamaldehyde" = "1.05 g mL"^(-1)#

Now, **molarity** is defined as the number of moles of solute present in

At this point, your strategy will be to pick a *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 **by mass**, which basically means that you get **for every**

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 **moles** of solute in

#color(darkgreen)(ul(color(black)("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.