Question #e8b23

1 Answer
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^@"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^@"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_"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 "1.0 L" of solution.

At this point, your strategy will be to pick a "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 "98 g" of cinnamaldehyde for every "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 "1.0 L" of solution, the volume of the sample, you can say that its molarity will be

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.