# Question #c0ec1

##### 1 Answer

#### Explanation:

I assume that you're interested in finding the *molarity* and the *molality* of the solution.

In both cases, you need to find the number of moles of solute, in your case sucrose, by using its molar mass.

#18.0color(red)(cancel(color(black)("g"))) * "1 mole sucrose"/(342.3color(red)(cancel(color(black)("g")))) = "0.05259 moles"#

Molarity is defined as number of moles of solute *per liters of solution*. To find the volume of the solution, you need to do a bit of research.

More specifically, you need to look up what the density of a solution that contains that much sucrose in that much water would be.

Find the percent concentration by mass of sucrose in the solution

#(18.0color(red)(cancel(color(black)("g"))))/((18.0 + 801.0)color(red)(cancel(color(black)("g")))) * 100 = "2.20%"#

The density of the solution can be approximated to be

http://homepages.gac.edu/~cellab/chpts/chpt3/table3-2.html

This means that the volume of the solution is

#(18.0 + 801.0)color(red)(cancel(color(black)("g"))) * "1 mL"/(1.0070color(red)(cancel(color(black)("g")))) = "813.3 mL"#

The molarity of the solution will thus be

#C = n/V#

#C = "0.05259 moles"/(818.3 * 10^(-3)"L") = color(green)("0.0647 M")#

Molality is defined as moles of solute per **kilograms** of solvent. In your case, the mass of the solvent, which is of course water, is equal to

This means that the solution's molality will be

#b = n/m#

#b = "0.05259 moles"/(801.0 * 10^(-3)"kg") = color(green)("0.0657 molal")#

Both values are rounded to three sig figs.