Question #c0ec1

1 Answer
Sep 20, 2015

Molarity: #"0.0647 M"#
Molality: #"0.0657 molal"#


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 #"1.0070 g/mL"#

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 #"801.0 g"#.

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.