A solution of a sugar with the chemical formula (c12h22o11) is prepared by dissolving 8.45 g in 250.0 ml of water at 25

c. the density of this solution is 1.21 g/cm3. calculate the concentration in terms of molarity, molality, weight percent and ppm. assume that the volume of the solution is equal to the volume of the solvent. show your work.

A. Concentration in terms of molarity

First, calculate for the molar mass of the given solution, C12H22O11.

m = 12 (12) + 22 (1) + 11 (16) = 232

Calculate for the number of moles.

n = 8.45 g / 232 g/mol = 0.036422 moles

Divide the determined number of moles by volume of solution in L.

V (solution) = (250 mL) (1 L/1000 mL) = 0.25 L

The concentration in molarity is determined by dividing the number of moles by the volume of solution in liters.

M = 0.03622 moles / 0.25 L = 0.146 moles/L

b. Concentration in terms of molality

Calculate the mass of the solvent in kilogram given the volume and its density.

m (solvent) = (1.21 g/cm3) (250 mL) (1 kg/1000 g) = 0.3025 kg

Divide the calculated number of moles by the mass of the solvent in kilograms.

molality = 0.036422 moles/0.3025 kg = 0.12 molal

c. Concentration in terms of weight percent.

The total weight of the solution is equal to 302.5 g.

weight percent of solute = (8.45 g / 302.5 g) (100%) = 2.79%

d. Weight percent in ppm.

weight percent of solute x (10000 ppm/1%)

= (2.79%) (10000 ppm/1%) = 27933.88 ppm