A 2.300×10^−2 M solution of glycerol (C3H8O3) in water is at 20.0°C. The sample was created by dissolving a sample of C3H8O3 in water and then bringing the volume up to 1.000 L. It was determined that the volume of water needed to do this was 999.0 mL . The density of water at 20.0°C is 0.9982 g/mL. Required:
a. Calculate the molality of the glycerol solution.
b. Calculate the mole fraction of glycerol in this solution.
c. Calculate the concentration of the glycerol solution in percent by mass.
d. Calculate the concentration of the glycerol solution in parts per million.

a) Based on the given information, the volume of the solvent given is 999 ml and the density of water given is 0.9982 gram per ml.  

The mass of solvent can be calculated by the formula mass = density * volume

mass = 0.9982 * 999 = 0.997 Kg

The molarity of the solution given is 2.300 * 10^-2 M

Molality of the glycerol solution can be calculated by using the formula,  

Molality = molarity/solvent (kg) = 2.300 * 10^-2 / 0.997 = 0.023 m

b) Molarity or the moles of the solute given is 2.300 * 10^-2 moles

The moles of solvent can be determined by using the formula, n = mass of solvent/mol.wt = 997/18 (mol.wt of solvent is 18 g/mol), now putting the values we get,  

n = 997/18 = 55.4

The mole fraction of the glycerol will be = 2.300 * 10^-2 M/(2.300 * 10^-2)+55.40

= 4.15 * 10^-4

c) The mass percent of glycerol can be determined by using the formula,  

mass of solute/mass of solution * 100% (i)

The mass of solute can be determined by using the formula,

n = mass of solute/mol.wt of solute

The n or the no. of moles is 2.300 * 10^-2 and the molecular weight of glycerol is 92.09 g/mol. Now putting the values we get,  

mass = 2.300 * 10^-2 * 92.09 = 2.12 grams

Now putting the values in equation (i) we get,  

mass percent = 2.12 / 997 * 100% = 0.21%

d) Based on the above calculation, the mass of solute (glycerol) is 2.12 g or 2.12 * 1000 mg

The volume of water is 999 ml or 999 * 10^-3 L

The concentration of the glycerol solution will be,  

Concentration = 2.12 * 10^3 mg/999 * 10^-3 L  

= 2.12/999 * 10^6 mg/L

= 2122.1 ppm

answer; talc and gypsum;