Question

A 2.250×10⁻² 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.

Part A:  

Calculate the molality of the glycerol solution.

Express your answer to four significant figures and include the appropriate units. Definitely not 0.0225

Part B: Calculate the mole fraction of glycerol in this solution.

Express the mole fraction to four significant figures. definitely not 55.40

Part C:  Calculate the concentration of the glycerol solution in percent by mass.

Express your answer to four significant figures and include the appropriate units. definitely not 0.002

Answer 1

Part A : molality of glycerol = 0.02256 m

Part B : mole fraction glycerol = 4.063 x 10^-4

Part C : concentration of the glycerol solution in percent by mass = 0.2074 %

Explanation

Part A :

Given : concentration of glycerol = 2.250 x 10^-2 M

volume of solution = 1.000 L

moles of glycerol = (concentration of glycerol) * (volume of solution)

moles of glycerol = (2.250 x 10^-2 M) * (1.000 L)

moles of glycerol = 2.250 x 10^-2 mol

volume of water = 999.0 mL

mass of water = (volume of water) * (density of water)

mass of water = (999.0 mL) * (0.9982 g/mL)

mass of water = 997.2 g

mass of water = 0.9972 kg

molality of glycerol = (moles of glycerol) / (mass of water in kg)

molality of glycerol = (2.250 x 10^-2 mol) / (0.9972 kg)

molality of glycerol = 0.02256 m

Part B

mass of water = 997.2 g

moles of water = (mass of water) / (molar mass of water)

moles of water = (997.2 g) / (18.0 g/mol)

moles of water = 55.353 mol

Total moles = (moles glycerol) + (moles water)

Total moles = (2.250 x 10^-2 mol) + (55.353 mol)

Total moles = 55.3756 mol

mole fraction glycerol = (moles glycerol) / (total moles)

mole fraction glycerol = (2.250 x 10^-2 mol) / (55.3756 mol)

mole fraction glycerol = 4.063 x 10^-4