Question

A 2.400×10−2 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.

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

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

Part D: Calculate the concentration of the glycerol solution in parts per million.

Answer 1

Given : Molarity of glycerol solution , M = 2.4 x 10⁻²  

Volume of solution = 1L

Volume of solvent = 999mL

Density of water , d = 0.9982g/mL

Part A :

Number of moles of glycerol = Molarity of solution x Volume of solution = 2.4 x 10⁻²   x 1 =  2.4 x 10⁻² mole

Mas of solvent = density x volume of solvent = 0.9982 x 999 = 997.202 g = 0.9972 Kg

Molality of glycerol solution = Number of moles of solute / Mass of solvent in Kg

= 2.4 x 10⁻² / 0.9972 = 2.4067 x 10⁻² = 0.0241 molal

Part B :

Mass of solvent (water) = 997.202g

Number of moles of water = 997.202 / 18 = 55.4 mole

Mole fraction of glycerol = No. of moles of glycerol / (no. of moles of glycerol + no. of moles of water)

= 0.024 / (0.024 + 55.4) = 0.000433 = 4.33 x 10⁻⁴

Part C:

Mass of glycerol = number of mole x Molar mass of glycerol

(Molar mass of glycerol = 3 x 12 + 8 x 1 + 3 x 16 =92)

Mass of solute (glycerol) = 2.4 x 10⁻²   x 92 = 2.208g

Mass of solution = mass of solute + mass of solvent = 2.208 + 997.202 = 999.41 g

Percent by mass of glycerol = ( mass of solute / mass of solution ) x 100

= (2.208 / 999.41) x 100 = 0.2209 %

Part D :

Parts per million of glycerol = (amount of solute / amount of solution ) x 10⁶

= (2.208 / 999.41) x 10⁶ = 2209.3 PPM

M