Question

Chemists often use molarity M, in moles/liter, to measure the concentration of solutions. Molarity is a common unit of concentration because the volume of a liquid is very easy to measure. However, the drawback of using molarity is that volume is a temperature-dependent quantity. As temperature changes, density changes, which affects volume. Volume markings for most laboratory glassware are calibrated for room temperature, about 20∘C. Fortunately, there are several other ways of expressing concentration that do not involve volume and are therefore temperature independent.

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 mLmL . The density of water at 20.0∘C is 0.9982 g/mL.
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.

Answer 1

A. 2.40×10⁻⁵ m

B. % mole fracion of glycerol → 4.33×10⁻⁴

C. 0.22% m/m

D. 2209 ppm

Step-by-step explanation:

This is our data:

Solute → 2.400×10⁻² M → Glycerol (C₃H₈O₃)

Solvent → 999.0 mL → water

Volume of solution → 1.000 L

A. To determine molality, we know that this sort of concentration indicates the moles of solute in 1kg of solvent. We need to know the moles of solute:

- As the volume of solution is 1L, and we have the molar concentration of glycerol we can notice, that in 1L of solution, we have 2.400×10⁻² moles of glyrcerol. Let's determine the mass of solvent, by density:

Density of water = Mass of water / Volume of water

0.9982 g/mL = Mass of water / 999 mL

Mass of water = 0.9982 g/mL . 999 mL = 997.2 g

We convert the mass to kg → 997.2 g . 1kg /1000g = 0.9972 kg

Molality = 2.400×10⁻² mol / 0.9972 kg = 2.40×10⁻⁵ m

B. Mole fraction is the moles of solute / total moles (moles of solute + solvent)

Moles of solute: 2.400×10⁻² mol

Moles of solvent: Mass of solvent / Molar mass of solvent

997.2 g / 18 g/mol = 55.4 mol

% mole fraction of glycerol = 2.400×10⁻² / (2.400×10⁻² + 55.4) = 4.33×10⁻⁴

C. Percent by mass is the concentration that indicates the mass of solute in 100 g of solution.

Mass of solute: 2.400×10⁻² mol . molar mass of glycerol

2.400×10⁻² . 92 g/mol = 2.208 g

Mass of solvent: 997.2 g

Mass of solution: solvent + solute → 2.208 g + 997.2 g = 999.408 g

%by mass = (2.208 g / 999.408g) . 100 = 0.22%

D. ppm are the mass of solvent in μg / grams of solution

We convert the mass of solvent: 2.208 g . 1×10⁶ μg / 1 g = 2.208×10⁶ μg

Mass of solution: 999.408 g

ppm = 2.208×10⁶ μg / 999.408 g → 2209 ppm