Question

What are the mole fraction (X) and molality (m) of ethanol (C₂H₅OH) in an aqueous solution that is 45.0% ethanol by volume? The density of water is 1.00 g/mL and that of ethanol is 0.789 g/mL.

mole fraction = ?????;

molality =

??? m

Answer 1

Step 1 : Calculation of the mass of ethanol and water present in 45.0 % ethanol by volume

45% ethanol by volume means that there are 45 ml ethanol for every 100 ml of the aqueous soluton.

The volume of water is therfore 100 - 45 = 55 ml

VOLUME OF ETHANOL = 45 mL

VOLUME OF WATER = 55 mL

Density = Mass / Volume

Mass = Volume x Density

Mass of Ethanol = Volume of Ethanol x Densty of Ethanol = 45 mL x 0.789 g/mL = 35.5 g

Mass of Ethanol = 35.5 g

Mass of Water= Volume of Water x Densty of Water = 55 mL x 1.00 g/mL = 55.0 g

Mass of Water = 55.0 g

Step 2: Calculation of number of moles of Ethanol and Water

Molar mass of Ethanlol C2H5OH = 2x12.0 + 6 x1.00 + 16.0 = 46.0 g/mol

Number of moles of Ethanol = Mass of Ethanol in gram /Molar mass = 35.5 g /46.0 g/mol = 0.772 mol

Number of moles of Ethanol = 0.772 mol

Molar mass of Water H2O = 2 x1.00 + 16.0 = 18.0 g/mol

Number of moles of Water = Mass of Water in gram /Molar mass = 55 g /18.0 g/mol = 3.06 mol

Number of moles of water = 3.06 mol

Step 3: Calculation of mole fraction of ethanol

Mole fraction of ethanol = Number of moles of ethanol / ( Number of moles of ethanol + Number of moles of water)

                                        = 0.772 mol / (0.772 mol+3.06 mol) = 0.2014

MOLE FRACTION OF ETHANOL = 0.2014

Step 4: Calculation of molality of the ethanol solution

Molality of a solution is the number of moles of the solute present in 1kg or 1000 g of solvent.

0.772 mol of Ethanol is present in 55 g of water

OR

55 g of water contains 0.772 mol of Ethanol

Therefore

1000 g of water will contain (0.772 mol x 1000g / 55g) mole of Ethanol = 14.0 mol

MOLALITY OF THE ETHANOL SOLUTION = 14.0 mol /1000g = 14.0mol / kg