A solution is made using 100.5 g of ethanol (molar mass = 46.07 g/mol) and 120.0 g of acetone (molar mass = 58.08 g/mol). What is the molality of the ethanol in the solution?

The molality of ethanol in the solution is calculated by dividing the number of moles of ethanol (2.18 moles, approx) by the mass of the solvent acetone in kilograms (0.120 kg), resulting in a molality of approximately 18.17 mol/kg.

Explanation:

To calculate the molality of the ethanol in the solution, we will follow these steps:

Calculate the number of moles of ethanol.
Convert the mass of the solvent (acetone) to kilograms.
Divide the moles of ethanol by the kilograms of solvent.

First, to find the number of moles of ethanol, we use the formula:

moles of ethanol = mass of ethanol / molar mass of ethanol

moles of ethanol = 100.5 g / 46.07 g/mol

= 2.18 moles of ethanol (approx).

Next, convert the mass of acetone to kilograms:

mass of acetone in kg = 120.0 g / 1000

= 0.120 kg

Last, we divide the moles of ethanol by the mass of acetone in kilograms to find the molality:

molality of ethanol = moles of ethanol / mass of acetone in kg

molality of ethanol = 2.18 moles / 0.120 kg

= 18.17 mol/kg (approx).

Therefore, the molality of the ethanol in the solution is approximately 18.17 mol/kg.