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Chapter 11: Problem 38

An aqueous antifreeze solution is 40.0\(\%\) ethylene glycol \(\left(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\right)\) by mass. The density of the solution is 1.05 \(\mathrm{g} / \mathrm{cm}^{3}\) . Calculate the molality, molarity, and mole fraction of the ethylene glycol.

Short Answer

Expert verified
The molality of the ethylene glycol in the solution is 10.73 mol/kg, its molarity is 6.76 mol/L, and its mole fraction is 0.162.

Step by step solution

01

Find the mass of ethylene glycol and water in the solution

Let's assume we have 100 g of the solution. Since it is 40.0% ethylene glycol by mass, it means there are 40.0 g of ethylene glycol and 60.0 g of water in the solution.
02

Convert mass of ethylene glycol to moles

Now, we will need to determine the number of moles of ethylene glycol in the solution. To do that, we should use the molar mass of ethylene glycol: Molar mass of C₂H₆O₂ = (2 × 12.01) + (6 × 1.01) + (2 × 16.00) = 62.07 g/mol Then, we can convert the mass of ethylene glycol to moles: Moles of ethylene glycol = (40.0 g) / (62.07 g/mol) = 0.644 mol
03

Calculate the molality of the solution

Molality is defined as the number of moles of solute per kilogram of solvent. Since we already know the moles of ethylene glycol and mass of water, we can calculate the molality: Molality = (moles of ethylene glycol) / (mass of water in kg) Molality = (0.644 mol) / (0.060 kg) = 10.73 mol/kg
04

Calculate the molarity of the solution

To calculate the molarity, we need to find the number of moles of ethylene glycol per liter of solution. We are given the density of the solution (1.05 g/cm³). First, let's convert the density to g/mL and find the volume of the 100 g solution: Density = (1.05 g/cm³) × (1 cm³/mL) = 1.05 g/mL Volume of 100 g solution = (100 g) / (1.05 g/mL) = 95.24 mL = 0.09524 L Now, we can calculate the molarity: Molarity = (moles of ethylene glycol) / (volume of solution in liters) Molarity = (0.644 mol) / (0.09524 L) = 6.76 mol/L
05

Calculate the mole fraction of ethylene glycol

Mole fraction is defined as the ratio of the number of moles of a component to the total number of moles in the solution. To find the mole fraction of ethylene glycol, we need to find the total number of moles in the solution: Moles of water = (60.0 g) / (18.02 g/mol) = 3.329 mol Total moles = moles of ethylene glycol + moles of water Total moles = 0.644 mol + 3.329 mol = 3.973 mol Now, we can calculate the mole fraction of ethylene glycol: Mole fraction = (moles of ethylene glycol) / (total moles) Mole fraction = (0.644 mol) / (3.973 mol) = 0.162 Thus, the molality of the solution is 10.73 mol/kg, the molarity is 6.76 mol/L, and the mole fraction of ethylene glycol is 0.162.

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