Question

An antifreeze solution is prepared from 222.6 g of ethylene glycol C₂H₆O₂ and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL^–1, then what shall be the molarity of the solution?

Answer 1

Step 1: Calculate the moles of each component.
Ethylene glycol (solute):
The molar mass of ethylene glycol \((C_2H_6O_2)\) = 62.07 g/mol
Moles of ethylene glycol = \(\frac{\text{mass}}{\text{molar mass}} =\frac{222.6 g}{62.07 g/mol} = 3.59 mol\)

Water (solvent):
The molar mass of water (\(H_2O\)) = 18.02 g/mol
Moles of water = \(\frac{\text{mass}}{\text{molar mass}} = \frac{200 g}{18.02 g/mol} = 11.11 \text{ mol}\)

Mass of solvent (water) = \(200 g\times\frac{1 kg}{1000 g}\) = 0.200 kg

Step 3: Calculate the molality.
Molality (m) = \(\frac{\text{moles of solute}}{\text{mass of solvent (in kg)}}\)
Molality (m) = \(\frac{3.59\ mol}{0.200\ kg}\) = 17.95 mol/kg

Step 4: Find the total volume of the solution.
Total mass of solution = mass of ethylene glycol + mass of water = 222.6 g + 200 g = 422.6 g
Density (ρ) = 1.072 g/mL
Volume (V) = \(\frac{\text{mass}}{\text{density}}\)
Volume (V) = \(\frac{422.6 g}{1.072 g/mL}\) = 394.2 mL (convert mL to L by dividing by 1000)
Volume (V) = 0.3942 L

Step 5: Calculate the molarity.
Molarity (M) = \(\frac{\text{moles of solute}}{\text{volume of solution (in L)}}\)
Molarity (M) = \(\frac{3.59\ mol}{0.3942\ L}\) = 9.11 M

So, the answer is 9.11 M.