Answer :
Final answer:
The molality of a 1.69 M ethanol solution with a density of 0.9731 g/mL is calculated by converting molarity to moles, finding the mass of the water in the solution using the density, and then dividing the moles of ethanol by the kilograms of water. The calculated molality for this ethanol solution is 1.888 m.
Explanation:
To calculate the molality of the solution, we first need to convert the Molarity (M) of the solution to molality (m). The molality is defined as the amount of solute expressed in moles divided by the mass of the solvent expressed in kilograms.
Molarity (M) is defined as the moles of solute divided by the liters of solution. In this situation, we have a 1.69 M solution of ethanol (C₂H₅OH). So, there are 1.69 moles of ethanol in every liter of solution.
Next, we can use the density of the solution (0.9731 g/mL) to find the mass of the solvent. Density is mass per volume. So, 1 liter (or 1000 mL) of the solution will have a mass of 0.9731 g/mL x 1000 mL = 973.1 g. So, the solution contains 973.1 g of both ethanol and water combined.
However, we need the weight of the water alone for the calculation of molality. Given that the molar mass of ethanol is about 46.07 g/mol, the weight of ethanol in the solution is 1.69 mol/L x 46.07 g/mol x 1 L = 77.838 g. Subtract this from the total mass to find the water weight: 973.1 g - 77.838 g = 895.262 g, or 0.895262 kg.
Finally, the molality of the solution is mol solvent/kg solute = 1.69 mol / 0.895262 kg = 1.888 m.
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