Answer:
= 0.316 mol/kg
Explanation:
Molality is defined as the number of moles of solute per kilogram of solvent. We need to find the molality of the solution prepared by dissolving 225 mg of glucose in 5.00 mL of ethanol.
First, we need to convert the mass of glucose to moles:
Molar mass of glucose (C6H12O6) = 180.16 g/mol
Mass of glucose = 225 mg = 0.225 g
Number of moles of glucose = mass/molar mass = 0.225 g/180.16 g/mol = 0.00125 mol
Next, we need to calculate the mass of ethanol used in the solution:
Volume of ethanol = 5.00 mL
Density of ethanol = 0.789 g/mL
Mass of ethanol = volume x density = 5.00 mL x 0.789 g/mL = 3.945 g
Now we can calculate the molality of the solution:
Molality = moles of solute / mass of solvent in kg
Mass of solvent in kg = mass of ethanol / 1000 = 3.945 g / 1000 = 0.003945 kg
Molality = 0.00125 mol / 0.003945 kg = 0.316 mol/kg
Therefore, the molality of the solution prepared by dissolving 225 mg of glucose in 5.00 mL of ethanol is 0.316 mol/kg.