Answer :
Final answer:
To prepare 250 mL of a 1.55 m solution of urea, one would need 23.28 grams or 23280 milligrams of urea, using its molecular weight of 60.06 g/mol.
Explanation:
To calculate the amount of urea required to prepare 250 ml of a 1.55 m (molal) solution, we will first convert the volume of the solution to liters since molality is expressed in moles of solute per kilogram of solvent. We have 250 ml which is equivalent to 0.250 liters. The molecular weight of urea is given as 60.06 g/mol.
Next, we will use the formula for molality, which is: molality (m) = moles of solute / kilograms of solvent . Since we want a 1.55 m solution, we need 1.55 moles of urea per kilogram of solvent. Assuming the density of the solution is close to that of water, we can approximate the mass of the solvent in kilograms by using the volume of the solution and the density of water (1 g/mL). Thus, for 250 mL (or 0.250 L), we would have approximately 0.250 kg of solvent.
To find the moles of urea required: moles of urea = molality x kilograms of solventmoles of urea = 1.55 m x 0.250 kg = 0.3875 moles .To convert moles to grams, we multiply by the molecular weight: grams of urea = moles of urea x molecular weightgrams of urea = 0.3875 moles x 60.06 g/mol = 23.28 g To express this mass in milligrams (mg): milligrams of urea = grams of urea x 1000 mg/gmilligrams of urea = 23.28 g x 1000 mg/g = 23280 mg
Therefore, to prepare 250 mL of a 1.55 m solution of urea, 23.28 grams (or 23280 milligrams) of urea are required.
