Final answer:
The concentration of a sucrose solution is 0.415 M in molarity, 0.4437 m in molality, 13.18% by mass, 0.0079 mole fraction, and 0.79% mole percent.
Step-by-step explanation:
To calculate the concentration of a sucrose solution in different units, we must first determine the amount of sucrose in moles. Sucrose has a molar mass of approximately 342.30 g/mol.
To find the molarity, divide moles of sucrose by the volume of the solution in liters:
Number of moles = 50.4 g / 342.30 g/mol = 0.1473 mol
Volume of solution = 355 mL = 0.355 L
Molarity (M) = 0.1473 mol / 0.355 L = 0.415 M
For molality, calculate moles of sucrose per kilogram of water:
Molality (m) = 0.1473 mol / 0.332 kg = 0.4437 m
The percent by mass is found by dividing the mass of solute by the total mass of the solution and multiplying by 100:
Percent by mass = (50.4 g / (50.4 g + 332 g)) * 100 = 13.18%
To calculate the mole fraction of sucrose, divide moles of sucrose by total moles in solution:
Total moles = Moles of water + Moles of sucrose
Moles of water = 332 g / 18.015 g/mol = 18.43 mol
Mole fraction of sucrose (χsucrose) = 0.1473 mol / (18.43 mol + 0.1473 mol) = 0.0079
Finally, to find the mole percent, multiply the mole fraction by 100:
Mole percent = 0.0079 * 100 = 0.79%