Question

a solution is prepared by dissolving 50.4 g sucrose (c12H22O11) in 0.384 kg of water. The final volume of the solution is 355 mL.

A. for this solution calculate molarity

B. For this solution Calulate molality

C. For this solution calculate percent by mass

D. For this solution calculate mole fraction

Answer 1

50.4 g 0.384 kg or Mass of Sucrose Mass of Water = 384 = 18.015 g/mol 342.296 g/mol Molar Mass of Sucrose Molar mass of water = = Moles of Sucrose = mass /molar mass 50.4 g Moles of water = mass / molar mass 18.015 g/mol 342.3 g/mol 384 g 21.316 mol 0.1472 mol = so total moles = 0.1472 21.3156 Volume of solution = ml 355 21.463 mol = moles volume in L 0.1472 mol Molarity of solution 0.355 I = Total mass= 0.355 L 50.4 + 384 434.4 g 0.415 M Molality of solution = moles -mass of water in Kg 0.384 kg or 0.1472 mol - 0.383 m Percent mass (Mass of sucrose Total mass) x 100 = 434.4 g 50.4 g 100 X 11.6 % Mole fraction of Sucrose = Moles of sucrose / total moles 0.1472 21.463 0.0069 Mole fraction of water = moles of water / total moles 21.463 21.316 0.993