Since pure water boils at 100.00 C, and since the addition of solute increases b
ID: 561651 • Letter: S
Question
Since pure water boils at 100.00 C, and since the addition of solute increases boiling point, the boiling point of an aqueous solution, Tb, will be Freezing-Point Depression and Boiling-Point Elevation 1h) °C = (100.00 Since pure water freezes at 0.00 °C, and since the addition of solute decreases freezing point, the freezing point of an aqueous solution, T, will be The changes in boiling point(7,) or freezing point (AR) in degrees Celsius from a pure solvent can be determined from the equations given here, respectively: Part A What is the boiling point of a solution made using 747 g of sucrose, C12 H220in 0.275 kg of water, H20? Express your answer in degrees Celsius using five significant figures. where m is the molality of the solution, and Kb and K are Hints constants for the solvent, respectively. For water K,-0.512 C kg solvent mo!solute- Submlt My Answers Give Up Part B What is the freezing point of a solution that contains 23.9 g of urea, CO(NH2)2, in 295 mL water, H O? Assume a density of water of 1.00 g/mL Express your answer numerically in degrees Celsius HintsExplanation / Answer
Part A
Answer
104.06
Explanation
Mass of Sucrose = 747g
Molar mass of Sucrose = 342.30g/mol
No of mole of Sucrose = 747g/342.30g/mol = 2.1823mol
Mass of Water =0.275g
Molality of solute = (2.1823mol/0.275g)×1kg =7.936m
Tb(solution) - Tb(solvent) = m × Kb
Tb(solution) - 100°C = 7.936m × 0.512/m
= 4.06
Tb(solution) =104.06
Part B
Answer
-2.51
Explanation
Mass of Urea = 23.9g
Molar mass of Urea = 60.06g/mol
No of mole of Urea = 23.9g/60.06g/mol = 0.3979
Mass of Water = 0.295kg
molality of urea =( 0.3979mol/0.295kg)×1kg = 1.3488m
Tf(solvent) - Tf(solvent) = 1.3488m × 1.86/m
0 - Tf(solution) = 2.51
Tf(solution) = -2.51