Question

A Review Constants Periodic Table The changes in boiling point (AT) or freezing point (AT) in degrees Celsius from a pure solvent can be determined from the equations given here, respectively: AT) = m x K = moles of solute XK K. kilograms of solvent Since pure water boils at 100.00 °C, and since the addition of solute increases boiling point, the boiling point of an aqueous solution, Th, will be T - (100.00+AT) 'C Since pure water freezes at 0.00 °C, and since the addition of solute decreases freezing point, the freezing point of an aqueous solution, Tf, will be Tf - (0.00 – AT) °C AT = mx K = - moles of solute kilograms of solvent XRf where m is the molality of the solution, and K) and Ky are the boiling-point-elevation and freezing-point-depression constants for the solvent, respectively. For water, Part A Kp = 0.512 "C. kg solvent mol solute K = 1.86 °C-kg solvent What is the boiling point of a solution made using 799 g of sucrose, C2H22011, in 0.300 kg of water, H120? Express your answer to five significant figures and include the appropriate units. View Available Hint(s) mol solute Value Units Submit

Answer 1

Atb = mx Kb Moly m = molality kg solvent mass ug molar mans (g/mol) m (molality) =- Kg solvent m = 799 & 342.381 mol 0.3ookg 7.78 mot/kg. m= Alb= = 7.78 x 0.512 3.98c The = loot 3.98 C. 103.981 Answer