Question

A chemistry student must calculate the boiling point of a solution composed of 10 g of glucose (C6H12O6) dissolved in 100 g of water at 1 atm pressure. The boiling point elevation constant for water (Kb) is equal to 0.512°C/m, and the normal boiling point of pure water is 100°C.

Answer 1

To calculate the boiling point of the solution, we need to find the boiling point elevation  Tb  and add it to the normal boiling point of pure water. The formula for boiling point elevation is:Tb = Kb  molalityFirst, we need to find the molality of the solution. Molality  m  is defined as the number of moles of solute per kilogram of solvent. 1. Calculate the moles of glucose  C6H12O6 :Molar mass of glucose =  6  12.01  +  12  1.01  +  6  16.00  = 72.06 + 12.12 + 96.00 = 180.18 g/molmoles of glucose = mass of glucose / molar mass of glucosemoles of glucose = 10 g / 180.18 g/mol = 0.0555 mol2. Convert the mass of water to kilograms:mass of water = 100 g = 0.1 kg3. Calculate the molality of the solution:molality = moles of glucose / mass of water in kgmolality = 0.0555 mol / 0.1 kg = 0.555 mol/kg4. Calculate the boiling point elevation  Tb :Tb = Kb  molalityTb = 0.512C/m  0.555 m = 0.284C5. Add the boiling point elevation to the normal boiling point of pure water:New boiling point = normal boiling point + TbNew boiling point = 100C + 0.284C = 100.284CThe boiling point of the glucose solution is 100.284C at 1 atm pressure.