Question

A chemistry student wants to determine the boiling point of a solution of 25 g of glucose in 500 mL of water. Calculate the boiling point of the solution if the boiling point constant of water is 0.512 °C/m and assume an ideal solution.

Answer 1

To calculate the boiling point of the solution, we need to first determine the molality of the glucose solution. Molality  m  is defined as the moles of solute  glucose  per kilogram of solvent  water .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 glucose = 25 g / 180.18 g/mol  0.1387 mol2. Convert the volume of water to mass:Assuming the density of water is 1 g/mL, the mass of 500 mL of water is 500 g.3. Calculate the molality of the glucose solution:Molality  m  = moles of glucose / mass of water  in kg  = 0.1387 mol / 0.5 kg = 0.2774 mol/kg4. Calculate the boiling point elevation using the boiling point constant of water  Kb :Tb = Kb  molality = 0.512 C/m  0.2774 mol/kg  0.1420 C5. Determine the boiling point of the glucose solution:The normal boiling point of water is 100 C. The boiling point of the glucose solution is the normal boiling point of water plus the boiling point elevation.Boiling point of glucose solution = 100 C + 0.1420 C  100.142 CThe boiling point of the glucose solution is approximately 100.142 C.