What are the freezing point and boiling point of an aqueous solution of 15.5 g glucose(C6H12O6) dissolved in 150 g of water?
sus. hope this helps
To determine the freezing point and boiling point of an aqueous solution, we need to know the molal concentration of the solution. The molal concentration (m) is the number of moles of solute per kilogram of solvent.
To calculate the molal concentration, we first need to find the number of moles of glucose (C6H12O6) in 15.5 g. The molar mass of glucose is 180.16 g/mol, so the number of moles can be calculated using the equation:
Number of moles of glucose = Mass of glucose / Molar mass of glucose
Number of moles of glucose = 15.5 g / 180.16 g/mol
Using a calculator, we find that the number of moles of glucose is approximately 0.086 moles.
Next, we calculate the molal concentration (m) using the equation:
m = Number of moles of solute / Mass of solvent in kg
In this case, the mass of the solvent (water) is given as 150 g, so converting it to kilograms:
Mass of solvent = 150 g / 1000 = 0.150 kg
Now we can substitute the values into the equation:
m = 0.086 moles / 0.150 kg
Using a calculator, we find that the molal concentration of the solution is approximately 0.573 mol/kg.
The freezing point depression (∆Tf) and boiling point elevation (∆Tb) can be calculated using the formulas:
∆Tf = Kf * m * i
∆Tb = Kb * m * i
Where Kf and Kb are the cryoscopic and ebullioscopic constants for water, respectively, and i is the van't Hoff factor, which represents the number of particles the solute dissociates into in the solution.
For glucose, which does not dissociate into ions in water, the van't Hoff factor (i) is 1.
The cryoscopic constant for water is approximately 1.86 °C/m, and the ebullioscopic constant for water is approximately 0.512 °C/m. Substituting the values into the formulas:
∆Tf = 1.86 °C/m * 0.573 mol/kg * 1
∆Tb = 0.512 °C/m * 0.573 mol/kg * 1
Using a calculator, we find that the freezing point depression (∆Tf) is approximately 1.06 °C and the boiling point elevation (∆Tb) is approximately 0.294 °C.
To determine the freezing point and boiling point of the solution, we need to add (∆Tf) to the normal freezing point of water and (∆Tb) to the normal boiling point of water.
The normal freezing point of water is 0 °C, so the freezing point of the solution would be:
Freezing point of solution = 0 °C - 1.06 °C ≈ -1.06 °C
The normal boiling point of water is 100 °C, so the boiling point of the solution would be:
Boiling point of solution = 100 °C + 0.294 °C ≈ 100.294 °C
Therefore, the freezing point of the aqueous solution of glucose is approximately -1.06 °C, and the boiling point is approximately 100.294 °C.