Calculate the boiling temperature of a 0.50 M solution of sucrose. Assume that the concentration is 0.50 m.
I'm assume the solution is one of sucrose and water.
delta T = Kb*m
Solve for delta T, then add to the normal boiling point to find the new boiling point.
To calculate the boiling temperature of a solution, we can use the equation:
ΔTb = Kb * m
Where:
ΔTb is the boiling point elevation (the difference between the boiling point of the solution and the pure solvent)
Kb is the molal boiling point elevation constant (a characteristic value for each solvent)
m is the molality of the solute in the solution (moles of solute per kilogram of solvent).
In this case, we are given the concentration of the solution in molarity (M), but we need to convert it to the molality (m) by using the formula:
m = Moles of solute / Mass of solvent (in kg)
First, let's convert the molarity to moles of sucrose:
Moles of sucrose = Molarity * Volume of solution
= 0.50 mol/L * 1.00 L
= 0.50 mol
Next, we need to convert the mass of the solvent to kg. Assuming the solution is water-based, we can use the density of water (1 g/mL) to convert the volume of the solvent to mass:
Mass of water = Volume of water * Density
= 1.00 L * 1000 g/L
= 1000 g
= 1.00 kg
Now, we can calculate the molality (m):
m = Moles of solute / Mass of solvent
= 0.50 mol / 1.00 kg
= 0.50 m
Finally, we can use the molality (m) and the boiling point elevation constant (Kb) for water (0.512 °C/m) to calculate the boiling temperature elevation (ΔTb):
ΔTb = Kb * m
= 0.512 °C/m * 0.50 m
= 0.256 °C
Add this boiling point elevation (ΔTb) to the boiling point of pure water (100 °C) to get the boiling temperature of the sucrose solution:
Boiling temperature = 100 °C + 0.256 °C
= 100.256 °C
Therefore, the boiling temperature of a 0.50 M solution of sucrose is approximately 100.256 °C.