What is the boiling point of a solution made by dissolving 1.0000 mole of sucrose in 1.0000 kg of water?
The change in Boiling Point of water can be calculated using this formula:
ΔTb = i * Kb * m
Where i is the van't hoff factor (the number of particles or ions), the kb is a constant (boiling point elevation constant) and m is the molality of the solution.
The kb for water is always 0.515 °C/m. Kb = 0.515 °C/m
The value for i in this case is 1. Since sucrose is a covalent compound and it doesn't dissociate into ions. i = 1
The molal concentration of the solution can be found using this formula:
molality = moles of sucrose/kg of water
molality = 1.000 mol / 1.000 kg of water
molality = 1 m
Now that we know all the values, we can use the formula to find the change in the boiling point of water:
ΔTb = i * Kb * m
ΔTb = 1 * 0.515 °C/m * 1 m
ΔTb = 0.515 °C
Finally, we are asked for the boiling point of the solution, not the change. The boiling point of water at atmospheric pressure is 100.00 °C. If the boiling point rises 0.515 °C when we prepare the solution. The boiling point of the solution is:
Boiling point solution = Boiling point of water + ΔTb
Boiling point solution = 100.000 °C + 0.515 °C
Boiling point solution = 100.515 °C
Answer: The boiling point of the solution is 100.515 °C.
