here is another question What is the freezing point of a solution that contains each of the following quantities of solute in 1.00 kg of water? Be sure each of your answer entries has the
correct number of significant figures.
Note: Reference the Molal freezing point depression and boiling point elevation constants table for additional information.
Note: The freezing point for water is 0.0 ℃.
Part 1 of 2
5.3 mol of fructose (C6H1206) molecules:
°C
x10
Part 2 of 2
5.0 mol of sodium sulfate (Na2SO4):
℃
write the correct significant figure and scientific notation 2 is the answer 1.86
Part 1 of 2:
- Using the Molal freezing point depression constant for water, which is 1.86 °C/kg/mol, and the number of moles of fructose (5.3 mol), we can calculate the freezing point depression as follows:
ΔTf = Kf * molality
ΔTf = 1.86 °C/kg/mol * 5.3 mol = 9.858 °C
Thus, the freezing point of the solution containing 5.3 mol of fructose molecules in 1.00 kg of water is -9.858 °C.
Part 2 of 2:
- Following the same process with the Molal freezing point depression constant for water and the number of moles of sodium sulfate (5.0 mol):
ΔTf = Kf * molality
ΔTf = 1.86 °C/kg/mol * 5.0 mol = 9.30 °C
Therefore, the freezing point of the solution containing 5.0 mol of sodium sulfate in 1.00 kg of water is -9.30 °C.