Part 1:
Using the molal freezing point depression constant for water (1.86 C/m), we can calculate the freezing point depression for a 3.6 molal solution of glucose in water:
ΔTf = i * Kf * m
ΔTf = 1 * 1.86 * 3.6 = 6.696
Freezing point = 0.0 C - 6.696 C = -6.7 C
Part 2:
Using the molal freezing point depression constant for water (1.86 C/m), we can calculate the freezing point depression for a 1.7 molal solution of sodium nitrate in water:
ΔTf = i * Kf * m
ΔTf = 2 * 1.86 * 1.7 = 6.372
Freezing point = 0.0 C - 6.372 = -6.4 C
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 C.
Part 1 of 2
3.6 mol of glucose (C6H1206) molecules:
℃
10
X
Part 2 of 2
1.7 mol of sodium nitrate (NaNO3):
1 answer