Five grams of glucose, C6H12O6, is dissolved in 500.0 grams of acetic acid.
What is the new freezing point and boiling point for the solution?
Kf acetic acid = 3.90, Kb acetic acid 3.07
(normal freezing point for acetic acid = 16.60 degrees C, boiling point 118.5 C)

This question involves the colligative properties of solutions. I will
first consider freezing point depression.

Finding the moles of glucose:
n = 5 g / 180 (g/mol)
n = 1/36 mol

Finding the molality of the solution:
m = moles solute per kg solvent
m = 1/36 mol / 0.5 kg
m = 1/18 molal

FREEZING POINT DEPRESSION
Let the change in freezing point, delta tf (?tf), be represented by x.

Equation:
x = i * Kf * m

i = 1.00 (since glucose is a molecular, non-ionizing compound)
Kf = 3.90 (given)
m = 1/18 molal (calculated above)

x = i * Kf * m
x = 1.00 * 3.90 * 1/18
x = 0.22 °C (degrees Celsius)

Therefore, the freezing point of the solution is 16.38 °C (degrees
Celsius).

Now I consider boiling point elevation.
BOILING POINT ELEVATION
Equation:
x = i * Kb * m
x = 1.00 * 3.07 * 1/18
x = 0.17 °C (degrees Celsius)

Therefore, the boiling point of the solution is 118.67 °C (degrees
Celsius).

Hope this helps.


Source

http://dbhs.wvusd.k12.ca.us/Solutions/BP-Elev-and-FP-Lower.html


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freezing point depression