freezing-point depression凝固点降低

Freezing-point depression and determination of the molecular weight of the solute:
Formula:
M B=k f*1000*w B/∆T f*w A
k f=1.855 K*kg*mol-1 ; w B is the mass of the solute; w A is the mass of the solvent; ∆T f is the value of the freezing-point depression.
When you cool a solvent, the temperature of this solvent should decrease equally until the crystal formed. When the crystal formed, it will give off the crystallization heat to cancel out the heat loss of the system, and at this point, the temperature of this system will not change until all the liquid solidified, then the temperature goes down equally. See the line (a)
In fact, the vapor pressure of the just formed crystal is bigger than the vapor pressure of the liquid at the same temperature, so we will always see this over-cooling phenomenon. See line (b)
And if super over-cooling, see line (c)
If we cool down a solution, because the formation of the solvent crystal, the concentration of this solution will increase, so the freezing-point of this system is gradually decrease along with the formation of the crystals. And the cooling curve should look like the line (d)
And of course, the over-cooling exists, see line (e)
As a matter of fact, we get the line (f)
Procedure:
Part I
1/ Add ice and salt and water into that 2000 mL beaker. Keep the temperature of the ice/water mixture at about -3 o C. (too high, the cooling process will be very slow; too low, the solution will have this super over-cooling, and the MW you get is smaller than the actual one)
2/ Add about 25 mL pure water in the sample tube, immerse it into the ice-salt bath directly to cool down quickly. (Take it out when crystal formed, wait for the crystal just melt, wipe the outer side with paper and then put the sample tube in the air-bath tube). Stirring gently, write down the temperature
(when should you start? one at every 40s, when you see the temperature suddenly goes up, you could write it down every 20s), draw the cooling curve.
Take the sample tube out, use your hand to warm it up until all the solid just melt, and then repeat the “put this into the air-bath” procedure.
Determine the freezing point 2 times whereas the difference between consecutive measurements is less than 0.006 o C.
3/ Pour down the pure water, use the prepared solution rinse the sample tube twice, and then add this prepared solution about 25 mL. Repeat step 2.
Determine the freezing point of the glucose solution 2-3 times, whereas the difference between consecutive measurements is less than 0.003 o C.
4/ Calculate the T f* by taking average of 2 measurements. Calculate the T f .
5/ Calculate the MW of solute.
Datasheet:
Draw a table write down all the temperature, and draw the cooling curve.
Part I
1/ Freezing point measurements (water):
(1) ______________o C; (2) ______________o C; (3) ______________o C;
2/ Average Freezing point of pure water: _____________________ o C;
Part II
1/ Freezing point measurements:
(1) ______________o C; (2) ______________o C; (3) ______________o C;
2/ Average Freezing point of Glucose solution: _____________________ o C;
3/ ∆T f _______________________________o C;
4/ MW (D-Glucose) ______________________ g/mol
Suggested topics to be discussed in your lab report:
1/ What are the factors contributed to the error of the MW measurement? How to optimize the experiment so the error can be reduced?
2/ Suppose you are not provided with k f = 1.86 o C/m, instead, you are provided with information of ∆H f = 3.341x105 J/kg, can you perform the experiment and determine the MW of D-Glucose using the data you collected in the experiment?
3/ In the experiment, we used notation m B. What does m stand for? What does subscript B stand for?What is the difference between molarity and molality?
4/ How to correct the effect of supercooling when you measure the freezing temperature?
"凝固点是指物质的固相纯溶剂的蒸汽压与它的液相蒸汽压相等时的温度。

纯水的凝固点又叫冰点，为273k。

此温度时水和冰的蒸汽压相等。

但在273k，水溶液的蒸汽压低于纯水的蒸汽压，所以，
水溶液在273k不结冰。

温度继续下降，因冰的蒸汽压下降率比水溶液大，当降到Tf时，冰和溶液
的蒸汽压相等，Tf就是溶液的凝固点。

溶液的凝固点总是比纯溶剂凝固点低。

这一现象叫做溶液
的凝固点降低。

和沸点升高一样，溶液的凝固点降低ΔTf也和Δp成正比，因此难挥发非电解质稀溶液的凝固点降低与溶液的质量摩尔浓度成正比，而与溶质本性无关。

即：ΔTf=Kf bB"
以上是物理化学的解释.
说白了就是水中有了盐离子, 因此表面上的水有一部分被盐给占了.因此造成了水的蒸气压下降.而冰的蒸气压是不变的.而凝固点要冰的蒸气压小于等于水的蒸气压,因此加入盐后,水的蒸气压变小,原来相等的蒸气压,就变成冰大了.这时由于冰的蒸气压下降快,因此要更低一些的温度,才能让冰的蒸气压等于冰的蒸气压.造成了凝固点降低.
葡萄糖质量比为20：1，下降约0.5度。