50.5g of glucose
glucose is 180.18 g/mol
50.5/180.18 = 0.280 mol glucose
0.280 mol glucose/0.475 kg H2O = 0.589 m which is molality
deltaTsubF = (-1.86 degress C/m)(0.589) = -1.10 degrees C
-1.10 degrees C + 0.00 degrees C = -1.10 degrees C as your freezing point seeing that 0.00 degrees is the standard freezing point of water
deltaTsubB = (0.512 degrees C/m)(0.589) = 0.301568
it asks for 6 sig figs so
0.301568 degrees C + 100.000 degrees C (boiling point H2O) = 100.302 degrees C
Mastering Chemistry sucks sometimes....
Freezing pt is zero. boiling pt is 100
Freezing doesn't effect the enzymes since freezing does not permanently affect enzyme structure. Boiling permanently changes the structure and can change the enzymes.
The boiling point of Carbonated water is 105°C because of the carbon dioxide gas it contains.
Infusion
When an enzyme is frozen, it only slows down activity. Unlike boiling an enzyme, it does not stop it from working.
it's a colligative property of solutions... when you add a higher boiling substance to a solution the boiling point increases and when you add anything that interferes with the intramolecular forces holding the solution together the freezing point decreases.
Yes, it is possible if the solution contain solutes.
The boiling point of the solution increases, and the freezing point of the solution decreases.
Raises the boiling point and lowers the freezing point.
nobody knows
The freezing point is lowered slightly as some energy is absorbed by the impurities.
Boiling and freezing points are colligative properties, meaning they depend on the number of solute particles dissolve in solution. Glucose is a molecular compound so it is one particle dissolved in solution. CaCl2 will dissociate into three particles in solution. There are three times as many particles present in solution when CaCl2 dissolves.
I believe that you have to convert 3.40g to molality and then use that and multiply it by the constant, .512 and that should give you the answer
1. Vapor pressure lowering: the decrease in vapor pressure with increasing the number of solute molecules in solution. 2. Boiling point elevation: the increase in boiling point with increasing number of solute molecules in solution. 3. Freezing point depression: the decrease in freezing point with increasing number of solute molecules in solution. 4. Osmotic pressure
boiling point is increased and freezing point is decreased
The phenomenon of boiling point elevation is analogous to freezing point depression
Higher boiling point and a lower freezing point. These are called colligative properties. When a solute is put into solution with the solvent, there is a change in the vapor pressure, osmotic pressure, elevation of the boiling point, and depression of the freezing point.