What is the change in the boiling point and freezing point of 500 grams of water if 45 grams of glucose C6H12O6 were dissolved in it?

Answer 1

This quantity is equivalent to 90 g glucose / kg water = 0.50 mole particles of solute / kg water, so with a 'molar cryoscopic constant' for water of -1.86 oC/kgthis lowers the freezing point to -0.93 oC.

Answer 2

The boiling point of water increases by 0.51 degrees Celsius, and the freezing point decreases by 1.86 degrees Celsius for a 1-molal solution of glucose in water. Since we have a 45-gram portion of glucose in 500 grams of water, we can calculate the molality and use it to determine the changes in the boiling and freezing points.