Question

Find the freezing point of the solution containing.520g glucose dissolve in 80.2 g water (kf=1.86)

Answer 1

Answer:

∆Tf=molality×kf ,,, here ∆Tf is freezing point depression and[ molality= weight of the glucose×1000/molar mass of glucose× weight of the water ]

∆Tf= 1.86×520×1000/180×80.2 after solving it ∆Tf= 66.99 means ∆Tf is 67°c and ∆Tf= freezing point temperature of solvent- freezing point temperature of solution { * we know freezing point of solvent (water) is 0°c then 67°c=0-Tsolution so Tsolution is -67°c ans