Question

5% solution of cane sugar in water has freezing point of 271 kelvin calculate the freezing point of 5% glucose in water is freezing point of pure water is 270 3.15 kelvin

Answer 1

Molar mass of cane sugar C12H22O11=342gmol−1C12H22O11=342gmol−1

Molality of sugar =5×1000342×1005×1000342×100

⇒0.146⇒0.146

ΔTfΔTf for sugar solution =273.15-271=2.15∘2.15∘

ΔTf=Kf×mΔTf=Kf×m

Kf=2.150.146Kf=2.150.146

Molality of glucose solution =5180×1000100=5180×1000100=0.278=0.278

ΔTfΔTf(Glucose)=2.150.1462.150.146×0.278×0.278

⇒4.09∘⇒4.09∘

∴∴ Freezing point of glucose solution =273.15−4.09∘=269.06K

Answer 2

Answer:

Sample 'B' will not freeze at 0°C because it is not pure water. At one atmospheric pressure, the boiling point of pure water is 100°C and the freezing point of pure water is 0°C.