Question

A 5% sol of cane sugar in water has freezing point of 217k. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 k

Answer 1

Answer:

269.06K

Explanation:

Freezing point of water = 217k

Glucose in water = 5%

Molar mass of cane sugar C12H22O11 =342gmol−1

Molarity of sugar = 5×1000/342×100

= 0.146

ΔTf for sugar solution = 273.15 - 271

= 2.15°

ΔTf = Kf × m

Kf = 2.15/0.146

Molarity of glucose solution = 5/180 × 1000/100

= 0.278

ΔTf(Glucose) = 2.150/146 × 0.278

= 4.09°

Freezing point = 273.15−4.09°

= 269.06K

Therefore, the freezing point is 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.