Question

Calculate the boiling point of a solution containing 12g of glucose C6h12O6 dissolved in 200g water (0.512-Kb of H20)

Answer 1

Explanation:

Boiling point elevation and freezing point depression are both colligative properties.

∆T = imK

∆T = change in temperature

i = van't Hoff factor = 1 for glucose since it does not ionize or dissociate. It is a single particle.

m = molality = moles solut/kg solvent = 43.8g/180 g/mol = 0.243 mol / 0.3599 kg = 0.676 m

K = boiling or freezing constant = 0.512 and 1.86 respectively

Boiling point elevation:

∆T = (1)(0.676)(0.512) = 0.35º

Freezing point depression:

∆T = (1)(0.676)(1.86) = 1.26º

Answer 2

Answer:100.171°C

Explanation:

m=no.of mole/no.of solvent

12g×(1/180.156g/mole)×(1/200g)×(1000g/1kg)≈0.333=m

Tb=Kb•m

Tb=0.512°C(0.333)

Tb≈0.171°C

100°C+0.171°C=100.171°C