Question

What is the boiling point of a solution of 0.250g of glycerol, C3H8O3, in 40g of H2O? Glycerol is a molecular compound, i = 1

Answer 1

Step-by-step explanation

Given:

mass of glycol = 0.250g

mass of solvent (water) = 40g = 0.04 Kg

i = 1

Required: The boiling point of the solution

Solution

`\Delta T\text{ = Kbm}`

where Kb is the boiling point elevation constant

m is the molal concentration (molality) of all species

Step 1: Find the moles of glycerol

n = m/M

n = 0.250g/92,09382 g/mol

n = 2.71x10^-3 mol

Step 2: Find the molality

molality (m) = moles of solute/Kg of solvent

molality = 2.71x10^-3/0.04Kg

molality = 0.068 m

Step 3: Find the change in temperature

`\Delta T\text{ = Kbm}`

Kb of water = 0.512 cm^-1

DT = 0.512 x 0.068

DT = 0.035 'C

Step 4: Find the boiling point

Boiling point of water = 100 'C

DT = 0.035 'C

New boiling point =100 + 0.035 = 100.035 'C

Answer

100.035 'C