Question

Ethylene glycol (C2H4(OH)2) , when dissolved in water, provides the standard ‘anti-freeze’ coolant for water-cooled engines. In order to depress the freezing point of water by 20 °C, how many grams of ethylene glycol would need to be dissolved in 15 kg of pure water? (The molal freezing point depression constant for water Kf = 1.86 K mol-1 kg and the relevant atomic masses are: C = 12g, H = 1g and O = 16g.) Note: ethylene glycol is an organic compound and does NOT break up or dissociate when it dissolves in water. Hint: first calculate the molality (m) of the ethylene glycol solution.

What is the boiling point of this same solution at atmospheric pressure? (The molal boiling point elevation constant for water Kb = 0.52 K mol -1 kg.)

Answer 1

146,575 grams of ethylene glycol would need to be dissolved in 15 kg of pure water

The boiling point of the solution is 181.95°C.

Step-by-step explanation:

Mass of ethylene glycol = x

Molar mass of ethylene glycol = 2 × 12 g/mol + 2 × 32 g/mol+ 4 × 1= 62g/mol

Moles of ethylene glycol =
`(x)/(62 g/mol)`

Mass of solvent that is water = 15 kg

The molal freezing point depression constant for water
`K_f=1.86 K kg/mol`

Molality of the solution:

`Molality=\frac{\text{Moles of compound}}{\text{Mass of solvent (kg)}}`

`Molality=m=(x)/(62 g/mol* 15 kg)`

Depression in freezing point of water =
`\Delta T_f=20^oC=293.15 K`

((T)°C =T+ 273.15 K)

`\Delta T_f=K_f* m`

`293.15 K=1.86 K kg/mol* (x)/(62 g/mol* 15 kg)`

x = 146,575 g

Boiling point of this solution =
`T_b`

The molal boiling point elevation constant for water
`K_b = 0.52 K kg/mol`

`\Delta T_b=K_b* m`

`\Delta T_b=0.52 K kg/mol* (146,575 g)/(62 g/mol* 15 kg)`

`\Delta T_b=81.95K`

Normal boiling point of water is = T = 373.15 K

`\Delta T_b=T_b-T`

`T_b=\Delta T_b+T=81.95K+373.15 K=455.1 K=181.95^oC`

The boiling point of the solution is 181.95°C.