Question

What is the freezing point of a solution of ethylene glycol, a nonelectrolyte, that contains 59.0 g of (CH2OH)2 dissolved in 543 g of water? Use molar masses with at least as many significant figures as the data given.

Answer 1

The expected freezing point of a 1.75 m solution of ethylene glycol is -3.26°C.

Step-by-step explanation:

`\Delta T_f=T-T_f`

`\Delta T_f=K_f* m`

`m=\frac{\text{mass of solute}}{\text{Molar mass of solute}* {Mass of solvent in kg}}`

where,

`T` = Freezing point of solvent

`T_f` = Freezing point of solution

`\Delta T_f` =depression in freezing point

`K_f` = freezing point constant

m = molality

we have :

Mass of ethylene glycol = 59.0 g

Molar mass of ethylene glycol = 62.1 g/mol

Mass of solvent i.e. water = 543 g = 0.543 kg ( 1 g = 0.001 kg)

`K_f` =1.86°C/m ,

`m =(59.0 mol)/(62.1 g/mol* 0.543 kg)=1.75 m`

`\Delta T_f=1.86^oC/m * 1.75m`

`\Delta T_f=3.26^oC`

Freezing point of pure water = T = 0°C

Freezing point of solution =
`T_f`

`\Delta T_f=T-T_f`

`T_f=T-\Delta T_f=0^oC-3.26^oC=-3.26^oC`

The expected freezing point of a 1.75 m solution of ethylene glycol is -3.26°C.