Question

What is the freezing point of a nonionizing antifreeze solution containing 388g ethylene glycol `C_2H_6O_2` and 409 g of water?

Answer 1

The freezing point = `- 28.42^oC`

Explanation:

`∆T_b` =`i*K_f*m`

• `∆T_b` = depression in freezing point
• `i`= vant hoff factor
• `K_f` = depression constant
• `m` = molality

mass of ethylene = `388 g`

molar mass =`62.07` `g`/`mol`

no. of moles = `388`/`62.07` = `6.25` moles

mass of solvent = `409 g` = `0.409 kg`

molality = `6.25` moles/`0.409 kg` = `15.28 m`

• `i`= `1`

• `T_i` =`0^oC`

• `K_f`= `1.86^ oC/m`

Therefore,

`∆T_f = i*K_f*m`
`∆T_f = 1.86^oC/m xx 15.28 m`

`∆T_f = 28.42^o C`

`T_f = 0^oC - 28.42^ oC = -28.42^ oC`