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

1 Answer

Answer:

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#