If I cool a solution containing 62 g of ethylene glycol in 250 g water to -9.3 °C, what mass of ice will separate out?

The formula for freezing point depression #ΔT_text(f)# is

#color(blue)(bar(ul(|color(white)(a/a)ΔT_text(f) = iK_text(f)bcolor(white)(a/a)|)))" "#

where

#i =# the van't Hoff #i# factor
#K_text(f) =# the freezing point depression constant for the solvent
#b =# the molality of the solute

We can rearrange this expression to get

#b = (ΔT_text(f))/(iK_text(f)#

In this problem,

#ΔT_text(f) = "9.3 °C"#
#i=1#, because ethylene glycol is a nonelectrolyte
#K_text(f) = "1.86 °C·kg·mol"^"-1"#

∴ #b = (9.3 color(red)(cancel(color(black)("°C"))))/(1 × 1.86 color(red)(cancel(color(black)("°C")))·"kg·mol"^"-1") = "5.00 mol·kg"^"-1"#

#"Moles of ethylene glycol" = 62 color(red)(cancel(color(black)("g EG"))) × "1 mol EG"/(62.07 color(red)(cancel(color(black)("g EG")))) = "0.999 mol EG"#

So, the allowed amount of water is

#"Mass of water" = 0.999 color(red)(cancel(color(black)("mol EG"))) × "1 kg water"/(5 color(red)(cancel(color(black)("mol EG")))) = "0.200 kg water" = "200 g water"#

You started with 250 g of water, so the other 50 g must be present as ice.