Ethylene glycol, CH_(2)OH*CH_(2)OH the major component of permanent antifreeze, effectively depresses the freezing point of water in an automobile radiator. What minimum weight of ethylene glycol must be mixed with 6 gallons of water to protect it from freezing at -24^(@)C. (1 gallon =3.785 lit, K_(f)=1.86) ?

Ethylene glycol, CH_(2)OH*CH_(2)OH the major component of permanent an

1.	Ethylene glycol, CH_(2)OH*CH_(2)OH the major component of permanent antifreeze, effectively depresses the freezing point of water in an automobile radiator. What minimum weight of ethylene glycol must be mixed with 6 gallons of water to protect it from freezing at -24^(@)C. (1 gallon =3.785 lit, K_(f)=1.86) ?
Answer» SOLUTION :We have, `DeltaT_(F)=K_(f)XX` molality `:.` molality `=(DeltaT_(f))/(K_(f))=(24)/(1.86)=12.90` (f.p. of `H_(2)O=0^(@)C`) Thus, `1kg` of `H_(2)O` should contain `12.90` moles of ethylene GLYCOL. `:.22.7kg` (i.e., 6 gallons) of `H_(2)O` should contain `12.90xx22.7` moles `=(12.90xx22.7xx62)g` `=18155g` `=18.155kg` of `CH_(2)OHCH_(2)OH` `(CH_(2)OHCH_(2)OH=62)`