If the radiator of an automobile contains `12 L` of water, how much would the freezing point be lowered by the addition of `5 kg` of prestone `("glycol" C_(2)H_(4)((OH)_(2))`. How many kg of Zeron (methyl alcohol) would be required to produce the same result?

Explanation

Given,
Weight of glycol`(W_(2))=5 kg =5000 g`
Molecular weight of glycol `(Mw_(2))62 g`
`:.` Weight fo water=12 kg (`:'`density =`V/W`)
Mole of glycol=`W_(2)/(Mw_(2))=5000/62 =80.6`
Molality of solution `(m) ="Mole"/"Weight of solvent" =80.6/12 =6.7`
Now,`DeltaT_(f)=K_(f)m=1.86 xx 6.7 =12 K`
`:.` Freezing point of solution = `-12^(@)C`
With methyl alcohol, to get the same freezing point depression same number of moles are required.
`:.` amount of methyl alcohol = `"Mole"xxM_(W)`
`=80.6 xx 32 =2579.2`
`=2.58 kg`