 # 1000 gm 1 molal solution of sucrose has f.p. -3.5c calculate wt. of ice that has been formed kf of h2o is 1.86 12 years ago

the given freezing point of the sucrose solution is -3.5 celsius.. the solvent in this sucrose solution is water.. the freezing point of water in 0.00 celsius.. hence the depression in freezing point is= [freezing point of solvent -freezing point of solution]=[0-(-3.5))=3.5 kelvin or 3.5 celsius.. we have got the formula

depression in freezing point=kf*molality

but here it has been asked to find the amount of ice formed..so we can assure a change in the molality of solution as some part of solvent is condensed to ice..

here is the calculation 3.5=1.86*m

m=3.5/1.86=1.88 molal..

a change in molality is observed.

intial molality is 1 molal in 1000g of water amount of sucrose(in moles) in it is=1

final molality is 1.88, amount of sucrose in remains unchanged but amount of solvent i.e water changes

1.88=1*1000/mass of solvent

mass of solvent =537 grams..

amount of water condensed is equal to 463 grams of water..463 grams of water condenses to form 463 grams of ice.. hence amount of ice formed is 463 gms and volume of ice formed is 514.44 millilitres