How many mmoles of sucrose should be dissolved in 500 gms of water so as to get a solution which has a difference of 103.57°C between boiling point and freezing point.

(Kf = 1.86 K Kg mol^–1, K_b = 0.52 K Kg mol^–1)

let x is the mole required of sucrose.
As we know the normal freezing and boiling point of water is 0 and 100 degreeC respectively.
molality of sucrose=(x/500)*1000=2x
Now calculate the final boiling point:
T1’.........=...........T1......+....Kb*2x
and the final freezing point:
T2’.........=...........T2......+....Kf*2x
where T1=100 degree
T2=0 degree
now, according to question,
T1’-T2’=103.57
now calculate value of x.

let x be milimole of sucrose dissolve in 500gm water ,so moles will be x*(10^-3)
molarity=x(10^-3) / 500(10^-3)  = x/500 from here Tb=100+.52 x/500
Tf-0 =1.86 x/500 from here  Tf=1.86 x/500
given Tb – Tf=103.57 =100+.52 x/500  +1.86 x/500
from here by solving u will get x=750 mili moles.