The freezing point of solution is –1° C, when 6 gram urea is dissolved in 100 gram water. If 342 gram sugar is dissolved in 1000 gram water, then the freezing point of the solution will be... .

Dear student,

Freezing point depression is given by Ebbing. The freezing point depression ΔT_f is a colligative property of the solution, and for dilute solutions is found to be proportional to the molal concentration c_m of the solution:

ΔT_f = K_fc_m

Where K_f is called the freezing-point-depression constant.

depression in frezing point = k_b m

6gm urea when dissolved in 100gm water , molality (m)= moles of solute/mass of solvent in kg

molar mass of urea = 60

      m = (6/60)/(0.1) = 1

depression in freezing point = (0-1) = 1= dT

now using fomula , dT = k_bm

     K_b = 1         .....................1

now for another solution using same formula ,

 molality of solution , moles/mass of solvent in kg = (342/342)/1 = 1

m = 1

molar mass of sugar = 342

now by using same formula , dT = k_bm = 1                    (use K_b from eq 1)

it means that frezing point of water is againg -1^oC...

approve if u like my ans