11 and 12 please And thanks a bunch Additional Problems to s

Additional Problems to solve: at is the freezing point of a solution of 5 grams of glycerol (C3H8O3) in 150 grams of water (Kwater = 1.86 oCrn)? 11. Wh 12. When 9.72 grams of sugar is dissolved in 175 grams of water, the freezing point of the solution is lowered to -0.302°C Calculate the molecular weight of the sugar. (Water freezes at 0°C and Kr water is 1.86 °C/m). 40

Solution

11)

dTf = i*Kf*m

DTf = Tf-TS c

Tf = freezingpoint of solvent = 0 c

Ts = freezingpoint of solution = x c

i= vanthoff factor of solute = 1

Kf of water = 1.86 c/m

m = molality = (w/M)*(1000/W)

    = (5/92)*(1000/150)

  
(0-x) = 1*1.86*(5/92)*(1000/150)

x = freezingpoint of solution = -0.674 c

12)

dTf = i*Kf*m

DTf = Tf-TS c

Tf = freezingpoint of solvent = 0 c

Ts = freezingpoint of solution = -0.302 c

i= vanthoff factor of solute = 1

Kf of water = 1.86 c/m

m = molality = (w/M)*(1000/W)

    = (5/M)*(1000/150)

  
(0-(-0.302)) = 1*1.86*(5/M)*(1000/150)

M = molarmass of solute = 205.3 g/mol