Calculate the freezing point and boiling point in each solution, assuming comple

Question

Calculate the freezing point and boiling point in each solution, assuming complete dissociation of the solute.

Part A - Calculate the freezing point of a solution containing 13.4 g FeCl3 in 172 g water.

Part B - Calculate the boiling point of a solution above.

Part C - Calculate the freezing point of a solution containing 6.5 % KCl by mass (in water).

Part D - Calculate the boiling point of a solution above.

Part E - Calculate the freezing point of a solution containing 0.151 m MgF2.

Part F - Calculate the boiling point of a solution above.

Explanation / Answer

Part A)

DTf = i*kf*m

Kf = depression in freezing point constant of water = 1.858 C/m

m = w/mwt*1000/wt ofsolvent in grams

w = weight of solute = 13.4 grams

Mwt = Molar mass of FeCl3 = 162.2040 g/mol

Wt = wt ofsolvent = 172 grams

= (13.4/162.2)*(1000/172) = 0.48 molal

i = vant hoff factor of FeCl3 = 4

DTf = T0- TF

Tf = freezing point of a solution = -3.57 C

t0 = freezing point of a solvent(water) = 0 c

0-Tf = 4*1.858*0.48

Tf = -3.57 C

Part B)

DTb = i*Kb*m

DTb = Tb-T0

Tb = boiling point of a solution = ?

T0 = boiling point of water = 100 C

Kb = elevation in boiling point constant of water = 0.521 C/m

Tb-100 = 4*0.521*0.48

Tb = 101.00032 C

Part C)

6.5 % KCl by mass (in water).

w = mass of KCL = 6.5 gram

Molar mass of KCl = 74.5513 g/mol

M = mass of solvent = 100 grams

molality = (6.5/74.5513)*(1000/100)

          = 0.872 molal

i = vanthoff factor = 2

DTf = i*Kf*m

0-Tf = 2*1.858*0.872

Tf = freezingpoint of solution = -3.24 C

Part D)

6.5 % KCl by mass (in water).

w = mass of KCL = 6.5 gram

Molar mass of KCl = 74.5513 g/mol

M = mass of solvent = 100 grams

molality = (6.5/74.5513)*(1000/100)

          = 0.872 molal

i = vanthoff factor = 2

DTb = i*Kb*m

Tb - 100 = 2*0.521*0.872

Tb = boiling point of solution = 100.9086 C

Part E)

DTf = i*kf*m

Kf = depression in freezing point constant of water = 1.858 C/m

molality of solution = 0.151 m

i = vant hoff factor of MgF2 = 3

DTf = T0- TF

Tf = freezing point of a solution = ?

t0 = freezing point of a solvent(water) = 0 c

0-Tf = 3*1.858*0.151

Tf = -0.84167 C

Part F)

molality = 0.151 molal

i = vanthoff factor of MgF2 = 3

DTb = i*Kb*m

Tb - 100 = 3*0.521*0.151

Tb = boiling point of solution = 100.236 C

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