Question

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

A. Calculate the freezing point of a solution containing 10.7 g FeCl3 in 164 g water.

B. Calculate the boiling point of the solution above

C. Calculate the freezing point of a solution containing 3.7 % KCl by mass (in water).

D. Calculate the boiling point of the solution above.

 

Answer 1

When we add a solute to a pure solvent, then the freezing point of the solution decreases from that of the original pure solvent. This decrease in freezing point is known as Depression in freezing point (ΔT_F) and it is calculated by the formula-

ΔT_F = K_F · m · i

where ΔT_F = Freezing point of pure solvent - Freezing point of solution = T_{F(puresolvent)} − T_F(solution).

m = molality of solution = moles of solute mass of solvent in Kg

i = Vant off factors = number of ions produces on dissociation of 1 mole of solute

Kf = freezing point constant for solvent

Similarly

When we add a solute to a pure solvent, then the Boiling point of the solution increases from that of the original pure solvent. This increase in Boiling point is known as Elevation in Boiling point (ΔT_b) and it is calculated by the formula-

ΔT_b = K_b · m · i

where ΔT_B = Boiling point of solution - Boiling point of pure solvent =  T_b(solution) - T_{b(puresolvent)}

m = molality of solution = moles of solute mass of solvent in Kg

i = Vant off factors = number of ions produces on dissociation of 1 mole of solute

Kb = Boiling point constant for solvent

Now putting the above concept-

1- freezing point of a solution containing 10.7 g FeCl3 in 164 g water.

mass of water = 164 g = 0.164 Kg

moles of solute = moles of FeCl3 = mass of FeCl3 taken / molar mass of FeCl3

= 10.7 g / 162.2 g/mol

= 0.0659 mols

So m= molality of solution = moles of FeCl3 / mass of water

= 0.0659 mols / 0.164 Kg

= 0.4 m

Now i = ions in FeCl3 = 4 ( 1 Fe+3 and 3 Cl-)

Kf for water = 1.86 °C/m

Now putting the above values-

ΔT_F = K_F · m · i

T_F(water) − T_F(solution). = 1.86 °C/m · 0.4 m · 4

0^oC − T_F(solution). = 2.976 °C

T_F(solution). = 0^oC - 2.976 °C

= - 2.976 °C

i.e freezing point of a solution containing 10.7 g FeCl3 in 164 g water is - 2.976 °C

2- Boiling point of a solution containing 10.7 g FeCl3 in 164 g water.

mass of water = 164 g = 0.164 Kg

moles of solute = moles of FeCl3 = mass of FeCl3 taken / molar mass of FeCl3

= 10.7 g / 162.2 g/mol

= 0.0659 mols

So m= molality of solution = moles of FeCl3 / mass of water

= 0.0659 mols / 0.164 Kg

= 0.4 m

Now i = ions in FeCl3 = 4 ( 1 Fe+3 and 3 Cl-)

Kb for water = 0.512 °C/m

Now putting the above values-

ΔT_B = K_B · m · i

T_b(solution) - T_{b(puresolvent)}. = 0.512 °C/m · 0.4 m · 4

T_b(solution) - 100°C = 0.8192‬ °C

T_b(solution). = 100^oC + 0.8192‬°C

= 100.8192‬ °C

i.e Boiling point of a solution containing 10.7 g FeCl3 in 164 g water is 100.8192‬ °C

3- Calculate the freezing point of a solution containing 3.7 % KCl by mass (in water).

mass of water = lets say 1000 g = 1kg

Then mass of KCl = 1000 g * 3.7 % = 37 g

moles of solute = moles of KCl = mass of KCl taken / molar mass of KCl

= 37 g / 74.5513 g/mol

= 0.496 mols

So m= molality of solution = moles of KCl / mass of water

= 0.496 mols / 1 Kg

= 0.496 m

Now i = ions in KCl = 1 ( 1 K+ and 1 Cl-)

Kf for water = 1.86 °C/m

Now putting the above values-

ΔT_F = K_F · m · i

T_F(water) − T_F(solution). = 1.86 °C/m · 0.496 m · 2

0^oC − T_F(solution). = 1.84512 °C

T_F(solution). = 0^oC - 1.84512 °C

= - 1.84512 °C

i.e freezing point of a solution containing 3.7 % KCl in water is - 1.84512 °C

4- Boiling point of a solution containing 3.7 % KCl by mass (in water).

Kb for water = 0.512 °C/m

Now putting the above values-

ΔT_B = K_B · m · i

T_b(solution) - T_{b(puresolvent)}. = 0.512 °C/m · 0.496 m · 2

T_b(solution) - 100°C = 0.5079‬ °C

T_b(solution). = 100^oC + 0.5079‬°C

= 100.5079‬ °C

i.e Boiling point of a solution containing 3.7 % KCl in water is 100.5079‬ °C