Question

- Part C Calculate the freezing point of a solution containing 61.6 % KCl by mass (in water). Express your answer using two significant figures. V ALQ R O ? Tf= Submit Request Answer Part D Calculate the boiling point of a solution above. O ACQ* o ? Tb = Submit Request Answer

Answer 1

Before solving any of the question, lets see what we need to solve them.

We are supposed to calculate the boiling point and freezing point of some aqueous solutions. Now, any solution in water has a different boiling point and freezing point than pure water.

These are called colligative properties where the deviation of boiling point and freezing points depend on the number of molecule of solute present in water i.e. the molality of the solution.

Generally, the boiling point of an aqueous solution is higher than the boiling point of pure water i.e. Elevation in boiling point.

The freezing point of an aqueous solution is lower than the freezing point of pure water, i.e depression in freezing point.

Hence, the deviations are calculated as follows:

AT, = ix K: xm

and

AT,=ix K x m

Where

AT; = Depression in freezing point

Ky = molal freezing point depression constant(for water, K; = 1.86°C kg mol-?)

Ký = molal boiling point elevation constant(for water, K = 0.512°C kg mol-?)

i = van't Hoff factor

m = molality in the units of moles of solute per kg of solvent.

Now, we can proceed to solve the questions.

Part A

We have 107 g FeCl₃ in 167 g water.

Molar mass of FeCl₃ = 162.2 g/mol

Hence, number of moles of FeCl₃ =

107 g mass molar mass 0.6597 mol 162.2 g/mol

Mass of solvent = 167 g = 0.167 kg

Hence, molality of the solution is

m = 0.6597 mol - = 3.95 mol kg 0.167 kg

Note that FeCl₃ dissociates into one Fe³⁺ ion and three Cl- ion in solution. Hence, van't Hoff factor is 1+3=4

Hence, the depression in freezing point can be calculated as

AT, = ix K: x m = 4 x 1.86 °C kg mol-1 x 3.95 mol kg-1 29.39°C

Note that the freezing point of pure water is 0 ^oC

Hence, the freezing point of the solution can be calculated as

Tf solution = Tf.water - AT, = 0.0°C - 29.39' – 29.4

Hence, the freezing point of the solution is
-29.4°C
.(Rounded to three significant figures)

Part B

Similar to above, the elevation in boiling point of the solution of FeCl₃ can be calculated as

AT, = ix Ko xm = 4 x 0.512°C kg mol-1 x 3.95 mol kg- 18.09°C

Boiling point of pure water = 100.0 ^oC

Hence, the boiling point of the solution can be calculated as

To,solution = To water + AT) = 100.0°C + 8.09 108.1°C

Hence, the boiling point of the solution is
108°C
. (Rounded to three significant figures).

Part C

We have a solution of KCl of 61.6 % by mass.

Hence, there are 61.6 g of KCl in 100 g of the solution.

Hence, mass of solvent = mass of solution - mass of solute KCl = 100.0 g - 61.6 g = 38.4 g = 0.0384 kg

Molar mass of KCl = 74.55 g/mol

Hence, number of moles KCl in the solution of 100 g is

mass molar mass 61.6 g 74.55 g/mol 0.8263 mol

Hence, the molality of the solution can be calculated as

m = moles KCI mass of water 0.8263 mol " = 21.52 mol kg-1 0.0384 kg

KCl dissociates as K+ and Cl- ion. Hence, van't Hoff factor of KCl solution is 1+1=2.

Hence, we can calculate the freezing point depression as follows:

AT, = i x K x m = 2 x 1.86 °C kg mol-1 x 21.52 mol kg- 1 80.0°C

Hence, the freezing point of the solution can be calculated as

Tf, solution = Tf water - AT = 0.0°C - 80.0°C -80,

Hence, the freezing point of the solution is
-80,
. (Rounded to 2 significant figures, note the zero before decimal is significant.)

Part D

We can calculate the boiling point elevation of the solution of KCl as follows:

AT, = ix Ko xm = 2 x 0.512°C kg mol-1 x 21.52 mol kg-1 22.0°C

Hence, the boiling point of the solution can be calculated as

To,solution = To water + AT) = 100.0°C + 22.0°C 122°C

Hence, the boiling point of the solution is
122
.

Part E

We have a solution of 3.14 m .

MgF

Hence, the molality of the solution is

3.14 m = 3.14 mol kg-1

MgF₂ dissociates into Mg²⁺ and two F^- ions. Hence, the van't Hoff factor of the MgF₂ solution is 3.

Now, we can calculate the depression in freezing point as follows:

AT, = ix K x m = 3 x 1.86 °C kg mol-1 x 3.14 mol kg-1 17.52°C

Hence, the freezing point of the solution can be calculated as

Tf solution = Tf.water – AT, = 0.0°C - 17.52' -17.5°C

Hence, the freezing point of the solution of MgF₂ is
-17.5°C
.

Part F

Similar to above, we can calculate the boiling point elevation as follows:

AT, = ix Ko x m = 3 x 0.512°C kg mol-1 x 3.14 mol kg- 14.82°C

Hence, the boiling point of the solution can be calculated as

To solution = T).water + AT, = 100.0°C + 4.82' 104.8°C 105°C

Hence, the boiling point of the solution is
105°C
( Rounded to three significant figures).