Calculate the boiling point and freezing point of the solutions

Question

1. 0.575 molal aqueous solution

2. 10 g of C₆H₁₂O₆dissolved in 259 g water.

3. 0.5 mole Br₂ dissolved in 507 g chloroform

Answer 1

We would need a bit more information to calculate the first question because the vant Hoff factor is an unknown at the moment.

But we can have a go at the rest!

First, we must consider the equation for freezing point depression

ΔT_f = T_{f solvent} - T_{f solution} = k_f*m*i

We want to rearrange that to solve for the new boiling point--the boiling point of the solution

ΔT_f = T_{f solvent} - T_{f solution} = k_f*m*i => T_{f solution} = T_{f solvent} - k_f*m*i

Next, we want to find the molality. The definition of molality is:

m = n_solute

kg_solvent

So, we must also consider the moles and molecular weight of glucose

MW = mass*n^-1 => n = mass*MW^-1

We can plug this back into our definition of molality

m = n_solute = (mass*MW^-1) = mass_solute

kg_solvent kg_solvent kg_solvent*MW_solute

Finally, we can plug this back into our original equation

T_{f solution} = T_{f solvent} - k_f*m*i = T_{f solvent} - k_f*(mass_solute/kg_solvent*MW_solute)*i

<=> 0°C - (1.86°C*m^-1)(0.0556 moles C₆H₁₂O₆/0.259kg H₂O)(1)

T_{f solution} = -0.399°C

Likewise, to find the boiling point we would follow a similar path

ΔT_b = T_{b solution} - T_{b solvent} = k_b*i*m => T_{b solvent} = k_b*i*m + T_{b solvent}

<=> 100°C + (0.512°C*m^-1)(0.0556 mol C_6H12O₆/0.259kg H₂O)(1)

T_{b solution} = 100.110°C

ΔT_f = T_{f solvent} - T_{f solution} = k_f*m*i => T_{f solution} = T_{f solvent} - k_f*m*i

<=> -63.5°C - (4.68 °C*m^-1)(0.5 mol Br₂/0.507kg CHCl₃)(1)

T_{f solution} = -68.1 °C

ΔT_b = T_{b solution} - T_{b solvent} = k_b*i*m => T_{b solvent} = k_b*i*m + T_{b solvent}

<=> 61.2°C + (3.63 °C*m^-1)(0.5 mol Br₂/0.507kg CHCl₃)(1)

T_{b solvent} = 64.8 °C

Answer 2

The boiling point will increase and the freezing point will decrease.

The equation we will use is

∆T = imK

∆T = change in boiling or freezing point

i = van't Hoff factor (see below)

m = molality = moles solute / kg solvent

K = boiling or freezing constant for the solvent being used (see below)

1. 0.575 molal aqueous solution

aqueous solution means water is the solvent, so K_boil = 0.512º/m and K_freeze = 1.86º/m

we don't know the nature of the solute so i (van't Hoff is not known), but we'll assume it to be 1

boiling point

∆T = imK = (1)(0.575)(0.512) = 0.29º

New boiling point = 100º + 0.29º = 100.29º

freezing point

∆T = (1)(0.575)(1.86) = 1.1º

New freezing point = 0º - 1.1º = -1.1ºC

2. 10 g of C₆H₁₂O₆ dissolved in 259 g water.

molar mass C₆H₁₂O₆ = 180 g / mol

m = mols/kg = 10 g / 180 g/mol = 0.055 mol and this is in 259 g or 0.259 kg so m = 0.055/0.259 = 0.215

i = 1 since C₆H₁₂O₆ is a non electrolyte and does not dissociate or ionize

boiling point

∆T = imK = (1)(0.215)(0.512) = 0.11º

New boiling point = 100º + 0.11º = 100.11º

freezing point

∆T = imK = (1)(0.215)(1.86) = 0.4º

new freezing point = 0º - 0.4º = -0.4ºC

3. 0.5 mole Br₂ dissolved in 507 g chloroform

look up Kb and Kf for chloroform

kg chloroform = 0.507

m = 0.5 mol / 0.507 kg = 0.986

i = 1 for Br₂ as it does not ionize of dissociate

Plug in the values and solve for ∆T for both boiling and freezing points.

Add the ∆T_boiling to the normal boiling point for chloroform

Subtract the ∆T_freezing from the normal freezing point for chloroform