Question

Assuming 100% dissociation, calculate the freezing point and boiling point of 2.67 m Na2SO4(aq). Constants may be found here.

Answer 1

Assuming 100% dissociation, calculate the freezing point and boiling point of 2.67m Na₂SO₄(aq).

The only information we need is the molality of the solution, the vant Hoff factor i for Na₂SO₄ (which we can estimate by counting how many ions the compound will produce in solution, in this case 2Na⁺ and 1SO₄^2-, so i =3) and the K_f and K_b constants for the solvent. It is important to note here that the solvent is water so we need to look up for K_f and K_b for water:

i = 3

m = 2.67 m

K_f = 1.86 °C/m

K_b = 0.512 °C/m

1) Freezing Point

Use the formula:

ΔT = iK_Fm

ΔT = 3(1.86 °C/m)(2.67m)

ΔT = 14.9°C

So, if the ΔT = 14.9°C, then the freezing point T_f will be:

T_f = T_f⁰ - ΔT

Where T_f⁰ is the freezing point of pure water (0°C).

T_f = T_f⁰ - ΔT

T_f = 0°C - 14.9°C

T_f = -14.9°C

2) Boiling point

Use the formula:

ΔT = iKbm

ΔT = 3(0.512 °C/m)(2.67m)

ΔT = 4.1°C

So, if the ΔT = 4.1°C, then the boiling point T_b will be:

T_b = T_b⁰ + ΔT

Where T_b⁰ is the boiling point of pure water (100°C).

T_b = T_b⁰ + ΔT

Tb = 100°C + 4.1°C

T_b = 104.1°C

So, the freezing point and boiling point for the solution will be -14.9°C and 104.1°C respectively.