Answer:
-2.79°C is the expected freezing point of a 0.50 m solution of lithium sulfate.
Explanation:
Freezing point of water = T =0°C
Melting point of sample = [tex]T_f[/tex]
Depression in freezing point = [tex]\Delta T_f[/tex]
Depression in freezing point is also given by formula:
[tex]\Delta T_f=i\times K_f\times m[/tex]
[tex]K_f[/tex] = The freezing point depression constant
m = molality of the sample
i = van't Hoff factor
We have: [tex]K_f[/tex] = 40°C kg/mol
m = 0.50 m
i = 3 (lithium sulfate is an ionic compound)
[tex]\Delta T_f=3\times 1.86^oC kg/mol\times 0.50 m=2.79^oC[/tex]
[tex]\Delta T_f=T- T_f[/tex]
[tex]T_f=T-\Delta T_f=0^oC-2.79^oC=-2.79^oC[/tex]
-2.79°C is the expected freezing point of a 0.50 m solution of lithium sulfate.
The freezing point of the solution is - 2.79∘c.
Using the formula;
ΔT = K m i
ΔT = freezing point depression
K = Freezing point constant
m = molality of solution
i = Van't Hoff factor
From the question;
K = 1.86 ∘c/m
m = 0.50 m
i = 3
ΔT = 1.86 ∘c/m × 0.50 m × 3
ΔT = 2.79∘c
But ΔT = Freezing point of pure solvent - freezing point of solution
Freezing point of pure solvent = 0∘c
Freezing point of solution = Freezing point of pure solvent - ΔT
Freezing point of solution = 0∘c - 2.79∘c
Freezing point of solution = - 2.79∘c
Learn more: https://brainly.com/question/5624100
