Colligative
properties calculations are used for this type of problem. Calculations are as
follows:
ΔT(freezing point) = (Kf)m
ΔT(freezing point)
= 1.86 °C kg / mol (0.50 mol/kg)
ΔT(freezing point) = 0.93 °C
Tf - T = 0.93 °C
T = -0.93 °C
Answer:
The expected freezing point of a 0.50 m solution of sodium sulfate is -0.93°C.
Explanation:
[tex]\Delta T_f=K_f\times m[/tex]
where,
[tex]\Delta T_f[/tex] =depression in freezing point =
[tex]K_f[/tex] = freezing point constant
m = molality
we have :
[tex]K_f[/tex] =1.86°C/m ,
m = 0.50 m
[tex]\Delta T_f=1.86^oC\times 0.50 m[/tex]
[tex]\Delta T_f=0.93^oC[/tex]
Freezing point of pure water = T = 0°C
Freezing point of solution = [tex]T_f[/tex]
[tex]\Delta T_f=T-T_f[/tex]
[tex]T_f=T-\Delta T_f=0^oC-0.93^oC=-0.93^oC[/tex]
The expected freezing point of a 0.50 m solution of sodium sulfate is -0.93°C.
