Answer:
The expected freezing point of a 0.50 m solution of sodium sulfate is -0.93°C.
Explanation:
\(\Delta T_f=K_f\times m\)
where,
\(\Delta T_f\) =depression in freezing point =
\(K_f\) = freezing point constant
m = molality
we have :
\(K_f\) =1.86°C/m ,
m = 0.50 m
\(\Delta T_f=1.86^oC\times 0.50 m\)
\(\Delta T_f=0.93^oC\)
Freezing point of pure water = T = 0°C
Freezing point of solution = \(T_f\)
\(\Delta T_f=T-T_f\)
\(T_f=T-\Delta T_f=0^oC-0.93^oC=-0.93^oC\)
The expected freezing point of a 0.50 m solution of sodium sulfate is -0.93°C.
