Benzene freezes at 5.45oC. The Kf for benzene is - 5.07 oC/m. What would be the freezing point of a 0.210m solution of octane in benzene
a) 1.06oC
b) 4.39oC
c) 6.51oC
d) 10.52oC
e) 5.80oC
delta T = dT = Kf*m
Substitute Kf and m and solve for dT. Then subtract from the normal freezing point to find the new freezing point.
answer is b
To find the freezing point of a solution, we can use the equation:
ΔT = Kf * m
where ΔT is the change in freezing point, Kf is the freezing point depression constant, and m is the molality of the solution (moles of solute per kilogram of solvent).
Given that the Kf for benzene is -5.07 °C/m, and we have a 0.210 m solution of octane in benzene, we can substitute these values into the equation:
ΔT = (-5.07 °C/m) * (0.210 m)
ΔT = -1.06 °C
Since ΔT represents the change in freezing point, to find the actual freezing point, we need to subtract ΔT from the freezing point of pure benzene.
Freezing point of pure benzene = 5.45 °C
Freezing point of the solution = 5.45 °C - 1.06 °C = 4.39 °C
Therefore, the freezing point of the 0.210 m solution of octane in benzene is 4.39 °C.
Therefore, the correct answer is option b) 4.39 °C.