The equation that represents the relation between freezing point depression and molality is,
Δ [tex] T_{f} = K_{f} m [/tex]
Where,
Δ[tex] T_{f} [/tex] is the change in the freezing point of the solvent.
[tex] K_{f} [/tex] of benzene is 5.12[tex] ^{0}C/m [/tex]
m is molality of the solution
Moles of solute = [tex] 22.0 g C_{8}H_{18} * \frac{1 molC_{8}H_{18}}{114.23 gC_{8}H_{18}} [/tex] = 0.193 mol
Mass of solvent in kg = [tex] 148.0 g * \frac{1 kg}{1000g} = 0.148 kg [/tex]
Molality = [tex] \frac{0.193 mol}{0.148 kg} [/tex]
= 1.30 m
Δ [tex] T_{f} = K_{f} m [/tex]
Δ [tex] T_{f} = ([tex] 5.12^{0}C/m [/tex]) (0.130 m) [/tex]
= 6.66[tex] ^{0}C [/tex]
Δ[tex] T_{f} = T_{f} (solvent) - T_{f} (solution) [/tex]
[tex] 6.66^{0}C =5.50^{0}C - Tf_{solution} [/tex]
[tex] -T_{f}(solution) = 6.66^{0}C -5.50^{0}C [/tex]
[tex] T_{f(solution) = -1.16^{0}C [/tex]
So, the correct answer is a. -1.16[tex] ^{0}C [/tex]
