To calculate the freezing point depression, we first need to calculate the molality of the solution:
$$molality = \frac{moles\ C_7H_{11}NO_7S}{kg\ H_2O}$$
$$molality = \frac{(25.5\ g \ C_7H_{11}NO_7S)}{(1.00\ x 10^2\ g\ H_2O)} \times \frac{1\ mol\ C_7H_{11}NO_7S}{239.25\ g \ C_7H_{11}NO_7S}$$
$$molality = 1.06\ m$$
Now, we can calculate the freezing point depression:
$$\Delta T_f = K_f \times molality$$
$$\Delta T_f = 1.86\ \frac{°C}{m} \times 1.06\ m$$
$$\Delta T_f = 1.98\ °C$$
Finally, we can calculate the freezing point of the solution:
$$Freezing\ point = 0\ °C - \Delta T_f$$
$$Freezing\ point = 0\ °C - 1.98\ °C$$
$$Freezing\ point = -1.98\ °C$$
Therefore, the freezing point of 25.5 g of C7H11NO7S in 1.00 x 10^2 g of H2O is -1.98 °C.
