Find the normal freezing point of 0.32 m
C12H22O11(aq). The freezing point constant
Of water is 1.86 K · kg/mol.
Answer in units of ◦C.

Using

T_{f(pure water)} - T_{f(C_{12}H_{22}O_{11})} = K_f m_B

Gives you:

273 - T_f = (1.86)(0.32)

Solve for T_f and convert into degrees Celsius.