Step 1: Write the freezing point depression relation.
The lowering of freezing point depends on molality $m$ through \[ \Delta T_f = K_f\, m \] where $K_f$ is the cryoscopic constant of the solvent.
Step 2: List the data.
Here $\Delta T_f = 1.0$ K, $K_f = 1.86$ K kg mol$^{-1}$, the solvent water is $0.5$ kg, and the solute molar mass is $78$ g mol$^{-1}$.
Step 3: Express molality through mass and molar mass.
Since molality is moles of solute per kg of solvent, \[ m = \frac{x/78}{0.5} \] where $x$ is the unknown mass of solute in grams.
Step 4: Substitute into the depression formula.
Putting this into $\Delta T_f = K_f\, m$ gives \[ 1.0 = 1.86 \times \frac{x/78}{0.5} \]
Step 5: Solve for $x$.
Rearranging, \[ x = \frac{1.0 \times 0.5 \times 78}{1.86} \] \[ x = \frac{39}{1.86} \] \[ x \approx 20.96 \text{ g} \]
Step 6: State the answer.
About $20.96$ g of the solid is needed to lower the freezing point by $1.0^{\circ}$C.
\[ \boxed{20.96 \text{ g}} \]