1. Use freezing point depression formula

   $\Delta T_f = K_f \cdot m$

   where:

   • $\Delta T_f$ = depression in freezing point
   • $K_f = 5.12\ ^\circ \mathrm{C/m}$
   • $m$ = molality of solution

2. Calculate moles of solute

   Solute is $C_4H_{10}$.

   Molar mass of $C_4H_{10}$: $4\times 12 + 10\times 1 = 48 + 10 = 58\ \mathrm{g/mol}$

   Moles of $C_4H_{10}$: $n = \frac{10}{58} \approx 0.1724\ \mathrm{mol}$

3. Calculate mass of solvent in kg

   Solvent is benzene, $200\ \mathrm{g} = 0.200\ \mathrm{kg}$.

4. Calculate molality

   $m = \frac{0.1724}{0.200} = 0.862\ \mathrm{m}$

5. Calculate depression in freezing point

   $\Delta T_f = 5.12 \times 0.862 \approx 4.41^\circ \mathrm{C}$

6. Calculate new freezing point

   Pure benzene freezes at $5.5^\circ \mathrm{C}$.

   Therefore, $T_f(\text{solution}) = 5.5 - 4.41 = 1.09^\circ \mathrm{C}$

7. Final integer answer

   $\boxed{1}$

So the solution freezes at approximately $1^\circ \mathrm{C}$.