\begin{align*}
\text{Given:} & \\
& \text{Molality of solution (m)} = 0.1 \, \text{m} = 0.1 \, \text{mol kg}^{-1} \\
& \text{Degree of dissociation } (\alpha) = 1.35\% = 0.0135 \\
& \text{Temperature } (T) = 25^\circ \text{C} = 25^\circ \text{C} + 273.15 = 298.15 \, \text{K} \\
& \text{Molarity of solution (M)} = 0.1 \, \text{M} \\
\text{To find:} & \\
1. & \text{Freezing point of solution} \\
2. & \text{Osmotic pressure of solution} \\
\text{Formulae:} & \\
1. & \alpha = \frac{i – 1}{n – 1} \\
2. & \Delta T_f = i \cdot K_f \cdot m \\
3. & \pi = i \cdot M \cdot R \cdot T \\
\text{Calculation:} & \\
& \text{Using formula (i),} \\
& \alpha = \frac{i – 1}{n – 1} = i – 1 \text{ because } n = 2 \\
& \therefore i = 1 + \alpha = 1 + 0.0135 = 1.0135 \\
& \text{Now, using formula (ii),} \\
& \Delta T_f = i \cdot K_f \cdot m = 1.0135 \cdot 1.86 \, \text{K kg mol}^{-1} \cdot 0.1 \, \text{mol kg}^{-1} = 0.189 \, \text{K} = 0.189^\circ \text{C} \\
& \text{Now, } \Delta T_f = T_f^0 – T_f \\
& \therefore T_f = T_f^0 – \Delta T_f = 0^\circ \text{C} – 0.189^\circ \text{C} = -0.189^\circ \text{C} \\
& \text{Now, using formula (iii),} \\
& \pi = i \cdot M \cdot R \cdot T = 1.0135 \cdot 0.1 \, \text{mol dm}^{-3} \cdot 0.08205 \, \text{dm}^3 \text{ atm K}^{-1} \text{ mol}^{-1} \cdot 298.15 \, \text{K} = 2.48 \, \text{atm} \\
& \therefore \text{The freezing point of the solution is } -0.189^\circ \text{C} \\
& \therefore \text{The osmotic pressure of the solution at } 25^\circ \text{C} \text{ is } 2.48 \, \text{atm}
\end{align*}