Answer:-
\begin{align*}
\text{Given:} & \quad \text{Density of a solution (\(d\)) = 1.063 g mL\(^{-1}\)} \\
& \quad \text{Osmotic pressure of solution (\(\pi\)) = 12.16 atm} \\
& \quad \text{Temperature (\(T\)) = 25 °C = 298.15 K} \\
& \quad \text{Freezing point of solution (\(T_f\)) = -1.03 °C} \\
\text{To find:} & \quad \text{Molar mass of a compound} \\
\text{Formulae:} & \\
1. & \quad \Delta T_f = K_f \cdot m \\
2. & \quad \pi = MRT \\
3. & \quad m = \frac{1000 \cdot W_2}{M_2 \cdot W_1} \\
\text{Calculation:} & \\
& \quad \text{Gas constant (\(R\)) = 0.08205 dm\(^3\) atm K\(^{-1}\) mol\(^{-1}\)} \\
& \quad \Delta T_f = T_f^0 – T_f = 0 °C – (-1.03 °C) = 1.03 K \\
& \quad K_f \text{ of water} = 1.86 K kg mol\(^{-1}\) \\
& \quad \text{Using formula (1):} \\
& \quad \Delta T_f = K_f \cdot m \\
& \quad m = \frac{\Delta T_f}{K_f} = \frac{1.03 \, \text{K}}{1.86 \, \text{K kg}^{-1}} = 0.554 \, \text{mol kg}^{-1} = 0.554 \, m \\
& \quad \text{Using formula (2):} \\
& \quad \pi = MRT \\
& \quad M = \frac{\pi}{RT} = \frac{12.16 \, \text{atm}}{0.08205 \, \text{dm}^3 \text{ atm K}^{-1} \text{ mol}^{-1} \times 298.15 \, \text{K}} = 0.497 \, \text{mol dm}^{-3} = 0.497 \, M \\
& \quad \text{Mass of solvent} = \frac{0.497 \, \text{mol dm}^{-3}}{0.554 \, \text{mol kg}^{-1}} \times 1 \, \text{dm}^3 = 0.897 \, \text{kg} = 897 \, \text{g} \\
& \quad \text{Mass of solution} = 1.063 \, \text{g mL}^{-1} \times 1000 \, \text{mL} = 1063 \, \text{g} \\
& \quad \text{Mass of solute} = 1063 \, \text{g} – 897 \, \text{g} = 166 \, \text{g} \\
& \quad \text{Using formula (3):} \\
& \quad m = \frac{1000 \cdot W_2}{M_2 \cdot W_1} \\
& \quad \therefore \, M_2 = \frac{1000 \, \text{g kg}^{-1} \times 166 \, \text{g}}{0.554 \, \text{mol kg}^{-1} \times 897 \, \text{g}} = 334 \, \text{g mol}^{-1} \\
& \quad \text{The molar mass of the compound is} \, 334 \, \text{g mol}^{-1}.
\end{align*}