Define Colligative Properties. Discuss any two colligative properties in detail with their relevant mathematical expressions. Also, solve the following numerical problem: Calculate the freezing point of an aqueous solution containing 10.5 g of $MgBr2$ ($Molar\ mass = 184\ g\ mol^{-1}$) in 200 g of water, assuming complete dissociation of $MgBr2$. ($Kf$ for water = $1.86\ K\ kg\ mol^{-1}$)

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Definition of Colligative Properties
Colligative properties are those properties of dilute solutions that depend only upon the number of solute particles (molecules or ions) present in the solution and do not depend upon the chemical nature of the solute. Important colligative properties include relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure. Elevation of Boiling Point ($\Delta Tb$)
The boiling point of a liquid is the temperature at which its vapour pressure becomes equal to atmospheric pressure. When a non-volatile solute is dissolved in a pure solvent, the vapour pressure of the solution decreases, so it must be heated to a higher temperature to make its vapour pressure equal to atmospheric pressure. Thus, the boiling point of the solution ($Tb$) is higher than that of the pure solvent ($Tb^0$). Mathematical Expression: $\Delta Tb = Tb - Tb^0 = Kb \cdot m$ Where $Kb$ is the ebullioscopic constant and $m$ is the molality of the solution. Depression of Freezing Point ($\Delta Tf$)
The freezing point of a substance is defined as the temperature at which the vapour pressure of the liquid form of the substance is equal to vapour pressure of its solid form. When a non-volatile solute is added to a solvent, the vapour pressure of the solution is lower than that of the pure solvent at freezing point. Consequently, freezing point gets depressed. Mathematical Expression: $\Delta Tf = Tf^0 - Tf = Kf \cdot m$ Where $Kf$ is the cryoscopic constant and $m$ is the molality of the solution. Numerical Solution: Given data: Mass of solute ($MgBr2$) $WB = 10.5\ g$ Molar mass of solute $MB = 184\ g\ mol^{-1}$ Mass of solvent (water) $WA = 200\ g$ $Kf = 1.86\ K\ kg\ mol^{-1}$ Step 1: Dissociation of $MgBr2$ and Van't Hoff factor ($i$): $MgBr2 ightarrow Mg^{2+} + 2Br^{-}$
Total number of particles = $1 + 2 = 3$.
Therefore, the Van't Hoff factor $i = 3$ (assuming 100% dissociation). Step 2: Calculate Molality ($m$): $m = \frac{WB 	imes 1000}{MB 	imes WA} = \frac{10.5 	imes 1000}{184 	imes 200} = \frac{10500}{36800} = 0.2853\ mol\ kg^{-1}$ Step 3: Calculate Depression in Freezing Point ($\Delta Tf$): $\Delta Tf = i \cdot Kf \cdot m$ $\Delta Tf = 3 	imes 1.86 	imes 0.2853 = 1.592\ K$ Step 4: Calculate Freezing Point of Solution ($Tf$):
Freezing point of pure water $Tf^0 = 0^\circ C$ or $273.15\ K$. $\Delta Tf = Tf^0 - Tf$ $1.592 = 273.15 - Tf$ $Tf = 273.15 - 1.592 = 271.558\ K$ (or $-1.59^\circ C$)