Q.Calculate the freezing point of an aqueous solution containing 10.5 g of Magnesium bromide in 200 g of water, assuming complete dissociation of Magnesium bromide. (Molar mass of Magnesium bromide = 184 g mol$^{-1}$, $K_f$ for water = 1.86 K kg mol$^{-1}$).

Concept and Intuition

For a non-volatile, non-electrolyte solute, freezing point depression is a colligative property:

where $m$ is the molality of the solution (mol solute per kg solvent). Since molality is defined in terms of moles of solute, and moles = mass/molar mass, rearranging this equation lets us solve directly for the unknown molar mass once $\Delta T_f$, $K_f$, the solute mass, and the solvent mass are known.

Step-by-Step Solution

1. Freezing point of pure water is $0^\circ C$; solution freezes at $-1.05^\circ C$, so $\Delta T_f = 0-(-1.05) = 1.05\ K$.
2. $m = \dfrac{\Delta T_f}{K_f} = \dfrac{1.05}{1.86} = 0.5645\ mol\,kg^{-1}$.
3. Mass of solvent = 20.4 g = 0.0204 kg.
4. Moles of solute $= m \times$ (kg solvent) $= 0.5645 \times 0.0204 = 0.01152\ mol$. …

Concept and Intuition

Freezing-point depression is a colligative property: for a non-volatile, non-electrolyte solute, the depression depends only on the number of solute particles per kg of solvent (molality), not their identity:

Once we know the molality, we know moles of solute per kg of solvent; combined with the actual mass of solute dissolved, we can back out the molar mass.

Step-by-Step Solution

1. Given: $\Delta T_f = 0.64$ K, $K_f(\mathrm{C_6H_6}) = 5.12$ K kg mol⁻¹, mass of solute $w = 1.95$ g, mass of solvent (benzene) $W = 100$ g $= 0.100$ kg.
2. Find molality: $m = \dfrac{\Delta T_f}{K_f} = \dfrac{0.64}{5.12} = 0.125\ \mathrm{mol\,kg^{-1}}$. …

Concept and Intuition

Freezing-point depression is a colligative property: $\Delta T_f = K_f \times m$, where $m$ is the molality of solute particles. For a non-electrolyte (non-dissociating) solute, this directly gives the number of moles of solute per kg of solvent, from which the molar mass follows once we know the actual mass dissolved.

Step-by-Step Solution

1. Freezing point depression: $\Delta T_f = 273.16\ \text{K} - 272.814\ \text{K} = 0.346$ K.
2. Molality from $\Delta T_f = K_f m$: $m = \dfrac{\Delta T_f}{K_f} = \dfrac{0.346}{1.86} = 0.186$ mol/kg.
3. Moles of solute present: since molality is per kg of solvent, and here there's $21.5\ \text{g} = 0.0215$ kg of water, $n = m \times 0.0215\ \text{kg} = 0.186 \times 0.0215 \approx 0.004\ \text{mol}$ …

Concept and Intuition

Freezing point depression is a colligative property: $\Delta T_f = K_f \times m$,

where $m$ is the molality of the solute (moles of solute per kg of solvent) and

$K_f$ is the molal depression (cryoscopic) constant of the solvent. Molecular

mass of the solvent (ethanol, given as $46.07\ g.mol^{-1}$) is extra/unneeded

information here — it would matter if converting between mole fraction and

molality, but molality is defined directly using the mass of solvent in kg.

Step-by-Step Solution

1. Moles of maltose $= \dfrac{72.4\ g}{342.3\ g.mol^{-1}} = 0.2115\ mol$.
2. Mass of solvent (ethanol) $= 1000\ g = 1\ kg$.
3. Molality $m = \dfrac{0.2115\ mol}{1\ kg} = 0.2115\ mol.kg^{-1}$. …

Concept and Intuition

Freezing-point depression is a colligative property depending only on the number of solute particles per kg of solvent (molality), not their identity: $\Delta T_f = K_f \times m$.

Step-by-Step Solution

1. $m = \Delta T_f / K_f = 0.4 / 5.12 = 0.078125\ mol\,kg^{-1}$.
2. Moles of solute $= m \times \text{kg of solvent} = 0.078125 \times 0.050 = 3.90625\times10^{-3}\ mol$.
3. Molar mass $= \dfrac{\text{mass of solute}}{\text{moles of solute}} = \dfrac{2\ g}{3.90625\times10^{-3}\ mol} = 512\ g\,mol^{-1}$. …

Concept and Intuition

Adding a solute to water lowers its freezing point and raises its boiling point (colligative properties). Ethylene glycol is favoured for automotive coolant because it's highly water-miscible, has a high boiling point of its own, is relatively cheap, and doesn't damage engine components — making the water/glycol mixture resistant to both freezing in winter and boiling over in summer.

Step-by-Step Solution

1. The functional requirement: a substance that mixes completely with water and substantially depresses its freezing point (and ideally raises its boiling point too).
2. Ethylene glycol ($HOCH_2CH_2OH$) is fully miscible with water and is specifically manufactured/used for exactly this application in vehicle radiators. …