The freezing point of one molal KCl solution, assuming KCl to be completely dissociated in water, is: ($K_f$ for water = 1.86 K kg mol$^{-1}$)

Question

$W$ g of a non-volatile electrolyte solid solute of molar mass $M\,\text{g mol}^{-1}$ when dissolved in $100\,\text{mL}$ water decreases vapour pressure of water from $640\,\text{mm Hg}$ to $600\,\text{mm Hg}$. If aqueous solution of the electrolyte boils at $375\,\text{K}$ and $K_b$ for water is $0.52\,\text{K kg mol}^{-1}$, then the mole fraction of the electrolyte $(x_2)$ in the solution can be expressed as ___.
(Given: density of water $=1\,\text{g mL}^{-1}$, boiling point of water $=373\,\text{K}$)

Answer 1

The freezing point depression, $\Delta T_f$, is determined by the formula: \[\Delta T_f = i \cdot K_f \cdot m\]In this calculation:- $i$ represents the van’t Hoff factor, which is 2 for KCl due to its dissociation into $ \text{K}^+ $ and $ \text{Cl}^- $.- $K_f$ is the freezing point depression constant for water, given as 1.86 K kg mol$^{-1}$.- $m$ is the molality of the solution, which is 1 mol/kg.Substituting these values, we get:\[\Delta T_f = 2 \cdot 1.86 \cdot 1 = 3.72^\circ C\]Since the normal freezing point of water is $0^\circ C$, the solution's freezing point will be:\[0^\circ C - 3.72^\circ C = -3.72^\circ C\] Consequently, the correct option is (A).

The freezing point of one molal KCl solution, assuming KCl to be completely dissociated in water, is: (\(K_f\) for water = 1.86 K kg mol\(^{-1}\))

Show Hint

The Correct Option is A

Solution and Explanation

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