The solution from the following with the highest depression in freezing point (lowest freezing point) is:

  • A
    $180 \ g$ of acetic acid dissolved in water
  • B
    $180 \ g$ of acetic acid dissolved in benzene
  • C
    $180 \ g$ of benzoic acid dissolved in benzene
  • D
    $180 \ g$ of glucose dissolved in water

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Similar Questions

$20\%$ of acetic acid is dissociated when its $5\ g$ is added to $500\ mL$ of water. The depression in freezing point of such water is $..... \times 10^{-3} \ { }^{\circ}C$. Atomic mass of $C$,$H$,and $O$ are $12$,$1$,and $16 \ \text{a.m.u.}$ respectively.
[Given: Molal depression constant and density of water are $1.86 \ K \ kg \ mol^{-1}$ and $1 \ g \ cm^{-3}$ respectively.]

Elevation in the boiling point for $1 \ m$ solution of glucose is $2 \ K$. The depression in the freezing point for $2 \ m$ solution of glucose in the same solvent is $2 \ K$. The relation between $K_b$ and $K_f$ is

The difference between the boiling point and freezing point of an aqueous solution of urea containing $10.0 \ kg$ of water is $100.2372 \ ^oC$. How many grams of urea are dissolved in the solution? (Given: $K_b = 0.513 \ K \ kg \ mol^{-1}$,$K_f = 1.86 \ K \ kg \ mol^{-1}$)

An aqueous solution freezes at $-2.55 \ ^oC$. What is its boiling point in $^oC$?
[$K_b(H_2O) = 0.52 \ K \ kg \ mol^{-1}, K_f(H_2O) = 1.86 \ K \ kg \ mol^{-1}$]

An aqueous solution of a solute $AB$ has a normal boiling point of $101.08\,^{\circ}C$ and a normal freezing point of $-1.80\,^{\circ}C$. Hence,$AB$:
Given: $AB$ is $100\%$ ionised at the boiling point of the solution,$(K_b / K_f)_{\text{water}} = 0.3$.

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