When $20 \ g$ of naphthoic acid $(C_{11}H_8O_2)$ is dissolved in $50 \ g$ of benzene,a freezing point depression of $2 \ K$ is observed. The van't Hoff factor $(i)$ is [$K_f = 1.72 \ K \ kg \ mol^{-1}$].

  • A
    $0.5$
  • B
    $1.0$
  • C
    $2.0$
  • D
    $3.0$

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Similar Questions

When substance $A$ is dissolved in solvent $B$,it associates to form $A_3$. The van't Hoff factor $i$ is:

$HA_{(aq)} \rightleftharpoons H^{+}_{(aq)} + A^{-}_{(aq)}$
The freezing point depression of a $0.1 \ m$ aqueous solution of a monobasic weak acid $HA$ is $0.20^{\circ} C$. The dissociation constant for the acid is. Given: $K_{f}(H_2O) = 1.8 \ K \ kg \ mol^{-1}$,molality $\equiv$ molarity.

Determination of the molar mass of acetic acid in benzene using freezing point depression is affected by

$A$ non-volatile solute '$A$' tetramerizes in water to the extent of $80\%$. $2.5 \text{ g}$ of '$A$' in $100 \text{ g}$ of water lowers the freezing point by $0.3^\circ \text{C}$. The molar mass of $A$ in $\text{g mol}^{-1}$ is ($K_f$ for water $= 1.86 \text{ K kg mol}^{-1}$)

$2$ molal solution of a weak acid $HA$ has a freezing point of $-3.885^{\circ} C$. The degree of dissociation of this acid is ........ $\times 10^{-3}$. (Round off to the Nearest Integer).
[Given: Molal depression constant of water = $1.85 \ K \ kg \ mol^{-1}$,Freezing point of pure water = $0^{\circ} C$]

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