(D) The boiling point of the solution is $T_b = T_b^o + \Delta T_b = T_b^o + i K_b m$.The freezing point of the solution is $T_f = T_f^o - \Delta T_f

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(D) The boiling point of the solution is $T_b = T_b^o + \Delta T_b = T_b^o + i K_b m$.The freezing point of the solution is $T_f = T_f^o - \Delta T_f = T_f^o - i K_f m$.The difference is given as $T_b - T_f = 75.7 \ ^\circ C$.Substituting the expressions: $(T_b^o + i K_b m) - (T_f^o - i K_f m) = 75.7$.$(80 + i \times 2.65 \times 0.2) - (5.5 - i \times 5.12 \times 0.2) = 75.7$.$(80 - 5.5) + i \times 0.2 \times (2.65 + 5.12) = 75.7$.$74.5 + i \times 0.2 \times (7.77) = 75.7$.$i \times 1.554 = 75.7 - 74.5 = 1.2$.$i = \frac{1.2}{1.554} \approx 0.77$.

Class 12 Chemistry Solutions Abnormal molecular mass

Medium

The difference between the boiling point and the freezing point of a $0.2 \ m$ solution of acetic acid in benzene is $75.7 \ ^\circ C$. Calculate the value of the van't Hoff factor $i$. (For benzene,$K_b = 2.65 \ K \ m^{-1}$,$K_f = 5.12 \ K \ m^{-1}$,$T_b^o = 80 \ ^\circ C$,$T_f^o = 5.5 \ ^\circ C$)

A
$1.44$
B
$0.64$
C
$0.83$
D
$0.77$

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Abnormal molecular mass

$HA_{(aq)} \rightleftharpoons H^{+}_{(aq)} + A^{-}_{(aq)}$
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.

MediumJEE MAIN 2025

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$1.7920 \ g$ of $K_2SO_4$ is present in $1000 \ mL$ of solution. If such solution has an osmotic pressure of $0.680 \ bar$ at $26 \ ^\circ C$,then find out the van't Hoff factor value.

Medium

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The measured freezing point depression for a $0.1 \ m$ aqueous $CH_{3}COOH$ solution is $0.19^{\circ} C$. The acid dissociation constant $K_{a}$ at this concentration will be (Given, $K_{f}$ the molal cryoscopic constant $= 1.86 \ K \ kg \ mol^{-1}$)

MediumWBJEE 2013

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If $\alpha$ is the degree of dissociation of $Na_2SO_4$,the Van't Hoff factor $(i)$ used for calculating the molecular mass is

MediumAIEEE 2005

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If $0.1 \ M$ solution of $NaCl$ is isotonic with $1.1 \ w \%$ urea solution, the degree of ionisation of $NaCl$ is $($Molar masses of urea and $NaCl$ are $60 \ g \ mol^{-1}$ and $58.5 \ g \ mol^{-1}$, respectively.$)$

DifficultTS EAMCET 2019

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$HA_{(aq)} \rightleftharpoons H^{+}_{(aq)} + A^{-}_{(aq)}$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.

$1.7920 \ g$ of $K_2SO_4$ is present in $1000 \ mL$ of solution. If such solution has an osmotic pressure of $0.680 \ bar$ at $26 \ ^\circ C$,then find out the van't Hoff factor value.

The measured freezing point depression for a $0.1 \ m$ aqueous $CH_{3}COOH$ solution is $0.19^{\circ} C$. The acid dissociation constant $K_{a}$ at this concentration will be (Given, $K_{f}$ the molal cryoscopic constant $= 1.86 \ K \ kg \ mol^{-1}$)

If $\alpha$ is the degree of dissociation of $Na_2SO_4$,the Van't Hoff factor $(i)$ used for calculating the molecular mass is

If $0.1 \ M$ solution of $NaCl$ is isotonic with $1.1 \ w \%$ urea solution, the degree of ionisation of $NaCl$ is $($Molar masses of urea and $NaCl$ are $60 \ g \ mol^{-1}$ and $58.5 \ g \ mol^{-1}$, respectively.$)$

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$HA_{(aq)} \rightleftharpoons H^{+}_{(aq)} + A^{-}_{(aq)}$
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.