(C) For the association reaction $nA \rightleftharpoons (A)_n$,let the degree of association be $x$.Initial moles: $A = 1$,$(A)_n = 0$.Moles at equili

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(C) For the association reaction $nA \rightleftharpoons (A)_n$,let the degree of association be $x$.Initial moles: $A = 1$,$(A)_n = 0$.Moles at equilibrium: $A = 1 - x$,$(A)_n = \frac{x}{n}$.Total moles at equilibrium = $(1 - x) + \frac{x}{n}$.The Van't Hoff factor $i$ is defined as the ratio of total moles of particles after association to the initial moles of solute.$i = \frac{1 - x + \frac{x}{n}}{1} = 1 - x + \frac{x}{n}$.

Class 12 Chemistry Solutions Abnormal molecular mass

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One mole of a solute $A$ is dissolved in a given volume of a solvent. The association of the solute takes place according to $nA \rightleftharpoons (A)_n$. The Van't Hoff factor $i$ is expressed as:

A
$i = 1 - x$
B
$i = 1 + \frac{x}{n}$
C
$i = 1 - x + \frac{x}{n}$
D
$i = 1$

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Abnormal molecular mass

When $1.0 \ g$ of $KCl$ is dissolved in $200 \ g$ water,the decrease in freezing point of such solution is $0.24 \ K$,calculate Van't Hoff factor $(i)$ for such solution. $K_f$ of water $=$ $1.86 \ K \ kg \ mol^{-1}$.

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

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$A$ $0.004 \ M$ solution of $Na_2SO_4$ is isotonic with a $0.010 \ M$ solution of glucose at the same temperature. The apparent degree of dissociation of $Na_2SO_4$ is $..... \ \%$.

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Pure benzene freezes at $5.3^oC$. $A$ solution of $0.223 \ g$ of $C_6H_5CH_2COOH$ in $4.49 \ g$ of benzene freezes at $4.47^oC$. Given $K_f = 5.12 \ K \ kg/mol$,what can be concluded from this?

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Pure benzene freezes at $5.3\,^oC$. $A$ solution of $0.223\, g$ of phenylacetic acid $(C_6H_5CH_2COOH)$ in $4.4\, g$ of benzene $(K_f = 5.12\, K\, kg\, mol^{-1})$ freezes at $4.47\,^oC$. From this,it can be concluded that phenylacetic acid ........

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One mole of a solute $A$ is dissolved in a given volume of a solvent. The association of the solute takes place according to $nA \rightleftharpoons (A)_n$. The Van't Hoff factor $i$ is expressed as:

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When $1.0 \ g$ of $KCl$ is dissolved in $200 \ g$ water,the decrease in freezing point of such solution is $0.24 \ K$,calculate Van't Hoff factor $(i)$ for such solution. $K_f$ of water $=$ $1.86 \ K \ kg \ mol^{-1}$.

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

$A$ $0.004 \ M$ solution of $Na_2SO_4$ is isotonic with a $0.010 \ M$ solution of glucose at the same temperature. The apparent degree of dissociation of $Na_2SO_4$ is $..... \ \%$.

Pure benzene freezes at $5.3^oC$. $A$ solution of $0.223 \ g$ of $C_6H_5CH_2COOH$ in $4.49 \ g$ of benzene freezes at $4.47^oC$. Given $K_f = 5.12 \ K \ kg/mol$,what can be concluded from this?

Pure benzene freezes at $5.3\,^oC$. $A$ solution of $0.223\, g$ of phenylacetic acid $(C_6H_5CH_2COOH)$ in $4.4\, g$ of benzene $(K_f = 5.12\, K\, kg\, mol^{-1})$ freezes at $4.47\,^oC$. From this,it can be concluded that phenylacetic acid ........

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