(A) For association,the van't Hoff factor $i$ is given by $i = 1 + (\frac{1}{n} - 1) \alpha$.Given $n = 3$ and $\alpha = 0.80$,we have $i = 1 + (\frac

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(A) For association,the van't Hoff factor $i$ is given by $i = 1 + (\frac{1}{n} - 1) \alpha$.Given $n = 3$ and $\alpha = 0.80$,we have $i = 1 + (\frac{1}{3} - 1) \times 0.80 = 1 - 0.5333 = 0.4667$.Depression in freezing point is $\Delta T_f = i \cdot K_f \cdot m$.$\Delta T_f = 0.4667 \times 5.12 \times 0.25 = 0.597 \ K \approx 0.6 \ K$.The freezing point of the solution $T_s = T_f^o - \Delta T_f$.$T_f^o$ of benzene in Kelvin is $273.15 + 5.5 = 278.65 \ K$.$T_s = 278.65 - 0.6 = 278.05 \ K$.

Class 12 Chemistry Solutions Abnormal molecular mass

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Three particles of a solute $A$ associate in benzene to form a trimer $A_3$. Calculate the freezing point of a $0.25 \ m$ solution if the degree of association of solute $A$ is $0.80$. The freezing point of benzene is $5.5 \ ^oC$ and $K_f$ is $5.12 \ K \ kg \ mol^{-1}$. (Answer in $K$)

A
$278.05$
B
$278.4$
C
$272.4$
D
$301.0$

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Abnormal molecular mass

$A$ solute $A$ undergoes association in a medium as $nA \rightleftharpoons A_n$. If the degree of association is $\alpha$,then the van't Hoff factor $i$ is given by:

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The freezing point of a solution containing $0.85 \ g$ of $ZnCl_2$ in $125.0 \ g$ of water is $-0.23^\circ C$. The degree of dissociation of this salt is .......... $\%$. (Given: $K_f$ of water $= 1.86 \ K \ kg \ mol^{-1}$,atomic mass of $Zn = 65.3$,$Cl = 35.5$)

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The molecular weight of benzoic acid dissolved in benzene is found to be ............ $g/mol$.

Medium

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$2 \,g$ of benzoic acid $(C_{6}H_{5}COOH)$ dissolved in $25 \,g$ of benzene shows a depression in freezing point equal to $1.62 \,K$. The molal depression constant for benzene is $4.9 \,K \,kg \,mol^{-1}$. What is the percentage association of the acid if it forms a dimer in the solution?

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Assertion : The molecular weight of acetic acid determined by depression in freezing point method in benzene and water was found to be different.
Reason : Water is polar and benzene is nonpolar.

MediumAIIMS 2005

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Three particles of a solute $A$ associate in benzene to form a trimer $A_3$. Calculate the freezing point of a $0.25 \ m$ solution if the degree of association of solute $A$ is $0.80$. The freezing point of benzene is $5.5 \ ^oC$ and $K_f$ is $5.12 \ K \ kg \ mol^{-1}$. (Answer in $K$)

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$A$ solute $A$ undergoes association in a medium as $nA \rightleftharpoons A_n$. If the degree of association is $\alpha$,then the van't Hoff factor $i$ is given by:

The freezing point of a solution containing $0.85 \ g$ of $ZnCl_2$ in $125.0 \ g$ of water is $-0.23^\circ C$. The degree of dissociation of this salt is .......... $\%$. (Given: $K_f$ of water $= 1.86 \ K \ kg \ mol^{-1}$,atomic mass of $Zn = 65.3$,$Cl = 35.5$)

The molecular weight of benzoic acid dissolved in benzene is found to be ............ $g/mol$.

$2 \,g$ of benzoic acid $(C_{6}H_{5}COOH)$ dissolved in $25 \,g$ of benzene shows a depression in freezing point equal to $1.62 \,K$. The molal depression constant for benzene is $4.9 \,K \,kg \,mol^{-1}$. What is the percentage association of the acid if it forms a dimer in the solution?

Assertion : The molecular weight of acetic acid determined by depression in freezing point method in benzene and water was found to be different.Reason : Water is polar and benzene is nonpolar.

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Assertion : The molecular weight of acetic acid determined by depression in freezing point method in benzene and water was found to be different.
Reason : Water is polar and benzene is nonpolar.