What is association? What is dissociation?
(N/A) Association: It is the process in which two or more molecules of a solute combine to form a larger single molecule. This leads to a decrease in the total number of particles in the solution,resulting in a van't Hoff factor $(i < 1)$. For example,the dimerization of acetic acid in benzene.
Dissociation: It is the process in which a solute molecule splits into two or more simpler ions or smaller molecules when dissolved in a solvent. This leads to an increase in the total number of particles in the solution,resulting in a van't Hoff factor $(i > 1)$. For example,the dissociation of $NaCl$ into $Na^+$ and $Cl^-$ ions in water.
Dissociation: It is the process in which a solute molecule splits into two or more simpler ions or smaller molecules when dissolved in a solvent. This leads to an increase in the total number of particles in the solution,resulting in a van't Hoff factor $(i > 1)$. For example,the dissociation of $NaCl$ into $Na^+$ and $Cl^-$ ions in water.
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When $12.2 \ g$ of benzoic acid is dissolved in $100 \ g$ of water,the freezing point of the solution was found to be $-0.93^{\circ} C$ $(K_{f}(H_{2}O) = 1.86 \ K \ kg \ mol^{-1})$. The number $(n)$ of benzoic acid molecules associated (assuming $100 \ \%$ association) is ........ .
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 $=$ $1.86 \ K \ kg \ mol^{-1}$.
Medium
View SolutionAssertion : 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.
Reason : Water is polar and benzene is nonpolar.
MediumAIIMS 2005
View SolutionBenzoic acid molecules undergo dimerisation in benzene. $2.44 \ g$ of benzoic acid when dissolved in $30 \ g$ of benzene caused a depression in freezing point of $2 \ K$. What is the percentage of association of it? (Given $K_f(C_6H_6) = 5 \ K \ kg \ mol^{-1}$; molar mass of benzoic acid $= 122 \ g \ mol^{-1}$)
$20 \ g$ of naphthoic acid $(C_{11}H_8O_2)$ dissolved in $50 \ g$ of benzene $(K_f = 1.72 \ K \ kg \ mol^{-1})$ shows a depression in freezing point of $2 \ K$. The Van't Hoff factor is?
Medium
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