What is abnormal molar mass?
(N/A) Abnormal molar mass is defined as the molar mass of a substance that is either higher or lower than the expected (theoretical) value calculated based on its chemical formula.
This phenomenon occurs due to the association or dissociation of solute particles in a solution.
$1$. Association: When solute particles combine to form larger aggregates,the number of particles decreases,leading to a higher observed molar mass than the theoretical value.
$2$. Dissociation: When solute particles break apart into ions,the number of particles increases,leading to a lower observed molar mass than the theoretical value.
The extent of this deviation is expressed using the van't Hoff factor $(i)$.
This phenomenon occurs due to the association or dissociation of solute particles in a solution.
$1$. Association: When solute particles combine to form larger aggregates,the number of particles decreases,leading to a higher observed molar mass than the theoretical value.
$2$. Dissociation: When solute particles break apart into ions,the number of particles increases,leading to a lower observed molar mass than the theoretical value.
The extent of this deviation is expressed using the van't Hoff factor $(i)$.
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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 ........ .
$A$ $0.5\%$ solution of potassium chloride was found to freeze at $-0.24^{\circ} C$. The percentage dissociation of potassium chloride is .... (Nearest integer)
(Molal depression constant for water is $1.80\, K\, kg\, mol^{-1}$ and molar mass of $KCl$ is $74.6\, g\, mol^{-1}$)
(Molal depression constant for water is $1.80\, K\, kg\, mol^{-1}$ and molar mass of $KCl$ is $74.6\, g\, mol^{-1}$)
DifficultJEE MAIN 2022
View SolutionThe 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,$K_b = 2.65 \ K \ m^{-1}$,$K_f = 5.12 \ K \ m^{-1}$,$T_b^o = 80 \ ^\circ C$,$T_f^o = 5.5 \ ^\circ C$)
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View SolutionWhen $20 \ g$ 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 [$K_f = 1.72 \ K \ kg \ mol^{-1}$].
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