(D) In water,benzoic acid $(C_6H_5COOH)$ undergoes dissociation to form benzoate ions and hydronium ions,which leads to a decrease in the observed mol

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

It dissociates in water and associates in benzene.

(D) In water,benzoic acid $(C_6H_5COOH)$ undergoes dissociation to form benzoate ions and hydronium ions,which leads to a decrease in the observed molecular weight.In benzene (a non-polar solvent),benzoic acid molecules form intermolecular hydrogen bonds,leading to the formation of a dimer. This association results in an increase in the observed molecular weight.Therefore,the correct option is $(D)$.

Class 12 Chemistry Solutions Abnormal molecular mass

Difficult

Benzoic acid has a higher molecular weight in benzene and a lower molecular weight in water because:

A
Water has a lower freezing point and a higher boiling point than benzene.
B
It dissociates to a greater extent in benzene than in water.
C
It associates in water and dissociates in benzene.
D
It dissociates in water and associates in benzene.

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Chemistry Questions

Chapter

Solutions

Subtopic

Abnormal molecular mass

$A$ solute $A$ dimerizes in water. The boiling point of a $2 \, M$ solution of $A$ is $100.52^{\circ} C$. The percentage association of $A$ is ..... $\%$.
(Round off to the Nearest integer)
[Use : $K_{b}$ for water $= 0.52 \, K \, kg \, mol^{-1}$,Boiling point of water $= 100^{\circ} C$]

MediumJEE MAIN 2021

View Solution

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 Solution

The Van't Hoff factor of $BaCl_2$ at a concentration of $0.01 \ M$ is $1.98$. The percentage dissociation of $BaCl_2$ at this concentration will be: (in $\%$)

Difficult

View Solution

Dimerisation of solute molecules in low dielectric constant solvent is due to

MediumKCET 2023

View Solution

When $0.0106 \text{ mole}$ of acetic acid is dissolved in $1 \text{ kg}$ of water, the observed freezing point depression is $0.0205 \text{ K}$. If the calculated freezing point depression is $0.0197 \text{ K}$, the Van't Hoff factor $(i)$ and the degree of dissociation $(\alpha)$ of acetic acid are respectively:

DifficultKCET 2026

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

Benzoic acid has a higher molecular weight in benzene and a lower molecular weight in water because:

Similar Questions

$A$ solute $A$ dimerizes in water. The boiling point of a $2 \, M$ solution of $A$ is $100.52^{\circ} C$. The percentage association of $A$ is ..... $\%$.(Round off to the Nearest integer)[Use : $K_{b}$ for water $= 0.52 \, K \, kg \, mol^{-1}$,Boiling point of water $= 100^{\circ} C$]

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 $=$ $1.86 \ K \ kg \ mol^{-1}$.

The Van't Hoff factor of $BaCl_2$ at a concentration of $0.01 \ M$ is $1.98$. The percentage dissociation of $BaCl_2$ at this concentration will be: (in $\%$)

Dimerisation of solute molecules in low dielectric constant solvent is due to

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$A$ solute $A$ dimerizes in water. The boiling point of a $2 \, M$ solution of $A$ is $100.52^{\circ} C$. The percentage association of $A$ is ..... $\%$.
(Round off to the Nearest integer)
[Use : $K_{b}$ for water $= 0.52 \, K \, kg \, mol^{-1}$,Boiling point of water $= 100^{\circ} C$]