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:
- A$1.041$ and $0.041$
- B$1.041$ and $0.1041$
- C$0.041$ and $0.041$
- D$0.041$ and $1.041$
Explore More
Similar Questions
$A$ molecule $M$ associates in a given solvent according to the equation $M \rightleftharpoons (M)_n$. For a certain concentration of $M,$ the van't Hoff factor was found to be $0.9$ and the fraction of associated molecules was $0.2.$ The value of $n$ is
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 $K$)
Medium
View SolutionIf the van't Hoff factor of a weak electrolyte $A_xB_y$ is $i$,then the degree of dissociation $(\alpha)$ is represented by which equation?
Medium
View SolutionThe measured freezing point depression for a $0.1 \ m$ aqueous $CH_{3}COOH$ solution is $0.19^{\circ} C$. The acid dissociation constant $K_{a}$ at this concentration will be (Given, $K_{f}$ the molal cryoscopic constant $= 1.86 \ K \ kg \ mol^{-1}$)
MediumWBJEE 2013
View SolutionDuring the determination of the molar mass of benzoic acid in benzene using the freezing point depression method,it corresponds to which of the following?
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo