$2$ molal solution of a weak acid $HA$ has a freezing point of $-3.885^{\circ} C$. The degree of dissociation of this acid is ........ $\times 10^{-3}$. (Round off to the Nearest Integer).
[Given: Molal depression constant of water = $1.85 \ K \ kg \ mol^{-1}$,Freezing point of pure water = $0^{\circ} C$]
[Given: Molal depression constant of water = $1.85 \ K \ kg \ mol^{-1}$,Freezing point of pure water = $0^{\circ} C$]
- A$50$
- B$60$
- C$55$
- D$65$
Explore More
Similar Questions
If $0.1 \ M$ solution of $NaCl$ is isotonic with $1.1 \ w \%$ urea solution, the degree of ionisation of $NaCl$ is $($Molar masses of urea and $NaCl$ are $60 \ g \ mol^{-1}$ and $58.5 \ g \ mol^{-1}$, respectively.$)$
DifficultTS EAMCET 2019
View SolutionCalculate the percent dissociation of $0.02 \ m$ solution if its freezing point depression is $0.046 \ K$. $\left[K_{f} \text{ for water } = 1.86 \ K \ kg \ mol^{-1} ; n=2\right]$ (in $\%$)
One mole of a solute $A$ is dissolved in a given volume of a solvent. The association of the solute takes place according to $nA \rightleftharpoons (A)_n$. The Van't Hoff factor $i$ is expressed as:
Easy
View SolutionDimerisation of solute molecules in low dielectric constant solvent is due to
MediumKCET 2023
View Solution$2 \ g$ of benzoic acid $(C_6H_5COOH)$ dissolved in $25 \ g$ of benzene shows a depression in freezing point equal to $1.62 \ K$. The molal depression constant for benzene is $4.9 \ K \ kg \ mol^{-1}$. The percentage of benzoic acid in dimeric form is ......... $\%$.
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