The resistance of a conductivity cell containing $0.01 \, M \, KCl$ solution at $298 \, K$ is $1750 \, \Omega$. If the conductivity of $0.01 \, M \, KCl$ solution at $298 \, K$ is $0.152 \times 10^{-3} \, S \, cm^{-1}$,then the cell constant of the conductivity cell is $.......... \, \times 10^{-3} \, cm^{-1}$.
- A$452$
- B$312$
- C$266$
- D$199$
Explore More
Similar Questions
An aqueous solution of $X$ is added slowly to an aqueous solution of $Y$ as shown in List-$I$. The variation in conductivity of these reactions is given in List-$II$. Match List-$I$ with List-$II$ and select the correct answer using the code given below the lists :
List-$I$ List-$II$ $P$. $\underset{X}{(C_2H_5)_3N} + \underset{Y}{CH_3COOH}$ $1$. Conductivity decreases and then increases $Q$. $\underset{X}{KI (0.1 \ M)} + \underset{Y}{AgNO_3 (0.01 \ M)}$ $2$. Conductivity decreases and then does not change much $R$. $\underset{X}{CH_3COOH} + \underset{Y}{KOH}$ $3$. Conductivity increases and then does not change much $S$. $\underset{X}{NaOH} + \underset{Y}{HI}$ $4$. Conductivity does not change much and then increases
Codes: $P \quad Q \quad R \quad S$
| List-$I$ | List-$II$ |
|---|---|
| $P$. $\underset{X}{(C_2H_5)_3N} + \underset{Y}{CH_3COOH}$ | $1$. Conductivity decreases and then increases |
| $Q$. $\underset{X}{KI (0.1 \ M)} + \underset{Y}{AgNO_3 (0.01 \ M)}$ | $2$. Conductivity decreases and then does not change much |
| $R$. $\underset{X}{CH_3COOH} + \underset{Y}{KOH}$ | $3$. Conductivity increases and then does not change much |
| $S$. $\underset{X}{NaOH} + \underset{Y}{HI}$ | $4$. Conductivity does not change much and then increases |
Codes: $P \quad Q \quad R \quad S$
The molar conductivities of $KCl$,$NaCl$,and $KNO_3$ are $152$,$128$,and $111 \ S \ cm^2 \ mol^{-1}$ respectively. The molar conductivity of $NaNO_3$ is ............. $S \ cm^2 \ mol^{-1}$.
Easy
View SolutionWhich of the following statements regarding the solution of an electrolyte is not correct?
Equivalent conductivity of $Cr_2(SO_4)_3$ is related to molar conductivity by the expression
Medium
View SolutionCalculate the degree of dissociation of $0.01 \ M$ acetic acid at $25^{\circ} C$ given that the molar conductivity at concentration $c$ is $\Lambda_{c} = 15.0 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ and the limiting molar conductivity is $\Lambda_0 = 300 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.
Vedclass 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