The resistance of 0.01 N NaCl solution at 25 | vedclass

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

Question

(A) Given: Resistance $(R)$ = $200 \ \Omega$,Cell constant $(G^*)$ = $1 \ cm^{-1}$,Normality $(N)$ = $0.01 \ N$.First,calculate the conductivity $(\ka

Vedclass · Accepted Answer

$5 	imes 10^2 \ \Omega^{-1} cm^2 eq^{-1}$

Vedclass · Answer

(A) Given: Resistance $(R)$ = $200 \ \Omega$,Cell constant $(G^*)$ = $1 \ cm^{-1}$,Normality $(N)$ = $0.01 \ N$.First,calculate the conductivity $(\kappa)$:$\kappa = \frac{G^*}{R} = \frac{1 \ cm^{-1}}{200 \ \Omega} = 0.005 \ \Omega^{-1} cm^{-1}$.Next,calculate the equivalent conductance $(\lambda_{eq})$:$\lambda_{eq} = \frac{\kappa 	imes 1000}{N} = \frac{0.005 	imes 1000}{0.01} = \frac{5}{0.01} = 500 \ \Omega^{-1} cm^2 eq^{-1}$.This can be expressed as $5 	imes 10^2 \ \Omega^{-1} cm^2 eq^{-1}$.

The resistance of $0.01 \ N$ $NaCl$ solution at $25 \ ^oC$ is $200 \ \Omega$. The cell constant of the conductivity cell is $1 \ cm^{-1}$. The equivalent conductance is:

Similar Questions

$A$ conductivity cell containing $5 \times 10^{-4} \ M \ NaCl$ solution develops a resistance of $14000 \ \Omega$ at $25^{\circ} C$. Calculate the conductivity of the solution if the cell constant is $0.84 \ cm^{-1}$.

The $INCORRECT$ statement about the conductivity of electrolytic (ionic) solution is

What is electrical resistance? Give a note on it.

Molar conductivities $\left(\Lambda_{m}^{\circ}\right)$ at infinite dilution of $NaCl$,$HCl$,and $CH_{3}COONa$ are $126.4$,$425.9$,and $91.0 \ S \ cm^{2} \ mol^{-1}$ respectively. $\Lambda_{m}^{\circ}$ for $CH_{3}COOH$ will be $:-$

What is the molar conductivity at infinite dilution of $CaCl_{2}$,if the molar conductivity of $Ca^{2+}$ ion and $Cl^{-}$ ion at infinite dilution is $119$ and $71 \ \Omega^{-1} \ cm^{2} \ mol^{-1}$ respectively?

Vedclass Test Series

Exam Paper Generator

Online Exam Module