SATHEE: Electrochemistry - Result Question 13

Electrochemistry - Result Question 13

####14. If $0.01 M$ solution of an electrolyte has a resistance of $40 o h m s$ in a cell having a cell constant of $0.4 c m^{- 1}$ , then its molar conductance in ohm $^{- 1} c m^{2} m o l^{- 1}$ is

(a) $10^{2}$

(b) $10^{4}$

(d) $10^{3}$

[1997]

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Solution:

(d) Given molarity $= 0.01 M$

Resistance $= 40$ ohm;

Cell constant $\frac{1}{A} = 0.4 c m^{- 1}$

Specific conductivity ( $κ$ )

$= \frac{cell constant}{resistance} = \frac{0.4}{40} = 0.01 o h m^{- 1} c m^{- 1}$

Molar conductance $(\land_{m}) = \frac{1000 κ}{Molarity}$

$= \frac{1000 \times .01}{.01} = 10^{3} o h m^{- 1} c m^{2} m o l^{- 1}$

Electrochemistry - Result Question 13

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Electrochemistry - Result Question 13

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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