Question

The conductivity of a 0.01M solution of a 1:1 weak electrolyte at 298 k is 1.5 × 10^ -4S cm-1 ..... 1. molar conductivity of the solution​

Answer 1

Answer:

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Answer 2

Answer:

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Explanation:

Given:

 

Resistance = 210 Ω

 

Concentration of solution = 0.1 M

 

Cell constant = 0.88 cm−1  

 

Specific conductance  

 

equals fraction numerator Cell space constant over denominator Resistance end fraction

equals fraction numerator 0.88 over denominator 210 end fraction

space equals 0.00419 space ohm to the power of negative 1 end exponent space cm to the power of negative 1 end exponent

 

Molar conductance  

 

equals Specific space conductance cross times 1000 over straight C

space equals 0.00419 cross times fraction numerator 1000 over denominator 0.01 end fraction

space equals space 419.04 ohm space cm squared space mole to the power of negative 1 end exponent

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