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Asked by Balbir 9th September 2019, 6:02 PM
Answered by Expert
Answer:
Given:
 
Molar specific conductance = straight lambda subscript straight m superscript straight C space equals space straight lambda subscript straight m superscript infinity space minus straight b square root of straight C
 
Molar specific conductance at infinite dilution =space straight lambda subscript straight m superscript infinity space
 
Concentration = C
 
1. Molar concentration of NaCl at infinite dilution;
 
     C = 4 × 10−4
 
    space straight lambda subscript straight m superscript straight C space = 107 ohm−1cm2mol−1  
 
 
straight lambda subscript straight m superscript straight C space equals space straight lambda subscript straight m superscript infinity space minus straight b square root of straight C

space straight lambda subscript straight m superscript infinity space equals space straight lambda subscript straight m space end subscript superscript straight C plus space straight b square root of straight C

space straight lambda subscript straight m superscript infinity space space equals space 107 space plus space straight b square root of 4 cross times 10 to the power of negative 4 end exponent space end root space space space

space straight lambda subscript straight m superscript infinity space space minus space 0.02 straight b space equals space 107 space space............. open parentheses 1 close parentheses
 
 
 
space straight lambda subscript straight m superscript infinity space equals space straight lambda subscript straight m space end subscript superscript straight C plus space straight b square root of straight C

space straight lambda subscript straight m superscript infinity space space equals space 97 space plus space straight b square root of 9 cross times 10 to the power of negative 4 end exponent space end root space space

straight lambda subscript straight m superscript infinity space minus space 0.03 straight b space space equals space 97 space space space............. open parentheses 2 close parentheses
 
From eq (1) and (2)
 
space space space space space space up diagonal strike space straight lambda subscript straight m superscript infinity end strike space minus 0.02 straight b space space space equals space 107
minus
space space space space space space up diagonal strike space space space straight lambda subscript straight m superscript infinity end strike space minus space 0.03 straight b space equals space 97
________ plus ______ minus ____________
space space space space space space space space space space space space space space space space space space space space 0.01 straight b space equals space 10

straight b space equals space 1000
 
Putting value of b in eq (1)
 
space straight lambda subscript straight m superscript infinity space space minus space 0.02 open parentheses 1000 close parentheses space equals space 107 space

space straight lambda subscript straight m superscript infinity space equals 127 space ohm to the power of negative 1 end exponent cm squared mol to the power of negative 1 end exponent space space space space space
 
Molar concentration of NaCl at infinite dilution is 127 ohm−1cm2mol−1  
 
2. Cell constant:
straight lambda subscript straight m superscript straight C space equals space straight lambda subscript straight m superscript infinity space minus straight b square root of straight C
 
       equals space 127 space minus space 1000 square root of 25 cross times 10 to the power of negative 4 end exponent end root
        = 127 − 50
 
straight lambda subscript straight m superscript straight C space = 77 
 
We know,
 
 
straight lambda subscript straight m superscript straight C space space equals space fraction numerator k cross times 1000 over denominator C end fraction

straight k space equals fraction numerator space straight lambda subscript straight m superscript straight C space cross times straight C over denominator 1000 end fraction

space space space space equals fraction numerator 77 cross times 25 cross times 10 to the power of negative 4 end exponent over denominator 1000 end fraction

straight k space equals space 1925 space cross times 10 to the power of negative 7 end exponent space ohm space cm to the power of negative 1 end exponent

straight k space equals space 1 over Resistance cross times Cell space constant

Cell space constant space equals space 1925 space cross times 10 to the power of negative 7 end exponent space cross times 1000

space space space space space space space space space space space space space space space space space space space space equals space 0.1925 space cm to the power of negative 1 end exponent
 
 
3. Molar conductance:
 
    Given:
 
    C = 5 × 10−3 N
 
   Resistance = 400 ohm
 
  Molar conductance = ?
 
  straight lambda subscript straight m superscript straight C space equals fraction numerator straight k cross times 1000 over denominator straight C end fraction

straight k space space equals space fraction numerator cell space constant over denominator resistance end fraction

straight lambda subscript straight m superscript straight C space space equals fraction numerator cell space constant cross times 1000 over denominator resistance cross times straight C end fraction

space space space equals fraction numerator 0.1925 cross times 1000 over denominator 400 cross times 2.5 cross times 10 to the power of negative 3 end exponent end fraction

straight lambda subscript straight m superscript straight C space space equals space 192.5 space ohm to the power of negative 1 end exponent cm squared mol to the power of negative 1 end exponent
 
 
 
Molar conductance is 192.5 ohm−1cm2mol−1  
Answered by Expert 10th September 2019, 12:38 PM
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