(a) The conductivity of 0.001 M solution of `CH_3COOH` is `3.905 xx 10^(-5) S cm^(-1)`. Calculate its molar conductivity and degree of dissociation (`alpha`). Given : `lambda^(@)(H^(+)) = 349.6 S cm^(2) mol^(-1)` and `lambda^(@)(CH_(3)COO^(-)) = 40.9 S cm^(2) mol^(-1)` (b) Define electrochemical cell. What happens if external potential applied becomes greater than `E_("cell")^(@)` of electrochemical cell ?
Text Solution
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The correct Answer is:
`Lambda_(0) = 390.5 S cm^(2) mol^(-1), alpha =0.1`
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U-LIKE SERIES|Exercise SELF ASSESSMENT TEST (SECTION C)|1 Videos
The conductivity of 0.001 "mol" L^(-1) solution of CH_(3)COOH " is " 3.905 xx 10^(-5) "S" cm^(-1) . Calculate its molar conductivity and degree of dissociation (alpha) . "Given" lambda^(@) (H^(+)) = 349.6 "S" cm^(2) "mol"^(-1) " and " lambda^(0) (CH_(3)COO^(-)) = 40.9 "S" cm^(2) "mol"^(-1) (b) Define electrochemical cell. What happens if external potential applied becomes greater than E_(cell)^(@) of electrochemical cell?
The conductivity of 0.001 mol L^(-1) solution of CH_(3)COOH is 3.905xx10^(-5)S cm^(-1) . Calculate its molar conductivity and degree of dissociation (alpha) . ("Given":lamda_((H^(+)))^(@)=349.65 S cm^(2)mol^(-1)andlamda^(@)(CH_(3)COO^(-))=40.9 D cm^(2)mol^(-1))
The conductivity of 0.001 " mol "L^(-1) solution of CH_(3)COOH is 4.95xx10^(-5)" S "cm^(-1) . Calculate its molar conductance and degree of dissociation (alpha). "Given" lambda_((H^(+)))^(@)=349.6" S "cm^(2)mol^(-1), lambda_((CH_(3)COO^(-)))^(@)=40.95 cm^(2)mol^(-1)
The conductivity of 0.02M solution of NaCl is 2.6 xx 10^(-2)" S cm"^(-1) . What is its molar conductivity ?
If conductivity of 0.1 mol / dm^3 solution of KCl is 1.3 * 10^(-2) S cm^(-1) at 298 K then its molar conductivity will be
U-LIKE SERIES-ELECTROCHEMISTRY -SELF ASSESSMENT TEST (SECTION D)
