`2.5 xx 10^(-4)` M मेथेनोइक अम्ल की चालकता `5.25 xx 10^(-5) S cm^(-1)` है। इसकी मोलर चालकता और वियोजन की मात्रा को परिकलित कीजिये।
दिया गया है: `Lamda^(@) (H^(+)) = 349.5 S cm^(2) mol^(-1)` और `Lamda^(@) (HCOO^(-)) = 50.5 S cm^(2) mol^(-1)`

Conductivity of 2.5 xx 10^(-4) M methonoic acid is 5.25 xx 10^(-5) S cm^(-1) . Calculate its molar Condcutivity and degree of dissociation. Given : lambda^(@) ( H^(+) = 349.5 S cm^(2) "mol"^(-1) and lambda^(@) ( HCOO^(-)) = 50.5 S cm^(2) "mol"^(-1) .

Conductivity of 2.5xx10^(-4) M methanoic acid is 5.25xx10^(-5)S com^(-1) . Calculate its molar conductivity and degree of dissociation. "Given ":lamda^(0)(H^(+))=349.5 S cm^(2) mol^(-1) and lamda^(0)(HCOO^(-))=50.5 S cm^(2) mol^(-1).

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))

Calculate the degree of dissociation (alpha) of acetic acid if its molar conductivity (^^_(m)) is 39.05 S cm^(2) mol^(-1) Given lamda^(@) (H^(+)) = 349.6 cm^(2) mol^(-1) and lamda^(2) (CH_(3)COO^(-)) = 40.9 S cm^(2) mol^(-1)

Calculate the degree of dissociation (alpha) of acetic acid if its molar conductivity (^^_(m)) is 39.05 S cm^(2) mol^(-1) Given lamda^(@) (H^(+)) = 349.6 cm^(2) mol^(-1) and lamda^(2) (CH_(3)COO^(-)) = 40.9 S cm^(2) mol^(-1)