Limiting molar conductances of electrolytes given below
`lamda^(@)(NaCl)=200Scm^(2)mol^(-1)`
`lamda^(@)(H_(2)O)=140Scm^(2)mol^(-1)`
`lamda^(@)(HCl)=180Scm^(2)mol^(-1)`
limiting molar conductance of `NaOH` will be

Calculate the degree of dissociation (alpha) of acetic acid if its molar conductivity is 39.05 Scm^(2)mol^(-1) . Given lamda^(@)(H^(+))=349.6S" "cm^(2)mol^(-1) and lamda^(@)(CH_(3)CO O^(-))=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)

Molar conductance of 1M solution of weak acid HA is 20 ohm^(-1) cm^(2) mol^(-1) . Find % dissocation of HA: (^^_(m)^(o)(H^(+)) =350 S cm^(2) mol^(-1)),(^^_(m)^(o)(A^(-)) =50 S cm^(2) mol^(-1))

For strong electrolytes the values of molar conductivities at infinite dilution are given below {:("Electrolyte",^^_(m)^(@) (Sm^(2 mol^(-1))),(BaCl_(2),280 xx 10^(-4)),(NaCl,126.5 xx 10^(-4)), (NaOH,48 xx 10^(-4)):} The molar conductance at infinite dilution for Ba(OH)_(2) is

A saturated solution in AgA (K_(sp)=3xx10^(-14)) and AgB (K_(sp)=1xx10^(-14)) has conductivity of 375xx10^(-10) S cm^(-1) and limiting molar conductivity of Ag^(+) and A^(-) are 60S cm^(2) "mol"^(-1) and 80 S cm^(2) "mol"^(-1) respectively , then what will be the limiting molar conductivity of B^(-) (in S cm^(2) "mol"^(-1) )?

The specific conductance of a 0.01 M solution of acetic acid at 298 K is 1.65 xx 10^(-4) "ohm"^(-1)cm^(-1) . The molar conductance at infinite dilution for H^(+) ion and CH_(3)COO^(-) ion are 349.1 "ohm"^(-1) cm^(2 )"mol"^(-1) and 40.9 "ohm"^(-1) cm^(2) "mol"^(-1) respectively. Calculate : (i) Molar conductance of solution (ii) Degree of dissociation of acetic acid (iii) Dissociation constant for acetic acid.

From the following molar conductivities at infinite dilution, Lamda_(m)^(@) " for " Al_(2) (SO_(4))_(3) = 858 S cm^(2) mol^(-1) Lamda_(m)^(@) " for " NH_(4)OH = 238.3 S cm^(2) mol^(-1) Lamda_(m)^(@) " for " (NH_(4))_(2)SO_(4) = 238.4 S cm^(2) mol^(-1) Calculate Lamda_(m)^(@) " for " Al(OH)_(3)

(a) Calculate the degree of dissociation of 0.0024 M acetic acid if conductivity of this solution is 8.0 xx 10^(-5)" S "cm^(-1) . Given . lambda_(H^(+))^(@) = 349.6" S " cm^(2)" mol"^(-1), lamda_(CH_(3)COO^(-))^(@)= 40.9" S " cm^(2) " mol "^(-1) (b) Solutions of two electrolytes ‘A’ and ‘B’ are diluted. The limiting molar conductivity of ‘B’ increases to a smaller extent while that of ‘A’ increases to a much larger extent comparatively. Which of the two is a strong electrolyte? Justify your answer.

At 0.04 M concentration the molar conductivity of a solution of a electrolyte is 5000Omega^(-1)cm^(2)mol^(-1) while at 0.01 M concentration the value is 5100Omega^(-1)cm^(2)mol^(-1) making necessary assumption (taking it as strong electrolyte) find the molar conductivity at infinite dilution and also determine the degree of dissociation of strong electrolyte at 0.04M.