The equivalent conductivity of N/10 solution of acitic acid at `25^(@)C` is `14.3 ohm^(-1) cm^(2) eq^(-1)`. Calculate the degree of dissociation of `CH_(3)COOH` if `Lambda_(oo CH_(3)COOH)` is 390.71.

The equivalent conductivity of N/10 solution of acetic acid at 25^@C is 15 "ohm"^(-1) "cm"^(-1) "equiv"^(–1​) . What is the Degree of dissociation of acetic acid ? (oversetoowedge_(CH_3COOH)=400 "ohm"^(-1) "cm"^2 "equiv"^(-1))

The degree of dissociation of 0.1N CH_(3)COOH is (K_(a)= 1xx10^(-5))

The degree of dissociation of 0.1N CH_(3)COOH is (K_(a)= 1xx10^(-5))

The equivalent conductivity of 0.1 N CH_3COOH at 25^@ C is 80 and at infinite dilute is 400ohm cm^(2) eq^(-1) . The degree of dissociation of CH_3COOH

The degree of dissociation of CH_3COOH Sol^n , can be increased by

The equivalent conductivity of 0.1 N CH_(3) COOH " at " 25^(@)C is 80 and at infinite dilution 400 ohm^(-1) . The degree of dissociation of CH_(3)COOH is:

The conducitivity of a 0.01 M solution of acetic acid at 298 K is 1.65 xx 10^(-4) S cm^(-3) . Calculate (i) Molar conductivity of the solution (ii) degree of dissociation of CH_3 COOH (iii) dissociation constant for acetic acid Given that lambda^(@) (H^(+)) = 349.1 and lambda^(@) (CH_(3) COO^(-)) = 40.9 S cm^(2) mol^(-1) .