1.

Molar conductivities at infinitedilution of KCl, HCl and CH_3COOK are 0.013o and 0.038 and 0.09 S m^(2) mol^(-1) respectively at 291 K . If conductivity of 0.001 M CH_3COOH is2.72 xx 10^(-3) Sm^(-1) then find % degree of ionization of CH_3COOH.

Answer»


Solution :`^^_(M)^(0)[{:(^^_M^(0)+ , ^^_M^(0)- , ^^_M^(0)),(CH_3COOK, HCl , KCl):}]=(0.009 + 0.038 - 0.013)`
`^^_M^(0)(CH_3COH)=0.034 xx 10^(4) = 340S -cm^(2) -"mole"^(-1)`
`^^_M^(0) = (K xx 10^(3))/(M)=(2.72 xx 10^(-3)xx 10^(-2))/(0.001) = 2.72 xx 10`
`^^_M = 2.72 xx 10^(32) = 27.2 , ALPHA = (^^_M)/(^^_M^(0))=(27.2)/(340)=0.08, alpha % = 8%`


Discussion

No Comment Found

Related InterviewSolutions