1.

The specific conductivity of an aqueous solution of a weak monoprotic acid is 0.00033 ohm^(-1) cm^(-1) at a concentration 0.02 M. If at this concentration the degree of dissociation is 0.043, then calculate the value of lambda_(0) ("in"ohm^(-1)cm^(2)//eqt):

Answer»

483
438
348
384

Solution :`alpha=0.043=lambda_m/lambda_oo` and`lambda_m=0.00033xx50xx10^3=33xx50xx10^(-2)`
so `lambda_oo=(33xx50xx10^(-2))/0.043 =383.72 ~~384`


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