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 concent
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)`:
A. 483
B. 438
C. 348
D. 384
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Correct Answer - D
`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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